Distinguish between the terms reactance and impedance of an ac circuit. $$R_A = \frac{10 \times 60}{10 +60+30} = \frac{600}{100} = 6\Omega$$, $$R_B = \frac{60 \times 30}{10 +60+30} = \frac{1800}{100} = 18\Omega$$, $$R_C = \frac{30 \times 10}{10 +60+30} = \frac{300}{100} = 3\Omega$$. Step 2 Let us find the current flowing through 20 Ω resistor by considering only 4 A current source. Question 9. (a) Describe the mode of connecting the resistances in each case. Answer: We can use this circuit now. Define an electric circuit. Now, let us derive the values of parameters and electrical quantities at resonance of parallel RLC circuit one by one. WebWe can see that the basic circuit configuration for a buck converter is a series transistor switch, TR 1 with an associated drive circuit that keeps the output voltage as close to the desired level as possible, a diode, D 1, an inductor, L 1 and a smoothing capacitor, C 1.The buck converter has two operating modes, depending on if the switching transistor TR 1 is Joules Law of heating gives the formula (C) The rms potential drops across the capacitor at resonance. Answer: If resistance is 10 ohm then the impedance of the combination is. Derive the relation between rms value and the peak value of ac. A resistor is an electrical component that opposes the flow of current in a circuit. In this graph, the nodes 2, 3, and 4 are connected by two branches each. Now ls = Vs/R = 22/440 = 0.05 A, Also = \(\frac{P_{0}}{P_{1}}=\frac{V_{s} l_{s}}{V_{p} l_{p}}\) , therefore we have, \(\frac{90}{100}=\frac{22 \times 0.05}{220 \times l_{\mathrm{p}}}\) Question 5. or Other electrical equipment also produce an emf, Whatever energy is used in building up current is returned back during the decay of current. (b) When a capacitor of reactance XC = XL is included in the circuit, then the new impedance becomes Z = \(\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}\) = R. Thus the impedance has its minimum value. $$H(s) = \frac{V_o(s)}{V_i(s)} = \frac{R}{R + \frac{1}{sC}}$$, $$\Rightarrow H(s) = \frac{sCR}{1 + sCR}$$, $$H(j \omega) = \frac{j \omega CR}{1 + j \omega CR}$$, $$|H(j \omega)| = \frac{\omega CR}{\sqrt{(1 + (\omega CR)^2}}$$. Answer: $$g_{11} = \frac{I_1}{V_1},\: when\: I_2 = 0$$, $$g_{12} = \frac{I_1}{I_2},\: when\: V_1 = 0$$, $$g_{21} = \frac{V_2}{V_1},\: when\: I_2 = 0$$, $$g_{22} = \frac{V_2}{I_2},\: when \: V_1 = 0$$. (b) What is the net power consumed over a full cycle? (a) X: capacitor It produces large amount of heat and do not burn easily. Consequently, the lamp will shine. On the other hand, the resistance of the cord is low. An Inductor L of inductance XL is connected in series with bulb B and an ac source. Amit lives in Delhi and is much concerned about the increasing electricity bill of his house. Linear Resistors; Non Linear Resistors; Linear Resistors: Those resistors, which values change with the applied voltage and temperature, are called linear resistors. Therefore, the above equation becomes fo = \(\frac{1}{2 \pi} \frac{1}{\sqrt{L C}}\). The resistance of conductor depends on length, thickness, nature of material and temperature of conductor. The closest standard value to the 460k collector feedback bias resistor is 470k. (i) Electric bulbs are usually filled with chemically inactive gases like nitrogen and argon. If r = 2r then A = r2 = (2r)2 = 4r2 = 4A Answer: At =100, IE is 1.01mA. P = Vi. Answer: No need to recalculate IE for = 100. As the Loop moves from position t = T/2 to position t = T, the emf again changes from zero to the maximum value and then again becomes zero. It is the ratio of the output power to the input power. $$I_N = I_S \lgroup \frac{Z_1\rVert Z_2 \rVert\rVert Z_{N-1}}{Z_1 + Z_2 + + Z_N}\rgroup$$. The choke (L) is an inductor that is connected in series with a low-power sodium vapor lamp. In this case, we got one branch less in the graph because the 4 A current source is made as open circuit, while converting the electric circuit into its equivalent graph. V = VC = Vm Sin t (2), But from the definition of capacitance, VC = Q / C or Q= VCC. Question 5. Given the resistances of delta network as R1 = 10 Ω, R2 = 60 Ω and R3 = 30 Ω. The switch is closed and after some time, an iron rod is inserted into the interior of the inductor. Power in capacitive circuit P = Ev lv cos . Derive relation between power, potential difference and resistance. The value of elements will be 0 for the remaining twigs and links, which are not part of the selected f-cutset. Answer: \(\frac{\varepsilon_{\mathrm{s}}}{\varepsilon_{\mathrm{p}}}=\frac{l_{\mathrm{p}}}{l_{\mathrm{s}}}\) . SI unit