Circuit Analysis: Theorems and Applications

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Questions and Answers

Find the current flowing through 20 Ω shown in figure using the superposition theorem, Thevenin's and Norton's theorem.

The current flowing through the 20 Ω resistor is 0.5 A. This answer can be found by applying the superposition theorem, Thevenin's theorem, and Norton's theorem.

For the given circuit shown in Figure, write node voltage equations and determine currents in each branch for given network.

The node voltage equations are:

Node 1: $V_1/10 + (V_1 - V_2)/20 = 4$
Node 2: $(V_2 - V_1)/20 + V2/5 + V_2/10 = 20/10$.

Solving these equations, we get:

$V_1$ = 88 V
$V_2$ = 60 V

Hence, the branch currents can be determined as:

Current through 10 Ω from Node 1: $I_1 = V_1/10 = 88/10 = 8.8 A$
Current through 20 Ω: $ I_2 = (V_1 - V_2)/20 = (88 - 60)/20 = 1.4 A$
Current through 5 Ω: $I_3 = V_2/5 = 60/5 = 12 A$
Current through 10 Ω from Node 2: $I_4 = V_2/10 = 60/10 = 6 A$

Using Thevenin's theorem calculate the range of current flowing through the resistance R when its value is varied from 6 W to 36 W.

The Thevenin equivalent of the circuit is:

Thevenin Voltage: $V_{th} = 50 V$
Thevenin Resistance: $R_{th} = 40 Ω$.

The current flowing through the resistance R when its value is varied from 6 Ω to 36 Ω is:

For R = 6 Ω: $I_R = V_{th}/(R + R_{th}) = 50/(6 + 40) = 1.11 A$
For R = 36 Ω: $I_R = V_{th}/(R + R_{th}) = 50/(36 + 40) = 0.63 A$.

Therefore, the range of current flowing through the resistance R is 1.11 A to 0.63 A.

What resistance should be connected across x-y in the circuit shown in figure such that maximum power is developed across this load resistance? What is the amount of this maximum power?

For maximum power transfer, the load resistance should be equal to the Thevenin equivalent resistance of the circuit seen from x-y. <ul> <li>Thevenin Resistance: $R_{th} = (1 +2)||(10 + 3) = 1.67 Ω$.</li> </ul> Therefore, the resistance that should be connected across x-y for maximum power transfer is $R_{th} = 1.67 Ω$. The maximum power transferred to the load is: <ul> <li>Maximum Power: $P_{max} = V^2_{th}/(4R_{th}) = 15^2/(4 \times 1.67) = 21.2 W$</li> </ul> Hence, the amount of maximum power is 21.2 W. Signup and view all the answers

Determine Vx in the circuit.

Applying Kirchhoff's voltage law around the left loop, we have: <ul> <li>Equation: $V_x + 10(5) = 60$ or</li> <li>Equation: $V_x + 50 = 60$ or $V_x = 60-50 = 10 V$</li> </ul> Therefore, Vx is 10 V. Signup and view all the answers

State Norton's theorem with a neat diagram of the Norton's equivalent circuit.

Norton's theorem states that any linear two-terminal network can be replaced by an equivalent current source in parallel with an equivalent resistance. Diagram: <img src="https://www.electronics-tutorials.ws/wp-content/uploads/2019/07/nortons-equivalent-circuit.gif" alt="Norton's equivalent circuit diagram"> Signup and view all the answers

Two batteries of EMF 2.05 V and 2.15 V having internal resistances of 0.05 Ω and 0.04 Ω, respectively are connected together in parallel to supply a load resistance of 1 Ω. Calculate using the superposition theorem, current supplied by each battery and also the load current.

The superposition theorem states that the total current in a circuit is the sum of the individual currents due to each independent voltage source. We can apply this theorem to calculate the current supplied by each battery and the load current. <ul> <li>Current due to 2.05 V battery: $I_1 = 2.05/(1 + 0.05) = 2 A$</li> <li>Current due to 2.15 V battery: $I_2 = 2.15/(1 + 0.04) = 2.06 A$.</li> </ul> The total current supplied by each battery is equal to the individual currents due to each independent battery: $I_1 + I_2 = 2 + 2.06 = 4.06 A$ Therefore, the current supplied by each battery is 2A and 2.06A respectively, and the load current is 4.06A. Signup and view all the answers

Calculate the voltage of the dependent source.

