(a) Consider an ac circuit consisting of a pure inductor connected to the terminals of an ac source. It has a single voltage source (VS) and three resistors having resistances of R1, R2 and R3. V = Vm sin t . As the iron rod is inserted, the magnetic field inside the coil magnetizes the iron rod thereby increasing the magnetic field inside it. Step 3 In this case, we will get two mesh equations since there are two meshes in the given circuit. Therefore, the voltage across inductor at resonance is $V_L = j QV$. So, negative polarity of the induced voltage is present at the dotted terminal of this primary coil. Therefore, inductor acts as a constant current source in steady state. After applying an input to an electric circuit, the output takes certain time to reach steady state. V = Vm sin t (1), Let VC be the instantaneous voltage drop across the capacitor, then by Kirchoffs loop rule, conventional current and electronic current: When a conductor AB is connected across the terminals of a cell , free electrons begin to drift or move from its end B (connected to the negative terminal of the cell) to the end A (connected to the positive terminal of the cell). The quality factor is defined as the ratio of the voltage developed across the capacitor or inductor to the applied voltage. Where, L is the inductance of an inductor and C is the capacitance of a capacitor. Using XL= 2 f L or L = \(\frac{X_{L}}{2 \pi f}\), therefore But, practically five time constants are sufficient. (ii) the number of turns in the primary and secondary windings. A part of the graph is called as a subgraph. The above circuit diagram consists of a voltage source, VS in series with two resistors R1 and R2. The current flowing through 20 Ω resistor can be found by doing the following simplification. How? (2012) In Mesh analysis, we will consider the currents flowing through each mesh. At = , the magnitude of transfer function is equal to 1. Why does electric current start flowing in a circuit the moment circuit is complete? Step 4 The above equation is in the form of $V_1 = AV_2 BI_2$. P = V I We know the following relations of the resistances of delta network in terms of resistances of star network. V L\(\frac{di}{dt}\) = 0 (2), Using equations (1) and (2) we have What is the main Difference between Potentiometer and Rheostats? Consider an ac source. Inductor current does not change instantaneously, when the switching action takes place. Step 1 We know that, the following matrix equation of two port network regarding Z parameters as, $\begin{bmatrix}V_1 \\V_2 \end{bmatrix} = \begin{bmatrix}Z_{11} & Z_{12} \\Z_{21} & Z_{22} \end{bmatrix} \begin{bmatrix}I_1 \\I_2 \end{bmatrix}$Equation 3, Step 2 We know that, the following matrix equation of two port network regarding Y parameters as, $$\begin{bmatrix}I_1 \\I_2 \end{bmatrix} = \begin{bmatrix}Y_{11} & Y_{12} \\Y_{21} & Y_{22} \end{bmatrix} \begin{bmatrix}V_1 \\V_2 \end{bmatrix}$$, $\begin{bmatrix}V_1 \\V_2 \end{bmatrix} = \begin{bmatrix}Y_{11} & Y_{12} \\Y_{21} & Y_{22} \end{bmatrix}^{-1} \begin{bmatrix}I_1 \\I_2 \end{bmatrix}$Equation 4, Step 4 By equating Equation 3 and Equation 4, we will get, $$\begin{bmatrix}Z_{11} & Z_{12} \\Z_{21} & Z_{22} \end{bmatrix} = \begin{bmatrix}Y_{11} & Y_{12} \\Y_{21} & Y_{22} \end{bmatrix}^{-1}$$, $$\Rightarrow \begin{bmatrix}Z_{11} & Z_{12} \\Z_{21} & Z_{22} \end{bmatrix} = \frac{\begin{bmatrix}Y_{22} & - Y_{12} \\- Y_{21} & Y_{11} \end{bmatrix}}{\Delta Y}$$, $$\Delta Y = Y_{11} Y_{22} - Y_{12} Y_{21}$$. One can find the voltage drop across a resistor simply by using Ohms law of current electricity. Answer: \(\frac{\varepsilon_{s}}{\varepsilon_{p}}=\frac{N_{s}}{N_{p}}\) and \(\frac{\varepsilon_{\mathrm{s}}}{\varepsilon_{\mathrm{p}}}=\frac{l_{\mathrm{p}}}{l_{\mathrm{s}}}\). Such a continuous and closed path on an electric current is called an electric circuit. Write KVL equation around the loop of the following circuit. $$\Rightarrow f^2 = \frac{1}{(2 \pi)^2 L C}$$, $$\Rightarrow f = \frac{1}{(2 \pi) \sqrt{LC}}$$, Therefore, the resonant frequency fr of series RLC circuit is. The magnitude of current in (I) does not become zero with the Passage of time. A bulb is rated at 200 V, 100 W. Calculate its resistance. Draw a graph showing the variation of a capacitive reactance with the frequency of the ac source. Now cos = \(\frac{R}{Z}\) The response in a particular branch could be either current flowing