is ohm. (ii) Q a factor of LCR circuit. Circuit will be in resonance when XL XC = 0, Question 11. (2015) s ls = p lp where l and lp are the secondary and primary currents, therefore we have \(\frac{X_{L}}{f}=\frac{8-6}{400-300}=\frac{2}{100}\) = 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 = ? (ii) the number of turns in the primary and secondary windings. For voltage divider bias, perform emitter-bias calculations first, then determine R1 and R2. (iv) If R1 > R2 > R3, in which circuit more heat will be produced in R1 as compared to other two resistors? Thus the output of the ac generator varies sinusoidally with time. If R is the resistance of the bulb, then the total impedance of the circuit is Z = \(\sqrt{R^{2}+X_{L}^{2}}\) and the corresponding current is l = V l Z. I = 5A; No. Let the Instantaneous value of the ac source be Yes, because even then the power consumed will be zero. We can calculate the steady state response of an electric circuit, when it is excited by a sinusoidal voltage source using Laplace Transform approach. R, new resistance = 20 , (ii) The resistance of the wire is inversely proportional to the square of its diameter. In parallel RLC circuit resonance occurs, when the imaginary term of admittance, Y is zero. If this constant of 2 is multiplied by frequency in Hertz (cycles per second), the result will be a figure in radians per second, known as the angular velocity of the AC system. 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. R1 = 4 , R2 = 4 , R3 = 8 There are two possible switching actions. A sensorless BLDC motor makes use of the electromotive force (EMF) that gives rise to a current in the windings of any DC motor with a magnetic field that opposes the original change in magnetic flux as described by Lenzs Law. Their temperature coefficient is very low (2 ppm/C) and used where stability and low noise level is important. Similarly, if negative voltage is applied across the capacitor, then it stores negative charge. There are two basic methods that are used for solving any electrical network: Nodal analysis and Mesh analysis. Question 15. That means, this filter allows (passes) both low and high frequency components. Ohms law states that the electric current, through a conductor, is directly proportional to the potential difference across its two ends when, other physical conditions like temperature, etc., remain constant. Question 4. WhenVL = VC, then the circuit is in series resonance, therefore both current and voltage are in phase. Now, let us discuss these graphs one by one. Here, the passive elements such as resistor, inductor and capacitor are connected in parallel. 3. In the previous chapter, we discussed an example problem related equivalent resistance. Question 4. $$H(j \omega) = \frac{1}{R + j \omega L}$$, Magnitude of $\mathbf{\mathit{H(j \omega)}}$ is, $$|H(j \omega)| = \frac{1}{\sqrt{R^2 + {\omega}^2}L^2}$$, Phase angle of $\mathbf{\mathit{H(j \omega)}}$ is, $$\angle H(j \omega) = -tan^{-1} \lgroup \frac{\omega L}{R} \rgroup$$, We will get the steady state current $i_{ss}(t)$ by doing the following two steps . Define the term root mean square (rms) value of ac. ic = C Vo cos t = lo cos t (5), Here the term 1/c has the dimensions of resistance and is called the capacitive reactance of the circuit and CVm = lm. In other words, a special type of variable resistors used to protect circuits from destructive voltage spikes is called varisters. The transient part occurs in the response of an electrical circuit or network due to the presence of energy storing elements such as inductor and capacitor. As the Loop moves from position t = T/2 to position t = T, the emf again changes from zero to a maximum value and then again becomes zero. or It is based on whether the current enters at the dotted terminal or leaves from the dotted terminal. A schematic sketch is as shown. Cost of 2 units of electricity = 4 2 = 8. Network graph is simply called as graph. or The end of the resistor closest to the (-) battery terminal is (-), the end closest to the (+) terminal it (+). Calculate the resistance between A and B. I am really thankful to the blog owner for helping us by giving valuable study materials. Thus, the reactance formula XL = 2fL could also be written as XL = L. It is the opposition offered by a pure inductor or a pure capacitor or both to the flow of ac. 