Applying Kirchhoff's voltage law around the loop, we have: <ul> <li>Equation: $6(4) + 15(4) + 9(4) + 0.9(4) + V_x =0$</li> <li>Equation: $24 + 60 + 36 + 3.6 + V_x = 0$</li> <li>Equation: $V_x = -123.6 V$</li> </ul> Therefore, the voltage of the dependent source is -123.6 V. Signup and view all the answers

State Superposition theorem. Using Thevenin's theorem calculate the range of current flowing through the resistance R when its value is varied from 6 Ω to 36 Ω.

The superposition theorem states that in a linear circuit with multiple independent sources, the total current or voltage at any point in the circuit is the algebraic sum of the currents or voltages produced by each source acting independently, with all other sources set to zero. To calculate the range of current flowing through the resistance R, when its value is varied from 6 Ω to 36 Ω, we can use Thevenin's theorem to simplify the circuit to a single voltage source and a single resistor. The Thevenin equivalent of the circuit is: <ul> <li>Thevenin Voltage: $V_{th} = 50 V$</li> <li>Thevenin Resistance: $R_{th} = 40 Ω$.</li> </ul> The current flowing through the resistance R when its value is varied from 6 Ω to 36 Ω is: <ul> <li>For R = 6 Ω: $I_R = V_{th}/(R + R_{th}) = 50/(6 + 40) = 1.11 A$.</li> <li>For R = 36 Ω: $I_R = V_{th}/(R + R_{th}) = 50/(36 + 40) = 0.63 A$.</li> </ul> Therefore, the range of current flowing through the resistance R is 1.11 A to 0.63 A. Signup and view all the answers

Find the equivalent resistance across the terminals A and B of the network shown in figure using Star-delta transformation.

To find the equivalent resistance across terminals A and B, we need to apply the star delta transformation: <ol> <li>Identify delta: The network has delta connection (A-B, B-C ,C-A).</li> <li>Calculate the equivalent star: We can find the equivalent star resistances by using the star delta transformation formulas.</li> </ol> <ul> <li>Ra: $R_a = (2 \times 2) / (2 + 2 + 3) = 4/7 Ω$</li> <li>Rb: $R_b = (1 \times 3) / (2 + 2 + 3) = 3/7 Ω$</li> <li>Rc: $R_c = (1 \times 2) / (2 + 2 + 3) = 2/7 Ω$</li> </ul> <ol start="3"> <li>Redraw the circuit replacing the delta: Now, we replace the delta network with the equivalent star connection.</li> <li>Simplify and Find equivalent: We can now easily calculate the equivalent resistance of the simplified circuit. The resistance between A and B is:</li> </ol> <ul> <li>Equivalent Resistance: $R_{AB} = [(3/7 + 2/7)||2/7] + 3/7 = 2/3 Ω$</li> </ul> Therefore, the equivalent resistance across the terminals A and B is $R_{AB} =2/3 Ω$. Signup and view all the answers

Find the voltage drop between the terminals a-e.

To find the voltage drop between the terminals a-e, we first need to determine the voltage drop across terminals a-b and b-e. <ul> <li>Voltage drop across a-b: Applying KVL in the loop a-b-d-a, we get:</li> <li>Equation: $V_{ab} = 15 - 3(2) = 9 V$</li> <li>Voltage drop across b-e: Applying KVL in the loop b-c-e-b, we get:</li> <li>Equation: $V_{be} = 10 - 2(5) = 0 V$.</li> </ul> Since there is no voltage drop across b-e, the voltage drop between terminals a-e is equal to the voltage drop across a-b: <ul> <li>Voltage drop: $V_{ae} = V_{ab} = 9 V$</li> </ul> Therefore, the voltage drop between the terminals a-e is 9 V. Signup and view all the answers

Calculate using Thevenin's theorem the current flowing through the 5 Ω resistor connected across terminals A and B as shown in figure.

To find the current flowing through terminal A and B, We need to find the Thevenin equivalent of the circuit at those terminals and then apply Ohm's Law: Thevenin Resistance: $R_{th} = 4Ω || (2Ω+3Ω) = 4Ω \times 5Ω ÷ (4Ω+5Ω) = 4Ω || 5Ω = 2.22 Ω$ Thevenin Voltage: First, we can find the voltage across the 5Ω resistor (between A and B) using the voltage divider rule. <ul> <li>Equation $V_{AB} = 15V \times 5Ω ÷ (2Ω+3Ω+5Ω) = 15V \times 5Ω ÷ 10Ω = 7.5V$</li> </ul> Now, to calculate the Thevenin voltage, we remove the 5 Ω resistor and find the open-circuit voltage across AB. <ul> <li>Thevenin Voltage: $V_{th} = V_{AB} = 7.5 V$</li> </ul> Therefore, the Thevenin equivalent of the circuit between terminals A and B is 7.5 V in series with 2.22 Ω. Now, applying Ohm's Law: <ul> <li>Equation: $I_R = V_{th} ÷ (R_{th} + R) = 7.5 ÷ (2.22 + 5) = 1.11 A$</li> </ul> Therefore, the current flowing through the 5 Ω resistor is 1.11 A. Signup and view all the answers

Find Req for the circuit shown in the following figure.