through that branch or voltage across that branch. A linear circuit may contain independent sources, dependent sources, and resistors. Typical tolerances are as follows: resistor 5%, beta 100-300, power supply 5%. The units of parameters, B and C, are Ohm and Mho respectively. or As the loop rotates from position t = 0 to position t T/2, the Induced emf changes from zero to maximum value and then becomes zero again as shown in the diagram For angles 0 and 1800 the instantaneous rate of magnetic flux is zero, hence induced emf Is zero. (i) Resistances are connected in series to obtain minimum current Step 1 Identify the meshes and label the mesh currents in either clockwise or anti-clockwise direction. (NCERT Exemplar) Question 15. Question 8. The symbol of capacitor along with current I and voltage V are shown in the following figure. (a). Sometimes, only a single branch may connect to the node. Potential difference, V = 220 volt; (NCERT) $$H(s) = \frac{V_o(s)}{V_i(s)} = \frac{sL + \frac{1}{sC}}{R + sL + \frac{1}{sC}}$$, $$\Rightarrow H(s) = \frac{s^2 LC + 1}{s^2 LC + sCR + 1}$$, $$H(j \omega) = \frac{1 - \omega^2 LC}{1 - \omega^2 LC + j \omega CR}$$, $$|H(j \omega)| = \frac{1 - \omega^2 LC}{\sqrt{(1 - \omega^2 LC)^2 + (\omega CR)^2}}$$. This phase angle of reactive opposition to current becomes critically important in circuit analysis, especially for complex AC circuits where reactance and resistance interact. Recalculate emitter current IE for standard value resistors if necessary. 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Answer: This conversion is known as two port network parameters conversion or simply, two-port parameters conversion. Therefore, by using less electricity he is indirectly causing less pollution. Explain, giving the necessary mathematical formula, the effect on the brightness of the bulb in case (a) and (b), when the frequency of the ac source is increased. Therefore, the glow of the light bulb decreases. (V Potential difference, I Current) Let us calculate the resistances of star network, which are equivalent to that of delta network as shown in the following figure. Examples: Voltage sources and current sources. Hence, derive the expression for the impedance of the circuit. V = VC = Vm Sin t (2). In the parallel case, of the three resistors, the 300 ohm resistor would see the lowest current. Answer: Hence, the network element is a Bilateral element. Solution: The graph of variation of XC with f is as shown. The above circuit diagram consists of a voltage source, VS in series with two resistors R1 and R2. Give reason for the following: It is the number of radians per second that the alternating current is rotating at, if you imagine one cycle of AC to represent a full circles rotation. In another article, we have discussed different ways to find the voltage drop across a capacitor as well. Answer 2: An ammeter is always connected in series in a circuit, so that entire current, which we wish to measure, flows through it. Energy consumed by an electrical appliance is given by H = Pt For the TV set: Power W = 250 W and time t = 1 hour = 3600 seconds So, energy consumed H = 250 3600 = 900000 J For the toaster: Power W = 1200 W and time t = 10 minutes = 600 seconds So, energy consumed H = 1200 600 = 720000 J Hence, TV set uses more energy than toaster. lrms = \(\frac{V_{r m s}}{R}=\frac{140}{50 \sqrt{2}}\) = 1.98 A, Question 21. So, in this case T parameters are the desired parameters and Z parameters are the given parameters. (i) square of current (I2). WebEach resistor draws the same current it would if it were the only resistor connected to the voltage source. At $\omega = \frac{1}{CR}$, the magnitude of transfer function is equal to 0.707. Consider the following connected subgraph of the graph, which is shown in the Example of the beginning of this chapter. We will discuss the third method in the next chapter. Step 3 Find the Nortons current IN by shorting the two opened terminals of the above circuit. We choose IC = 1mA, typical of a small-signal transistor circuit. So yes, the resistor does reduce the current. bulb; (b) the peak voltage of the source; and (c) the rms current through the bulb. It is useful for analyzing complex electric circuits by converting them into network graphs. If we represent these phase angles of voltage and current mathematically in the form of complex numbers, we find that an inductors