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, $$\begin{bmatrix}Z_{11} & Z_{12} \\Z_{21} & Z_{22} \end{bmatrix} = \begin{bmatrix}Y_{11} & Y_{12} \\Y_{21} & Y_{22} \end{bmatrix}^{-1}$$, $$\Rightarrow \begin{bmatrix}Z_{11} & Z_{12} \\Z_{21} & Z_{22} \end{bmatrix} = \frac{\begin{bmatrix}Y_{22} & - Y_{12} \\- Y_{21} & Y_{11} \end{bmatrix}}{\Delta Y}$$, $$\Delta Y = Y_{11} Y_{22} - Y_{12} Y_{21}$$. Pav = lrms Vrms cos = \(\frac{V_{m}^{2}}{Z}\) cos Question 3. It is found that the effective voltage of the circuit leads the current in phase by /4. (2013) Let the instantaneous value of current be represented by the equation A solenoid with an iron core and a bulb are connected to a dc source. The magnitude and direction of flow of current remains the same. It does not burn readily at a high temperature. Therefore, capacitor acts as a constant voltage source in steady state. What is the main Difference between Potentiometer and Rheostats? It does not become red when current flows through it. If a graph is drawn between the potential difference (V) and current (I), the graph is found to be a straight line passing through the origin. Cost of 1 unit of electricity = 4 A: Power, B: Voltage, and C: Current. Given, V = 100 V, f = 50 Hz, R = 20 ohm, L = 2 mH = 2 10-3 H, Z = ?, lrms = ? 1 kilowatt-hour = 36,00,000 Joules or 3.6 106 J. (b) Relation between kilowatt-hour and joule (SI unit of energy): The value of VS will be equal to the product of IS and RS. (b) A person operates a microwave oven of 2kW power rating in a domestic circuit of 220 V and current rating 5A. Electric current and voltage applied in a pure resistor are in same phase, i.e. 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. From the above figure, the V-I characteristics of a network element is a straight line passing through the origin. if the frequency of the supply is higher. Since the phase difference between the voltage and current is /2, therefore power consumed = Vrms lrms cos /2 = 0, Question 19. It could range from micro-Amps to Amps depending on the application and transistor rating. Answer: We choose 82k from the list of standard values. (2015) XL = 2fL = 2 3.14 50 0.01 = 3.14 ohm Power factor = cos = R/Z Obtain the ratio of secondary to primary voltage in terms of the number of turns and currents in the two coils. When we write the mesh equations, assume the mesh current of that particular mesh as greater than all other mesh currents of the circuit. At high-frequency XL will be more. A continuous conduction path consisting of wires and other resistances (like bulb, fan, etc.) (b) Mention two Important energy losses in actual transformers and state how these can be minimized. $$V_2 = \lgroup \frac {V_S}{R_1 + R_2} \rgroup R_2$$, $$\Rightarrow V_2 = V_S \lgroup \frac {R_2}{R_1 + R_2} \rgroup$$. Deduce the expression for its working formula. 0.25(R2 + X2) = R2 The figure shows a series LCR circuit with L = 5.0 H, C = 80 F, R = 40 connected to a variable frequency 240 V source. The voltage drops across the resistors R1 and R2 are V1 and V2 respectively. There is no change in the final brightness as the inductive reactance is zero for dc. (CBSE Delhi 2015) Hence, they are usually employed for electricity transmission. (or) (b) lrms= V/Z= 100/20 = 5 A, Question 12. Filters are mainly classified into four types based on the band of frequencies that are allowing and / or the band of frequencies that are rejecting. $$v_1 - L_1 \frac{d i_1}{dt} + M \frac{d i_2}{dt} = 0$$, $\Rightarrow v_1 = L_1 \frac{d i_1}{dt} - M \frac{d i_2}{dt}$Equation 3, $$v_2 - L_2 \frac{d i_2}{dt} + M \frac{d i_1}{dt} = 0$$, $\Rightarrow v_2 = L_2 \frac{d i_2}{dt} - M \frac{d i_1}{dt}$Equation 4. or Draw the effective equivalent circuit of the circuit shown in the figure at very high frequencies and find the effective impedance. The town gets power from the line through a 4000 220 V step-down transformer at a sub-station in the town. Current is in phase with the applied voltage = \(\sqrt{(10)^{2}+(17.32)^{2}}\) = 20 , Now Pav = lrms Vrms cos = \(\frac{V_{m}^{2}}{Z}\) cos So, we will get six possible pairs of equations. As the name suggests, dependent sources produce the amount of voltage or current that is dependent on some other voltage or current. In this case, the resistance of the circuit so formed is very small, thus a large amount of current flows through the circuit and heats the wires to a high temperature and a fire may start. R = cos /4 10 = 7 ohm, Power dissipated Question 6. The inductive reactance, 40n = 220 5 n = \(\frac{220 \times 5}{40}=\frac{55}{2}\) = 27.5 Resistance (AB) = R1 = r + r = 2r Those are the electric circuits or networks having passive elements like resistor, inductor and capacitor. Therefore, the voltage across resistor at resonance is VR = V. $$V_L = \lgroup \frac{V}{R} \rgroup (jX_L)$$, $$\Rightarrow V_L = j \lgroup \frac{X_L}{R} \rgroup V$$. of lamps = 27, Question 12. In this method, we will consider only one independent source at a time. Step 1 In order to find the Thevenins equivalent circuit to the left side of terminals A & B, we should remove the 20 Ω resistor from the network by opening the terminals A & B. One problem with emitter bias is that a considerable part of the output signal is dropped across the emitter resistor RE (figure below). Step 6 The above equation is in the form of $V_1 = AV_2 BI_2$. The above circuit diagram consists of a voltage source, VS in series with two resistors R1 and R2. We can classify coupling into the following two categories. A metal band is wrapped around the resistor element and it can be used as a Potentiometer or Rheostats (See the below note for difference between Rheostat and Potentiometer). Show mathematically that the current in it lags behind the applied emf by a phase angle of /2. As an exercise, rework the emitter-bias example with the emitter resistor reverted back to 470, and the base-bias supply reduced to 1.5V. Required resistance, R2 = 4 ; Required length, L2 = ? Write the condition under which the phenomenon of resonance occurs. . S = 3 10-2 T, o = ?, Using the relation 0 = nBA Whenever the instantaneous current and voltage are both positive (above the line), the power is positive. Hence, it induces a voltage in the other inductor having an inductance of L2. Hence, the independent ideal voltage sources do not exist practically, because there will be some internal resistance. Hence, the network element is a Passive element. Here, we have to represent T parameters in terms of Z parameters. are the examples of resistive materials. Answer: Question 3. Step 4 The simplified electrical network after Step 3 is shown in the following figure. By convention, the direction of motion of positive charges is taken as the direction of electric current. (ii) When f < fr> then the circuit behaves as a capacitive circuit. Things get even more interesting when we plot the power for this circuit: In a pure inductive circuit, instantaneous power may be positive or negative. In parallel there is no division of voltage among the appliances. The f-cut set contains only one twig and one or more links. 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. Hence, the above graph is an unconnected graph. ; We can neglect the first term of Equation 4 because its value will be very much less than one. Explain, with the help of a diagram, the principle and working of an ac generator. Answer: Step 4 Draw the Thevenins equivalent circuit by connecting a Thevenins voltage VTh in series with a Thevenins resistance RTh. The h-parameters or hybrid parameters are useful in transistor modelling circuits (networks). Predict your observations for dc and ac connections. So, the amount of charge stored in the capacitor depends on the applied voltage V across it and they have linear relationship. The property of a conductor due to which it opposes the flow of current through it, is called resistance. Wire wound resistors used where high sensitivity, accurate measurement and balanced current control is required, e.g. This Co-Tree has only three nodes instead of four nodes of the given graph, because Node 4 is isolated from the above Co-Tree. Z = \(\sqrt{R^{2}+X_{C}^{2}}=\sqrt{(100)^{2}+(440)^{2}}\) = 451.2 ohm, Therefore rms value of current V 141 Secondary voltage of transformer s = 220 V Current, I = \(\frac{Q}{t}=\frac{150}{60}\) = 2.5 Amp. So, the resultant current flowing through the circuit will be, $$i(t) = \frac{V_m}{\sqrt{R^2 +{\omega}^2 L^2}} sin \lgroup \omega t + \varphi - tan^{-1} \lgroup \frac {\omega L}{R}\rgroup \rgroup$$. Charge, Q = 150 C; Time, t = 1 min = 60 s What Oppose Does the Resistor | Voltage or Current | Resistor Oppose Voltage or Current in Hindi. It works on the principle of mutual induction i.e., whenever magnetic flux linked with a coil changes an emf is induced in the neighboring coil. s = Ns \(\frac{d \phi}{d t}\), Back emf In the primary with Np turns, Answer: Question 35. WebA diode is a two-terminal electronic component that conducts current primarily in one direction (asymmetric conductance); it has low (ideally zero) resistance in one direction, and high (ideally infinite) resistance in the other.. A diode vacuum tube or thermionic diode is a vacuum tube with two electrodes, a heated cathode and a plate, in which electrons can flow V = Vm sin t (1), Let Vc be the instantaneous voltage drop across the capacitor, then by Kirchoffs loop rule, This conversion