To find the equivalent resistance of the circuit, we need to combine the resistors step-by-step, using the rules for series and parallel combinations. <ol> <li> The 2Ω and 5Ω resistors are in series. Their equivalent resistance is $R_{12} = 2+5 = 7Ω$. </li> <li> Now, the 6Ω and 3Ω resistors are in parallel. Their equivalent resistance is $R_{34} = 6\ast3/(6+3) = 2 Ω$. </li> <li> The 1Ω and 4Ω resistors are in parallel. Their equivalent resistance is $R_{56} = 1\ast4/(1+4) = 0.8Ω$. </li> <li> The 8Ω and 0.8Ω resistors are in series. Their equivalent resistance is $R_{78} = 8+0.8 = 8.8Ω$. </li> <li> Finally, the 7Ω, 2Ω, and 8.8Ω resistors are in parallel. Their equivalent resistance is $R_{eq}= 1/(1/7 + 1/2 + 1/8.8) = 1.34 Ω$. </li> </ol> Therefore, the equivalent resistance Req for the circuit is 1.34 Ω Signup and view all the answers

State Norton's theorem with a neat diagram of the Norton's equivalent circuit. Two batteries of EMF 2.05 V and 2.15 V having internal resistances of 0.05 Ω and 0.04 Ω, respectively are connected together in parallel to supply a load resistance of 1 Ω. Calculate using the superposition theorem, current supplied by each battery and also the load current.

Norton's theorem states that any linear circuit can be replaced by an equivalent circuit consisting of a current source in parallel with a resistor. The current supplied by each battery is 4.20 A and 4.25 A. The load current is 8.45 A. Signup and view all the answers

For the circuit shown in Figure, find: (a) v₁ and v2, (b) the power dissipated in the 3kQ and 20kΩ resistors, and (c) the power supplied by the current source.

(a) v1 = 9 V, v2 = 18 V. (b) Power dissipated in the 3 kΩ resistor is 27 W, and the power dissipated in the 20 kΩ resistor is 32.4 W. (c) Power supplied by the current source is 60 W. Signup and view all the answers

Find the Thevenin equivalent circuit of the circuit shown in Figure, to the left of the terminals a-b. Then find the current through RL = 6, 16, and 36 Ω.

Vth = 8.57 V and Rth = 1.43 Ω. The current through RL = 6 Ω is 1.07 A, the current through RL = 16 Ω is 0.43 A, and the current through RL = 36 Ω is 0.19 A. Signup and view all the answers

Using delta to star transformation, determine the resistance between terminals a and b and the total power drawn from the supply in the circuit shown in Figure.

The resistance between terminals a and b is 7 Ω. The total power drawn from the supply in the circuit shown in Figure is 25 W. Signup and view all the answers

Calculate the current, I supplied by the battery in the circuit shown in Figure.

The current, I supplied by the battery in the circuit shown in Figure is 2 A. Signup and view all the answers

In the circuit shown in the Figure, Determine the value of RL for which the maximum power will be transfer, also find the maximum power transferred to the load.

The value of RL for which the maximum power will be transferred is 3Ω. The maximum power transferred is 36 W. Signup and view all the answers

Transform the circuit given in Figure into Norton equivalent Circuit across terminal a-b and determine the current across the load resistance taking RL= 6 ohm.

The Norton equivalent circuit across terminal a-b has IN = 4 A and RN = 3 Ω. The current across the load resistance taking RL = 6 ohm is 2 A. Signup and view all the answers

Three equal resistors are connected as shown in Figure. Find the equivalent resistance between points A and B.

The equivalent resistance between points A and B is R/3. This is due to combining three resistors in series and then finally combining those into parallel. Signup and view all the answers

Find the voltage VAB in the circuit shown in Figure.

The voltage VAB in the circuit shown in Figure is 6 V. Signup and view all the answers

Figure shows two batteries connected in parallel, each represented by an emf along with its internal resistance. A load resistance of 6 Ω is connected across the ends of the batteries. Calculate the current through each battery and the load.

Current through battery 1 is 1.41 A, Current through battery 2 is 1.44 A and current through load is 2.85 A. Signup and view all the answers

Calculate the value of load resistance, R₁ for which maximum power will be transferred from the source to the load and the value of the maximum power. Also, calculate the maximum power transfer efficiency.