opposition to current has a phase angle, too: Current lags voltage by 90 in an inductor. (2014) tan = \(\frac{R}{Z}=\frac{X_{L}-X_{C}}{R}=\frac{2 \pi f L-1 / 2 \pi f C}{R}\). We choose 82k from the list of standard values. (ii) When f < fr> then the circuit behaves as a capacitive circuit. The resistor passively opposes the flow of current, and the capacitor is two parallel plates separated by an insulator. (iii) All the electrical appliances like bulbs, fans and sockets, etc. However, because the current and voltage waves are 90 out of phase, there are times when one is positive while the other is negative, resulting in equally frequent occurrences of negative instantaneous power. When the circuit is broken anywhere (or switch is turned OFF), the current stops flowing ant the bulb does not glow. This increases the brightness of the bulb. It determines the sharpness of the resonance. Example Problems 1. A radian is a unit of angular measurement: there are 2 radians in one full circle, just as there are 360 in a full circle. The s-domain circuit diagram (network) of High pass filter is shown in the following figure. Substitute the values of V2 and R in the above equation. So, we got the resistances of delta network as R1 = 10 Ω, R2 = 60 Ω and R3 = 30 Ω, which are equivalent to the resistances of the given star network. Following a bumpy launch week that saw frequent server trouble and bloated player queues, Blizzard has announced that over 25 million Overwatch 2 players have logged on in its first 10 days. The main difference is the use and circuit operation, i.e. When an electric current flows through a conductor it becomes hot. Loss of energy in the primary and secondary due to Joule heating. Now, place the Nortons equivalent circuit to the left of the terminals A & B of the given circuit. Types of Transducers and Applications, What is Potential Transformer (PT)? whereas VL = l XL and VC = l XC. When the frequency changes, so does an inductor or capacitors opposition to the flow of electricity. (i) From the electric pole two insulated wires come to our houses. Reducing C will increase reactance and the lamp will shine less brightly than before. Superposition theorem is based on the concept of linearity between the response and excitation of an electrical circuit. (b) How many 176 resistors in parallel are required to carry 5 A on a 220 V line? (iii) Fuse wire is placed in series with the device. If positive voltage is applied across the capacitor, then it stores positive charge. At high frequency, the capacitive reactance is small while the inductive reactance is large. Varisters are voltage dependent Resistors (VDR) which is used to eliminate the high voltage transients. The node voltage V1 and Thevenins voltage VTh are labelled in the above figure. In this tutorial, we will consider positive sign when the current leaves a node and negative sign when it enters a node. ic = lo sin ( t + /2) (6). In series circuit, all the electrical appliances have only one switch due to which they cannot be turned off or turned on separately. Given P = 2 kW = 2000 W, Vrms = 223 Z = \(\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}\). Passive Elements cant deliver power (energy) to other elements, however they can absorb power. Answer: Y parameters are called as admittance parameters because these are simply, the ratios of currents and voltages. 2. The SI unit of potential difference is joule/coulomb or volt. The rows and columns of the above matrix represents the twigs and branches of given directed graph. (CBSE Delhi 2011) Current, I = \(\frac{25}{100}\) = \(\frac{1}{10}\) = 0.1 A. The twigs d, e & f are represented with solid lines and links a, b & c are represented with dotted lines in the following figure. Both are variable resistors. The coefficients of independent variables, I1 and I2 are called as Z parameters. In this case, the equivalent inductance has been increased by 2M. Energy stored in the inductor at time t is The 33k base resistor is a standard value, emitter current at = 100 is OK. In this chapter, let us solve an example problem by considering both series and parallel combinations of similar passive elements. Question 5. Apply Thevenins Theorem to yield a single Thevenin equivalent resistance Rth and voltage source Vth. Here, $V_i(s)$ and $V_o(s)$ are the Laplace transforms of input voltage, $v_i(t)$ and output voltage, $v_o(t)$ respectively. lm = \(\sqrt{2}\) lrms = 1.4.1 1.04 = 1.47 A. But what does negative power mean? So, there will be three f-loops, since there are three links. Here, we have to represent T parameters in terms of Z parameters. Hence, substitute t = 0 & i(t) = 0 in Equation 3 in order to find the value of constant, K. $$0 = Ke^{-\lgroup \frac{0}{\tau} \rgroup} + \frac{V_m}{\sqrt{R^2 +{\omega}^2 L^2}} sin \lgroup \omega (0) + \varphi - tan^{-1} \lgroup \frac {\omega L}{R}\rgroup \rgroup$$, $$\Rightarrow 0 = K + \frac{V_m}{\sqrt{R^2 +{\omega}^2 L^2}} sin \lgroup \varphi - tan^{-1} \lgroup \frac {\omega L}{R}\rgroup \rgroup$$, $$\Rightarrow K = - \frac{V_m}{\sqrt{R^2 +{\omega}^2 L^2}} sin \lgroup \varphi - tan^{-1} \lgroup \frac {\omega L}{R}\rgroup \rgroup$$, $i(t) = - \frac{V_m}{\sqrt{R^2 +{\omega}^2 L^2}} sin \lgroup \varphi - tan^{-1} \lgroup \frac {\omega L}{R}\rgroup \rgroup e^{-\lgroup \frac{t}{\tau} \rgroup} + \frac{V_m}{\sqrt{R^2 +{\omega}^2 L^2}} sin \lgroup \omega t + \varphi - tan^{-1} \lgroup \frac {\omega L}{R}\rgroup \rgroup$Equation 4. Therefore, the current flowing through 20 resistor of given circuit is 2 A. In a torch, a switch provides a conducting link between a battery (a number of cells placed in proper order) and a bulb. It must be understood that this angular velocity is an expression of how rapidly the AC waveforms are cycling, a full-cycle being equal to 2 radians. The equivalent resistance is: (b) When a series combination of R1 and R2 is connected in parallel with R3 [Fig (b)]. R = \(\frac{12 \times 6}{12+6}=\frac{72}{18}\) = 4, Question 1. Thus the output of the ac generator varies sinusoidally with time. These are same as that of principal nodes in the electric circuit. The Co-Tree branches a, b & c are represented with dashed lines. (ii) the value of Q factor in the circuit. These equations represent the dependent variables in terms of independent variables. Inductors behave like the opposite of capacitors: an inductor opposes high frequency signals meaning it passes low frequency audio signals more easily. Required fields are marked *. SI unit is ohm. (ii) Three resistors are connected in an electrical circuit as shown. Mathematically, it can be written as. Amit lives in Delhi and is much concerned about the increasing electricity bill of his house. Let the current flowing through the resistor is I amperes and the voltage across it is V volts. This circuit diagram is shown in the following figure. Wing span (L1) = 20 m, Vertical component of Earths magnetic field Therefore, the above equation becomes fo = \(\frac{1}{2 \pi} \frac{1}{\sqrt{L C}}\). 2. (a) X: capacitor (CBSE AI 2018, Delhi 2018) IR = IR1 + IR2 + IR3 IR = I(R1 + R2 + R3) V = \(\frac{W}{Q}\) 15 = \(\frac{W}{2}\) Question 15. (c) Identify the device X. L = \(\frac{1}{C}\) The reactance of an Inductor, therefore, is proportional to the frequency, being given by L. This connected subgraph contains all the four nodes of the given graph and there is no loop. He took some steps to save electricity and succeeded in doing so. Those are the electric circuits or networks having passive elements like resistor, inductor and capacitor. (CBSE Delhi 2016) Q-factor of an LCR Circuit: Q-factor of the LCR circuit is the ratio of the potential difference across inductance (or capacitance) at resonance to the applied voltage. Answer: W is the electrical energy and it is measured in terms of Joule. WebDuring the negative half-cycle, the voltage in negative, so to is the current resulting in the power being positive, as a negative times a negative equals a positive. The switch is closed and after some time, an iron rod is inserted into the interior of the inductor. The available power rating of these resistors is 3 to 200 Watts. It is denoted by lrms. Find the collector voltage VC. (iv) If R1 > R2 > R3, in which circuit more heat will be produced in R1 as compared to other two resistors? (CBSE AI 2011C) (iv) The frequency of current in India is 50 Hz means the direction of current in India changes 100 times in 1 second as current direction changes twice in one cycle. (iii) Fuse wire is placed in series with the device because when large current passes through the circuit the fuse wire gets heated up and melts and whole circuit breaks and the device is protected from the damage. WhenVL = VC, then the circuit is in series resonance, therefore both current and voltage are in phase. Answer: Step 3 In this