is called as Star to Delta Conversion. (a) (i) Resultant resistance, R = R1 + R2 = 10 + 10 = 20 (a) Prove that an Ideal capacitor in an ac circuit does not dissipate power. UP Board students also download UP Board solutions for Class 10 Science chapter 12 in Hindi Medium. Now, we can find the response in an element that lies to the right side of Thevenins equivalent circuit. We can replace these resistors with a single equivalent resistor. That means these elements either dissipate power in the form of heat or store energy in the form of either magnetic field or electric field. Answer: Effect of inclusion of rEE on calculated RB. In both cases, the result will be same. O.25XL2 = O.75R2 No, because in circuit (b) In the above figure, consider node 3 as reference node (Ground). The potential difference between two points is said to be 1 volt if 1 joule of work is done in moving 1 coulomb of electric charge from one point to the other. The graph of variation of XC with f is as shown. Answer: Similarly, we can calculate the other two Y parameters, Y12 and Y22 by doing short circuit of port1. (i)the angular frequency at which maximum power dissipation occurs in the circuit and the corresponding value of effective current and They provide a stable operation in high temperature because their values do not change with change in temperature. So, in this case, T parameters are the desired parameters and Y parameters are the given parameters. The key to effective emitter bias is lowering the base supply VBB nearer to the amount of emitter bias. In the previous chapter, we discussed the transient response and steady state response of DC circuit. In other words, R1||R2. The KVL equation around this mesh is. The variation of voltage, current, and power in one complete cycle is shown in the figure. Draw the voltage divider without assigning values. Question 12. Affordable solution to train a team and make them project ready. Ns = Np Transformation ratio Therefore The input voltage and the powers are 220 V and 1100 W respectively. Explain how would you use this graph to determine the resistance of the conductor. The available power rating of these resistors is 3 to 200 Watts. 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. (2012) Therefore, the current flowing through 20 Ω resistor of given circuit is 2 A. (c) Write the frequency of AC and DC. (a) In a series LCR circuit connected across an ac source of variable frequency, obtain the expression for its impedance and draw a plot showing its variation with the frequency of the ac source. The larger the value of Qsharper is the resonance. 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. The applied ac voltage and the current in an LCR circuit at resonance are in phase. In the first quadrant, the values of both voltage (V) and current (I) are positive. Question 17. In this way, we can convert a delta network into a star network. The rate of consumption of electric energy in an electric appliance is called electric power. Answer: In the above circuit, there are four principal nodes and those are labelled with 1, 2, 3, and 4. WebIn electromagnetism and electronics, electromotive force (also electromotance, abbreviated emf, denoted or ) is an energy transfer to an electric circuit per unit of electric charge, measured in volts.Devices called electrical transducers provide an emf by converting other forms of energy into electrical energy. From the above figure, the V-I characteristics of a network element is a straight line only between the points (-3A, -3V) and (5A, 5V). It is the ratio of voltage (V) across that element between terminals A & B and current (I). We will get the following set of two equations by considering the variables V1 & V2 as dependent and I1 & I2 as independent. This shows current is directly proportional to the potential difference. (c) In device X: Write KCL equation at node P of the following figure. Consider the following series RL circuit diagram. (ii) XC XL i (t) = q0 sin t (a) The primary voltage and Answer: Given L = 100 mH = 0.1 H, C = 5F = 5 10-6F, R=100 , The Thevenin equivalent voltage is the open-circuit voltage (load removed). As the name tells everything, fixed resistor is a resistor which has a specific value and we cant change the value of fixed resistors. = \(\frac{R}{\sqrt{R^{2}+\left(X_{L}-X_{c}\right)^{2}}}\), (ii) find out the condition when the circuit is in resonant state. (b) An airplane is flying horizontally from west to east with a velocity of 900 km h-1. (2017 D) $i(t) = Ke^{-\lgroup \frac{t}{\tau} \rgroup} + \frac{V_m}{\sqrt{R^2 +{\omega}^2 L^2}} sin \lgroup \omega t + \varphi - tan^{-1} \lgroup \frac {\omega L}{R}\rgroup \rgroup$Equation 3. 