R₁ = 4 Ω, Pmax = 121 W, efficiency = 50%. Signup and view all the answers

By using the superposition theorem calculate the current flowing through the 10 Ω resistor in the network shown.as shown in Figure is

The current flowing through the 10 Ω resistor is 0.7 A. Signup and view all the answers

Determine the current through the 6Ω resistance connected across the terminals A and B in the electric circuit shown in Figure.

The current through the 6 Ω resistance connected across the terminals A and B in the electric circuit shown in Figure is 1 A. Signup and view all the answers

Find the value of R if 122 resistor draw 1 A current as shown in Figure. Also find the power absorbed in the R resistor.

The value of R is 12 Ω. The power absorbed in the R resistor is 12 W. Signup and view all the answers

Find the value of E, the current in the 12 ohm is 5 A as shown below.

The value of E is 107 V. Signup and view all the answers

Using Thevenin's theorem to calculate the current flowing through the 5 Ω resistor in the circuit shown in Figure.

The current flowing through the 5 Ω resistor in the circuit shown in Figure is 1.2 A. Signup and view all the answers

By applying Thevenin's as well as Norton's theorem show that current flowing through the 16 Ω resistance in the following network is 0.5 A.

The current flowing through the 16 Ω resistance in the following network is 0.5 A. Signup and view all the answers

State ohm's law. State and explain maximum power transfer theorem for DC circuits with suitable example.

Ohm's law states that the current (I) flowing through a conductor is directly proportional to the voltage (V) across its terminals and inversely proportional to its resistance (R). The maximum power transfer theorem states that maximum power is delivered to the load when the load resistance (R) is equal to the internal resistance (r) of the source. This means that the power delivered to the load will be maximum. Signup and view all the answers

Find the value of the current i for the circuit shown in Figure. Calculate the power delivered by 8A current source.

The value of the current i is 6 A. The power delivered by the 8 A current source is 96 W. Signup and view all the answers

Compute the power absorbed by the 3-ohm resistor in the circuit of Figure using any method of your choice.

The Power absorbed by the 3-ohm resistor in the circuit of Figure is 72 W. Signup and view all the answers

Find Rab. (R=900)

R1 = 900 Ω. Rab is 50 Ω. Signup and view all the answers

In the circuit shown, find the voltage Vx(in volts)

The Voltage Vx is 1 V. Signup and view all the answers

For the circuit shown in figure, find the thevenin's equivalent voltage in volts across teriminals a-b.

The Thevenin's equivalent voltage across terminals a-b is 8 V. Signup and view all the answers

Find the value of R₁ for maximum power transfer and calculate maximum power in the given circuit shown in Figure.

The value of R₁ for maximum power transfer is 2 Ω. The maximum power in the given circuit shown in the Figure is 25 W. Signup and view all the answers

Find current in 6Ω resistor using Norton's theorem for the network shown in Figure

The current in the 6 Ω resistor using Norton's theorem for the network shown in Figure is 0.5 A. Signup and view all the answers

Refer to the circuit shown in Figure below. Calculate (i) i₁(0+), vc(0+) and VR(0+) (ii) di₁(0+)/dt, dvc(0+)/dt and dvr(0+)/dt (iii) i₁(∞), vc(∞) and vr(∞).

(i) i1(0+) = 2A, Vc(0+) = 0V, Vr(0+) = 0V, (ii) di1(0+)/dt = 2, dVc(0+)/dt = 20, dVr(0+)/dt = 20, (iii) i1(∞) = 0, Vc(∞) = 10V, Vr(∞) = 10V. Signup and view all the answers

Determine the total current drawn from the supply by the series-parallel circuit shown in Figure. Also calculate the power factor of the circuit.

The total current drawn from the supply is 7.4 A. The power factor is 0.99. Signup and view all the answers

A circuit having a resistance of 12 Ω, an inductance of 0.15 H and a capacitance of 100 µF in series, is connected across a 100 V, 50 Hz supply. Calculate: (a) the total impedance; (b) the current drawn; (c) the voltages across R, L and C; (d) the phase difference between the current and the supply voltage.

(a) Z = 16.4 Ω, (b) I = 6.1 A, (c) VR = 73.2 V, VL = 57.9 V , VC = 193 V, (d) phase difference = 47.8° Signup and view all the answers

Calculate the value of R₁ such that the circuit will resonate.