case, we will get two nodal equations, since there are two principal nodes, 1 and 2, other than Ground. Since is positive therefore voltage leads current by the above phase. For this reason, is sometimes expressed in units of electrical radians per second rather than (plain) radians per second, so as to distinguish it from mechanical motion. An increase in frequency decreases the value of capacitive reactance. Hence, it induces a voltage in primary coil. Non-Linear Elements are those that do not show a linear relation between voltage and current. Similarly, we can calculate the other two Z parameters, Z12 and Z22 by doing open circuit of port1. For AC amplifiers, a bypass capacitor in parallel with RE improves AC gain. The inclusion of rEE in the calculation results in a lower value of the base resistor RB as shown in Table below. Substitute the value of V in the above equation. Answer: V = IR (or) R = \(\frac{V}{I}\), (b) Resistance, R = 176 , di = \(\frac{V_{m}}{L}\) sin t dt (4), Integrating the above equation we have Add the phase angles of input sinusoidal voltage and $H(j \omega)$. (c) How does the reactance of device X vary with the frequency of the ac? Give two reasons. Therefore, the current flowing through resistor at resonance is $\mathbf{\mathit{I_R = I}}$. (a) Draw a circuit diagram to show the connections. Draw the voltage divider without assigning values. But, we can easily understand the above waveform of current flowing through the circuit from Equation 6 by substituting a few values of t like 0, , 2, 5, etc. g-parameters are called as inverse hybrid parameters. The value of element will be +1 for the twig of selected f-cutset. In the previous chapter, we discussed an example problem related equivalent resistance. V = Vmsin t (1), But dQ/dt = lC, therefore the above equation becomes, But this time it reverses Its direction. What is the core motive of chapter 12 (Electricity) of 10th standard Science? If the resistivity of the metal is 4.8 10-7 ohm-metre then calculate the area of cross-section of the wire. (a) From the graph we find that for a frequency of 100 Hz the capacitive reactance is 6 ohm, therefore Working principle: Electromagnetic induction. (ii) Which alternative source of energy would you suggest Amit to use? WebA flyback diode is any diode connected across an inductor used to eliminate flyback, which is the sudden voltage spike seen across an inductive load when its supply current is suddenly reduced or interrupted. There are six branches in the above graph and those are labelled with a, b, c, d, e & f respectively. A light bulb is in turn connected in a series (a) across an LR circuit, (b) across an RC circuit, with an ac source. The live wire goes through ON/OFF switch. $$A = \frac{V_1}{V_2}, \: when \: I_2 = 0$$, $$B = -\frac{V_1}{I_2}, \: when \: V_2 = 0$$, $$C = \frac{I_1}{V_2}, \: when \: I_2 = 0$$, $$D = -\frac{I_1}{I_2}, \: when \: V_2 = 0$$. Derive an expression for the impedance of a series LCR circuit connected to an ac supply of variable frequency. This concept is illustrated in the following figure. Define the term root mean square (rms) value of ac. V I or \(\frac{V}{I}\) = Constant = R or V = IR Copper Loss, flux Loss, (d) Calculate the current drawn by the primary of a 90% efficient transformer which steps down 220 V to 22 V If the output resistance Is 440 . Based on the type of sources that are present in the network, we can choose one of these three methods. (b) When all these elements are connected in series across the same source, Answer: Replace any two terminal linear network or circuit to the left side of variable load resistor having resistance of RL ohms with a Thevenins equivalent circuit. Step 1 Consider the circuit diagram by opening the terminals with respect to which the Thevenins equivalent circuit is to be found. 