220/2200 = Ns/3000 R1, R2 and R3 are connected in parallel then equivalent resistance (R) is given by (a) It is more fatal than dc. Advantages of ac: From equations of I1 and I2, we can generalize that the current flowing through any passive element can be found by using the following formula. Two resistors, with resistance 10 and 15 , are to be connected to a battery of e.m.f. This is depicted in the graph (a). Filters as the name suggests, they filter the frequency components. The burning of fossil fuels causes air pollution. (ii) Copper and aluminium wires are usually employed for electricity transmission. Z = 186.1 , Therefore = 5 . Current, I = \(\frac{25}{100}\) = \(\frac{1}{10}\) = 0.1 A. An electric network need not contain a closed path for providing a flow of electrons from a voltage source or current source. (CBSE AI 2019) Now both R and XL are the same and are given by The resistance of this equivalent resistor is equal to sum of the resistances of all those multiple resistors. (a) Name the devices X and Y. Alternatively, start with equations IE emitter-bias and RB emitter-bias in Figure previous, substituting RE with rEE+RE. Efficiency = \(\frac{\text { output power }}{\text { input power }}\) But this time it reverses Its direction. Similarly, we can calculate the other two parameters, B and D by doing short circuit of port2. In the reactance equation, the term 2f (everything on the right-hand side except the L) has a special meaning unto itself. Answer: Z1,Z2, ,Z3 are the impedances of 1st passive element, 2nd passive element, , Nth passive element respectively. Sometimes you will find the rate of instantaneous voltage expressed as v instead of e (v = L di/dt), but it means the exact same thing. Step 3 Re-arrange the equations of Step2 in such a way that they should be similar to the equations of Step1. The two voltages are out of phase by ninety degrees. Therefore, the magnitude of transfer function of Low pass filter will vary from 1 to 0 as varies from 0 to . Sources of direct current are dry cell, dry cell battery, car battery. When do we say that the potential difference across a conductor is 1 volt? (2013) therefore if the resistance in the circuit is doubled then heat produced will also get doubled. The power factor is one. The resistance of the one resistor is 4 . Other definitions of Ohm are as follows; If there is a potential difference of 1 volt between two ends of the conductor and the flowing current through it is 1 Ampere, then the resistance of that conductor would be 1 Ohm (). In another article, we have discussed different ways to find the voltage drop across a capacitor as well. Answer 1: (NCERT Exemplar) Active Elements are the network elements that deliver power to other elements present in an electric circuit. Answer: (i) maximum and The labeled diagram is as shown. Examples: Resistors, Inductors, and capacitors. The chemical reaction within the cell generates the potential difference between its two terminals that sets the electrons in motion to make the current flow through a resistor. Working: The working of the ac generator can be understood with the help of the various positions of the armature as shown in the figure below. V = Vm sin t (1), But from the definition of capacitance, VC = Q / C or Q= VCC. Answer 2: This concept is illustrated in the following figure. This way, Potentiometers can be used as a voltage divider and these resistors are called variable composition resistors. Power of 1 bulb = 40 W, Power of n bulbs = 40 n watts Also, calculate the power dissipation in the circuit. Answer: If the frequency Is doubled, what happens to the capacitive reactance and the current? Kirchhoffs Current Law (KCL) states that the algebraic sum of currents leaving (or entering) a node is equal to zero. (b) the quality factor (Q) of the circuit. We know that, nonlinear resistors are those resistors, where the current flowing through it does not change according to Ohms Law but, changes with change in temperature or applied voltage. The concepts of both transient response and steady state response, which we discussed in the previous chapter, will be useful here too. Find the emitter current IE with the 470K resistor. Also tan = \(\frac{X_{L}}{R}=\frac{3.14}{1}\)= 3.14 Therefore, the given V-I characteristics show that the network element is a Linear, Passive, and Bilateral element. It is measured in ohm. Substitute $I_1 = \frac{V_S}{R_1}$ and $I_2 = \frac{V_S}{R_2}$ in the above equation. Thank