The value of R₁ such that the circuit will resonate for maximum power transfer to the load is 57.6 Ω. Signup and view all the answers

Consider a linear time inverse system given by d²y(t)/dt² + 7 dy(t)/dt + 10y(t) = dx(t)/dt + 6x(t) x(t) = e-2t u(t) for initial condition: y(0) = 6, dy(0)/dt = -4. Find the natural response, forced response, and total response

The natural response is y_n(t) = (5 + 1)e-2t + (1 - 6)e-5t, The forced response is y_f(t) = 2e-2t, The total response is y(t) = (5 + 1)e-2t + (1 - 6)e-5t + Ce-2t. Signup and view all the answers

Calculate the RMS value, average value and form factor of a half-rectified square voltage.

The RMS value is 5√2 V, the average value is 5V, and the form factor is √2. Signup and view all the answers

A variable resistance R and an inductance L of value 100 mH in series are connected across at 50 Hz supply. Calculate at what value of R the voltage across the inductor will be half the supply voltage.

The value of R for the voltage across the inductor to be half the supply voltage is 50 Ω. Signup and view all the answers

In a series R-L-C circuit, the following values are known. V = 230 v, f = 50 Hz, L = 20mH, R = 20Ω, C = 0.01 µF. Find impedance Z, Current I, power factor and Power consumed P.

The impedance is Z = 28.3 Ω, the Current is I = 8.14 A, the power factor is 0.707, and the power consumed is P = 1400 W. Signup and view all the answers

For the circuit shown in Fig. calculate the total current drawn from the supply. Also calculate the power and power factor of the circuit also draw the phasor diagram.

The current I = 1.3 A, power = 338 W, the power factor is 0.866. Signup and view all the answers

A series RLC circuit with L =160 mH, C = 100 µF, and R = 40.0Ω is connected to a sinusoidal voltage V (t) = 40sinot, with w = 200 rad/s (i) What is the impedance of the circuit? (ii) Let the current at any instant in the circuit be I (t) = Io sin (ωt -φ). Find Io. (iii)What is the power factor?

(i) Z = 50 Ω, (ii) Io = 0.8 A, (iii) Power factor = 0.8 Signup and view all the answers

For the first order circuit shown in the Figure, determine i(t) for t >0.

The solution of the first order circuit is i(t) = 1.33e-1/6t + 0.67A. Signup and view all the answers

A capacitor has a capacitance of 30 µF. Find its capacitive reactance for frequencies of 25 and 50 Hz. Find in each case the current if the supply voltage is 440 V.

The capacitive reactance of the capacitor is Xc = 212.2 Ω, the current at 25 Hz is 2.08A, and the current at 50 Hz is 4.16A. Signup and view all the answers

A 100 µF capacitor is connected across a 230 V, 50 Hz supply. Determine (i) the maximum instantaneous charge on the capacitor and (ii) the maximum instantaneous energy stored in the capacitor.

(i) Qmax = 3.29 × 10^-3 C, (ii) Wmax = 0.37 J Signup and view all the answers

An inductive coil having negligible resistance and 0.1 Henry inductance is connected across a 200 V, 50 Hz supply. Find I. Inductive reactance, II. Rms value of current, III. Power, IV. Power factor, and V. Equations for voltage and current.

(i) Xl = 31.42Ω, (ii) Irms = 6.37A, (iii) Power = 804.24W, (iv) Power factor = 0, (v) Equations for voltage and current are V(t) = 200√2sin(ωt) Volt, i = 6.37sin(ωt + 90°) Ampere. Signup and view all the answers

The voltage and current through a circuit element are v = 50 sin(314t + 55° )V i = 10sin(314t + 325°)A Find the value of power drawn by the element.

The power drawn by the element is 176.78 W. Signup and view all the answers

A 52 load at 0.8 PF connected across single phase, 240 V AC supply as shown in Figure. Calculate the reactive power drawn by the load.

The reactive power drawn by the load is 144 VAR. Signup and view all the answers

Flashcards

Superposition Theorem

A theorem that states that the current flowing through any linear resistive network can be determined by considering the effects of each independent source acting alone, with all other sources being replaced by their internal resistances.

Equivalent Resistance (Req)

This analysis involves finding the equivalent resistance (Req) of a circuit network, representing the total resistance between two specified terminals.

Thevenin's Theorem

Thevenin's theorem allows you to simplify a complex circuit into a simpler equivalent circuit consisting of a voltage source (Vth) in series with a resistor (Rth). This simplifies analysis, especially when dealing with different load resistances.

Norton's Theorem

Norton's theorem represents a circuit with an equivalent current source (In) in parallel with a resistor (Rn). This allows you to focus on the current flowing through the circuit.