2014 Because, every Tree will be having one Fundamental cut set matrix. R = 4 + 4 = 8. This can be converted into a practical voltage source as shown in the figure. In the previous chapter, we discussed about the conversion of delta network into an equivalent star network. The plot is as shown. However, in some situations, it is difficult to simplify the network by following the previous approach. As you have learned in the text, the two voltages are not in the same phase. $i(t) = Ke^{-\lgroup \frac{t}{\tau} \rgroup} + i_{ss}(t)$Equation 2. By convention, the direction of motion of positive charges is taken as the direction of electric current. Factors on which the resistance of a conductor depends, 10. (2012) Will the current in the circuit lag, lead, or remain in phase with the applied voltage when (i) f > fr (ii) f < fr Explain your answer in each case. Disadvantages of ac: Now slope of the graph is Find Show mathematically that the current in it lags behind the applied emf by a phase angle of /2. It is from positive terminal to negative terminal. The friction of the air is created as it meets and passes over an aeroplane and its components. The current through an LCR circuit is given by the expression lv = \(\frac{E_{v}}{Z}\), where Z is the impedance. Mesh equation is obtained by applying KVL first and then Ohms law. Q= r L/R = (103 1)/10 = 100. V = V1 + V2 + V3 Students should practice all 13 examples of this chapter. The given V-I characteristics of a network element lies in the first and third quadrants. Draw the effective equivalent circuit of the circuit shown in the figure at very high frequencies and find the effective impedance. The calculated RB = 39k is a standard value resistor. Hence, the inductance of the coil increases. The reactance of the capacitor in figure (c) is given by XC = \(\frac{1}{\omega C}=\frac{1}{2 \pi f C}\). The mesh currents I1 and I2 are considered in clockwise direction. Similarly, we can calculate the other two Y parameters, Y12 and Y22 by doing short circuit of port1. Their performance is well in overload conditions. Question 18. Resistance, R1 = 1 ; Calculate the amount of work done in shifting a charge of 2 coulombs from a point A to B having potentials +10V and -5V respectively. $$A' = \frac{V_2}{V_1}, \: when\: I_1 = 0$$, $$B' = -\frac{V_2}{I_1}, \: when\: V_1 = 0$$, $$C' = \frac{I_2}{V_1}, \: when\: I_1 = 0$$, $$D' = -\frac{I_2}{I_1}, \: when \: V_1 = 0$$. It is the ratio of the output power to the input power. What is meant by resistance of a conductor? Effective voltage in the circuit Step 2 Verifying the network element as active or passive. The Transfer function of the above circuit is. or S = 3 10-2 T, o = ?, Using the relation 0 = nBA There are two types of resonances, namely series resonance and parallel resonance. (a) When R1 is connected in series with the parallel combination of R2 and R3 [Fig (a)]. Consider an ac circuit consisting of a capacitor connected to an ac source. A part of the source energy in maintaining the current may be consumed into useful work and rest of the source energy may be expended in heat. Give your answer with reasons. At high-frequency XL will be more. Current is in phase with the applied voltage (c) The XL f graph is as shown below. The graph showing the variation of impedance Z of a series LCR circuit with the frequency f of the applied ac source is shown below. A power transmission line feeds input power at 2200 V to a step-down transformer with its primary windings having 3000 turns. $$\Rightarrow V - IR - I(j X_L) - I(-j X_C) = 0$$, $$\Rightarrow V = IR + I(j X_L) + I(-j X_C)$$, $\Rightarrow V = I[R + j(X_L - X_C)]$Equation 1. It is measured in terms of Volt. Question 16. (c) power in capacitor is zero. (b) Explain how laminating the core of a transformer helps to reduce eddy current losses in it. If too many electrical appliances of high power rating are switched on at the same time, they draw extremely large quantity of current from the circuit. But the heating element becomes hot due to its high resistance (H=IRt) and begins to glow. (b) What Is the time lag between the voltage maximum and the current maximum? $$V_2 = \lgroup \frac {V_S}{R_1 + R_2} \rgroup R_2$$, $$\Rightarrow V_2 = V_S \lgroup \frac {R_2}{R_1 + R_2} \rgroup$$. Suppose the voltage and current remain constant for a small-time dt. Moreover, Wire wound resistors are generally used in high power rating devices and equipment, Testing and measuring devices, industries, and control equipment. Published under the terms and conditions of the. Here = tan-1 \(\frac{1}{CR}\). $$V_{30 \Omega} = \lgroup \frac{14}{5} \rgroup 30$$. If the current flows in one direction only, it is called direct current (DC). Heat produced due to production of eddy currents. tan = \(\frac{X_{L}}{R}=\frac{0.5 \times 314}{100}\) = 1.571 Z = \(\frac{V_{\mathrm{rms}}}{l_{\mathrm{rms}}}=\frac{200}{0.5}\) = 400 , Now Z = \(\sqrt{R^{2}+X_{L}^{2}}\) , therefore Let VL be the instantaneous voltage drop across