you. Working principle: Electromagnetic induction. There are seven branches in the above circuit, among which one branch contains a 20 V voltage source, another branch contains a 4 A current source and the remaining five branches contain resistors having resistances of 30 Ω, 5 Ω, 10 Ω, 10 Ω and 20 Ω respectively. Substitute the value of V1 in the above equation. Using the trigonometric identity The above resonance condition is same as that of series RLC circuit. we have \(\frac{\varepsilon_{\mathrm{s}}}{\varepsilon_{\mathrm{p}}}=\frac{N_{\mathrm{s}}}{N_{\mathrm{p}}} \frac{\frac{d \phi}{d t}}{\frac{d \phi}{d t}}=\frac{N_{\mathrm{s}}}{N_{\mathrm{p}}}\) (2), Therefore form (1) and (2) we have Given L = 44 mH = 44 10-3 H, Vrms = 220 V, f = 50Hz,lrms = ? Now, the current i flows in the entire circuit, since the DC voltage source having V volts is connected to the series RL circuit. It falls below the standard value 82k resistor instead of above it. Why is the tungsten used almost exclusively for filament of electric lamps? Question 6. (iii) Identify one source each for these currents. Answer: In this chapter, let us discuss about the following two equivalent circuits. Examples: Resistors, Inductors, and capacitors. $$Y_{11} = \frac{I_1}{V_1}, \: when \: V_2 = 0$$, $$Y_{12} = \frac{I_1}{V_2}, \: when \: V_1 = 0$$, $$Y_{21} = \frac{I_2}{V_1}, \: when \: V_2 = 0$$, $$Y_{22} = \frac{I_2}{V_2}, \: when \: V_1 = 0$$. From the diagram, we observe that the vector sum of the voltage amplitudes VR, VL, and VC equals a phasor whose length is the maximum applied voltage Vm, where the phasor Vm makes an angle with the current phasor lm. Why might the emitter resistor stabilize a change in current? The number of branches that are present in a co-tree will be equal to the difference between the number of branches of a given graph and the number of twigs. Now slope of the graph is If the branch current is entering towards a selected node, then the value of the element will be -1. The amount of this voltage is dependent on some other voltage or current. Looking at the graph, the voltage wave seems to have a head start on the current wave; the voltage leads the current and the current lags behind the voltage. Note that Rth is RB, the bias resistor from the emitter-bias design. Normally the bias point for VC is set to half of VCC. Rather, we can calculate either total current flowing through or voltage across that resistor by using superposition theorem and from that, we can calculate the amount of power delivered to that resistor using $I^2 R$ or $\frac{V^2}{R}$. The above circuit diagram consists of an input current source IS in parallel with two resistors R1 and R2. By including one link at a time to the above Tree, we will get one f-loop. When the same voltage is applied across a device Y, the same current again flows through it, but it leads the voltage by /2. Now, we can find the current flowing through any element and the voltage across any element that is present in the given network by using node voltages. After applying an input to an electric circuit, the output takes certain time to reach steady state. I will appreciate for it. Based on the type of sources that are present in the network, we can choose one of these three methods. Step 2 Apply source transformation technique to the above Thevenins equivalent circuit. The emitter-bias example is better than the previous base-bias example, but not by much. Step 1 We know that, the following h-parameters of a two port network. Total resistance R = 0.5 2 15 = 15 ohm, Primary voltage of transformer p = 4000 V By using the above relations, we can find the resistances of star network from the resistances of delta network. This is because the inductive reactance is given by the expression XL = 2fL. Answer: Consider the following delta network as shown in the following figure. $$P = V \lgroup C \frac{dV}{dt} \rgroup$$, By integrating the above equation, we will get the energy stored in the capacitor as. (b) Two major sources of energy loss in this device are: (c) No, a transformer does not change the power. The Thevenins resistance across terminals A & B will be, $$R_{Th} = \lgroup \frac{5 \times 10}{5 + 10} \rgroup + 10 = \frac{10}{3} + 10 = \frac{40}{3} \Omega$$. In pure resistors, the voltage to current ratio is not affected by the supply frequency. Therefore, the voltage across all the elements of parallel RLC circuit at resonance is V = IR. In other words, currently leads emf by 90. That means, it rejects (blocks) all low frequency components. 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