the inductor, then Kirchoffs loop rule when applied to the circuit gives V + VL = 0 Answer: The above Tree contains three branches d, e & f. Hence, the branches a, b & c will be the links of the Co-Tree corresponding to the above Tree. Obtain the ratio of secondary to primary voltage in terms of the number of turns and currents in the two coils. Show graphically. lrms = \(\frac{V_{\text {rms }}}{x_{L}}=\frac{V_{\text {rms }}}{2 \pi f L}=\frac{220}{2 \times 3.14 \times 50 \times 44 \times 10^{-3}}\) The Thevenins resistance across terminals A & B will be, $$R_{Th} = \lgroup \frac{5 \times 10}{5 + 10} \rgroup + 10 = \frac{10}{3} + 10 = \frac{40}{3} \Omega$$. We know that Nortons resistance, RN is same as that of Thevenins resistance RTh. Omrons AC and DC Relays for Energy Applications, Download the full Panasonic Test & Measurement Guide, Inductive reactance can be calculated using this formula: X. (2015) Answer: \(\frac{X_{L}}{f}=\frac{8-6}{400-300}=\frac{2}{100}\) = 0.02, Therefore L is L = \(\frac{X_{L}}{2 \pi f}=\frac{0.02}{2 \times 3.14}\) = 0.0032 H, (b) NowR = 8 ohm, f = 300 Hz, Z = ? Average power dissipated in Z1, Z2, ,ZN are the impedances of 1st branch, 2nd branch, , Nth branch respectively. Note KVL is independent of the nature of network elements that are present in a loop. Thus the ratio \(\frac{V}{I}\) remains constant. In the above figure, the current (I) is flowing from terminals A to B through a passive element having impedance of Z Ω. Z = \(\sqrt{R^{2}+X_{L}^{2}}\), where XL is the inductive reactance. From the above equations, we can conclude that there exists a linear relationship between voltage across capacitor and current flowing through it. Step 2 Replace the part of the circuit, which is left side of terminals A & B of the given circuit with the above Thevenins equivalent circuit. (a) With the help of a labeled diagram, describe briefly the underlying principle and working of a step-up transformer. document.getElementById( "ak_js_1" ).setAttribute( "value", ( new Date() ).getTime() ); Enter Your Email Address to Subscribe to this Blog and Receive Notifications of New Posts by Email. It is that value of steady current (dc) which when passed through a resistor in a given time produces the same heat as is produced by the ac. Answer: Derive the relation between rms value and the peak value of ac. This shows that microwave requires more electrical energy than the energy provided by the circuit. Convert this previous emitter-bias example to voltage divider bias. Write a KVL equation about the loop containing the battery, RC, RB, and the VBE drop. As the frequency (or alternator shaft speed) is increased in an AC system, an inductor will offer greater opposition to the passage of current, and vice versa. Step 3 Re-arrange the equations of Step2 in such a way that they should be similar to the equations of Step1. Using the relation The modified circuit diagram is shown in the following figure. The resistance of a wire of 0.01 cm radius is 10 . Could you please send me this basic concept chapters. Given L= 5.0 H, C= 80 F , R = 40 , Vrms = 240 V o = ?, lrms = ?, VC = ? This concept is illustrated in the following figure. The graph between the potential difference (V) and the corresponding current (I) is a straight line passing through the origin. The h-parameters or hybrid parameters are useful in transistor modelling circuits (networks). Answer: (i) Which of the two circuits has more resistance? In other hand, they are less stable means their temperature coefficient is very high. The capacitor for an audio amplifier covering 20Hz to 20kHz would be: Note that the internal emitter resistance rEE is not bypassed by the bypass capacitor. Substitute the values of R1, R2 and R3 in the above equations. The arrangement of lights and various other electrical appliances in parallel circuits is used in domestic wiring because of following advantages: Question 5. I wish to control it with a variable from very low to standard high speed operation and to incorporate the potent within the front housing I do not have a wiring diagram so am using pos and neg only I do not think the armature is load protected. Resonance occurs in an LCR circuit when cadmium sulfide, lead sulfide etc. l = \(\frac{V}{Z}\). Question 7. p = v I = 220 2.2 = 484 W. Question 16. Given Ps = 60 W, ls = 0.54 A Now, let us discuss about some of the two port parameter conversions. As a result the current flowing through the circuit increases. The consumption of energy (electric) is given at the rate of 840 W at Voltage 220 V. Now Z = \(\sqrt{R^{2}+X_{L}^{2}}\) . (NCERT) So, the fundamental loop matrix will have b-n+1 rows and b columns. (a) the resistor P = l2R = (1.5)2 100 = 225 W Another device has twice the values for R, XC, XL. The current source present in the Nortons equivalent circuit is called as Nortons equivalent current or simply Nortons current IN. Determine Using the trigonometric identity cos t = sin ( t + /2) equation (5) can be written as (b) Define the efficiency of a transformer. (c) State any two factors that reduce the efficiency of a transformer. L = 25.48 mH, and C = 796 F. Z = \(\sqrt{(20)^{2}+\left(2 \times 3.14 \times 50 \times 2 \times 10^{-3}\right)^{2}}\) (a) number of turns of the inductor Is reduced? (a) A The value of elements will be 0 for the remaining links and twigs, which are not part of the selected f-loop. How are the answers affected? An electric iron draws 2.2 amperes of current from a 220 V source. What is an electric circuit? Rounding that is emitter current times emitter resistor: IERE = (1mA)(470) = 0.47V. So, positive polarity of the induced voltage is present at the dotted terminal of this inductor. (iii) time (t), for which current is passed. If the potential difference across the C, R is 120 V and 90 V respectively, and if rms current of the circuit is 3 A, calculate the (i) impedance and (ii) power factor of the circuit. Question 8. Write its SI units. When current and voltage are in phase then XL = XC, Question 11. Now XC and R are equal and are given by The above circuit has only one mesh. (i)the angular frequency at which maximum power dissipation occurs in the circuit and the corresponding value of effective current and it breaks or open the circuit. Find the inductance of the inductor used in series with a bulb of resistance 10 ohms connected to an ac source of 80 V, 50 Hz. Step 2 We know that the following set of two equations of two port network regarding T parameters. (NCERT) That means, each node in the connected graph will be having one or more branches that are connected to it. $$R_{CB} = \frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4 \Omega$$. So, the number of branches and/or nodes of a subgraph will be less than that of the original graph. We know that there are two practical sources, namely, voltage source and current source. The given graphs (i) and (ii) represent the variation of the opposition offered by the circuit element to the flow of alternating current, with the frequency of the applied emf. or Show this variation graphically. So, the resultant current flowing through the circuit will be, $$i(t) = \frac{V_m}{\sqrt{R^2 +{\omega}^2 L^2}} sin \lgroup \omega t + \varphi - tan^{-1} \lgroup \frac {\omega L}{R}\rgroup \rgroup$$. 2. Therefore, the magnitude of transfer function of Band pass filter will vary from 0 to 1 & 1 to 0 as varies from 0 to . In the previous chapter, we got the transient response of the current flowing through the series RL circuit. q (t) = q0 cos t An Inductor L of inductance XL is connected in series with bulb B and an ac source. Potential difference, V1 = ? Find the resistance between points A and B in the circuit diagram given below: (2013) (1), Efficiency of transformer = \(\frac{V_{s} I_{s}}{V_{p} I_{p}}\). Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes? XL = 17.32 ohm, Now using XL = 2fL we have \(\frac{N_{\mathrm{s}}}{N_{\mathrm{p}}}=\frac{l_{\mathrm{p}}}{l_{\mathrm{s}}}\) = k. Efficiency of transformer = \(\frac{V_{s} I_{s}}{V_{p} I_{p}}\). Step 2 We know that the following set of two equations of two port network regarding Y parameters. Answer: Now, R1 = 6 ; R2 = 9 ; R3 = 18 Let at any instant t, q be the charge on the capacitor and l be the current through the inductor An alternating emf Is applied across a capacitor. It is called a direct current. Which component opposes the flow of current? The labeled diagram is as shown. This voltage is independent of the amount of current that is flowing through the two terminals of voltage source. Because, every Tree will be having one Fundamental loop matrix. This is known as current division principle and it is applicable, when two or more passive elements are connected in parallel and only one current enters the node. 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