9 Important Solutions On Circuit Theory

[Specially picked questions for GATE, JEE, NEET]

In the Circuit theory series, we have come across some fundamental yet essential rules, formulas, and methods. Let us find out some applications of them and understand them more clearly. The problems will be mainly on – KCL, KVL, Thevenin’s theorem, Norton’s theorem, Superposition theorem, Maximum Power Transfer Theorem.

Helping Hands for problem Solving on Circuit Theory:

1. Kirchhoff’s Laws: KCL, KVL
2. Pure AC Circuits
3. Thevenin’s Theorem
4. Norton’s Theorem
5. Superposition Theorem
6. Maximum Power Transfer Theorem
7. Millman’s Theorem
8. Star & Delta Connection

Circuit Theory: 1. Find out the maximum power which can be transferred to the load R_L for the below-given circuit. Apply required theorems of Circuit Theory.

• Solution: Remove the load resistor from the circuitry and voltage source to find out Equivalent Resistance.

So, the resistance or the impedance (AC Circuit) of the circuit through the open terminal:

Z_TH = 2 || j2 = (2 x j2) / (2 +j2) = j2 / (1 +j)

Or, Z_TH = 2 ∠90^o / √2 ∠45^o

Or, Z_TH = √2 ∠45^o

Now, we will calculate the current through the j2 ohms resistor.

I = 4 ∠0^o / (2 +j2)

Or, I = 2 / (1+j) = √2 ∠ – 45^o

The Thevenin’s equivalent voltage comes as V_TH = I * j2.

Or, V_TH = 2√2 ∠45^o V

Now we can redraw the circuit in Thevenin’s equivalent circuit.

Now, from power transfer theorem, R_L = | Z_TH| = √2 ohm for full power.

Now, The current through the load I_L = V_TH / (R_TH + R_L)

Or, I_L = 2√2 ∠45^o / (√2 + √2 ∠45^o)

OR, I_L = 2 ∠45^o / (1 + 1∠45^o)

OR, I_L = 2 ∠45^o / [ 1 + (1 + √2) + (j / √2)]

OR, I_L = 1.08 ∠22.4^o A

|I_L| = 1.08 So, the maximum power is: |I_L²| R_L = (1.08 x 1.08) x √2 = 1.65 W.

Kirchhoff’s Laws: KCL, KVL

Circuit Theory: 2. Find out Norton’s equivalent resistance at terminal AB, for the below-given circuit.

• Solution: At first, we will apply a voltage source at the open circuit at the AB terminal. We name it V_DC and assume I_DC flows from it.

Now, we apply Kirchhoff’s Current Law to do nodal analysis at node a. We can write,

(V_dc – 4I) / 2 + (V_dc / 2) + (V_dc / 4) = I_dc

Here, I = V_dc / 4

Or, 4I = V_dc

Again, (V_dc – V_dc) / 2 + V_dc / 2 + V_dc / 4 = I_dc

Or, 3V_dc / 4 = I_dc

And, V_dc / I_dc = R_N

Or, R_N = 4/3 = 1.33 ohm.

So, the Norton’s Equivalent Resistance is 1.33 ohms.

Circuit Theory: 3. Find out the value of R1 in Delta equivalent circuit of the given star connected network.

• Solution: This problem can be solved easily, using the star’s conversion formula to delta connection.

Let us assume, that R_a = 5 ohms, R_b = 7.5 ohms, and R_c = 3 ohms.

Now, applying the formula,

R₁ = R_a + R_c + (R_a * R_c / R_b)

Or, R₁ = 5 + 3 + (5 x 3) / 7.5

Or, R₁ = 5 + 3 + 2 = 10 ohms.

So, the R1 Delta Equivalent resistance is: 10 ohms.

Circuit Theory: 4. Find out the current flowing through the R2 resistor for the circuit given below.

• Solution: We have to use the idea of source transformation and Kirchhoff’s Voltage Law to find out the answer.

Let us assume ‘I’ Ampere current flows through the R2 (1 kilo-ohm resistor). We can say current through 2-kilo ohm resistance will be (10 – I) Ampere (As current from 10 A source will be 10 A). Similarly, current from 2 A quotation will be 2 A and thus current through 4-kilo ohm resistance will be (I – 2) Ampere.

Now, we apply Kirchhoff’s voltage law in the loop. We can write

I x 1 + (I – 2) x 4 + 3 x I – 2 x (10 – I) = 0

Or, 10I – 8 – 20 = 0

Or, I = 28/10

Or, I = 2.8 mA

So, the current through the R2 resistor is 2.8 mA.

Circuit Theory: 5. If the equivalent resistance for the infinite parallel ladder given in the below image is R_eq, calculate R_eq / R. Also find the value of R_eq when R = 1 ohm.

• Solution: To solve the problem, we must know the equivalent resistance of the infinite parallel ladder. It is given by R_E = R x (1 + √5)/2.

So, we can replace the circuit in the following one.

The equivalent resistance comes here: R_eq = R + R_E = R + 1.618R

Or, R_eq / R = 2.618

And when R = 1 ohm, R_eq = 2.618 x 1 = 2.618 ohm.

Circuit Theory: 6. A source voltage supplies voltage, V_s(t) = V Cos100πt. The source has an internal resistance of (4+j3) ohm. Find out the resistance of a purely resistive load, for transferring maximum power.

• Solution: We know that the power transmitted for a purely resistive circuit is the average power transferred.

So, R_L = √ (R_s² + X_s²)

Or, R_L = √ (4² + 3²)

Or, R_L = 5 ohm.

So, the load will be of 5 ohms.

Circuit Theory: 7.  Find out the Thevenin’s equivalent impedance between node 1 and 2 for the given circuit.

• Solution: To find the Thevenin’s equivalent impedance, we need to connect a voltage source of 1 volt in the place of node 1 and 2. Then we will calculate the current value.

So, Z_TH = 1 / I_TH

Z_TH is the desired resistance we have to find. I_TH is the current flowing due to the voltage source.

Now applying Kirchhoff’s Current law at node B,

i_AB + 99i_b – I_TH =0

Or, i_AB + 99i_b = I_TH ——- (i)

Applying KCL at node A,

i_b – i_A – i_AB = 0

or, i_b = i_A + i_AB ——- (ii)

From equation (i) and (ii) we can write,

i_b – i_A + 99i_b = I_TH

Or, 100i_b – i_A = I_TH ——- (iii)

Now, we apply Kirchhoff’s Voltage law at the outer loop,

10 x 103i_b = 1

Or, i_b = 10^-4 A.

And also,

10 x 103i_b = – 100i_A

Or, i_A = – 100i_A

From equation (iii), we can write,

100i_A + 100i_b = I_TH

Or, I_TH = 200i_b

Or, I_TH = 200 x 10^-4 = 0.02

So, Z_TH = 1 / I_TH = 1 / 0.02 = 50 ohms.

S, the impedance in-between node 1 and 2 is 50 ohms.

Circuit Theory: 8. A complex circuit is given below. Let us assume that both the voltage source of the circuit is in phase with each other. Now, the circuit is divided virtually in two-part A and B by the dotted lines. Calculate R’s value in this circuit for which maximum power is transferred from Part A to Part B.

• Solution: The problem can be solved in a few steps.

First, we find the current ‘i’ through R.

Or, i = (7 / (2 – R) A

Next, current through the 3V source,

i₁ = i – (3 / -j)

Or, i₁ = i – 3j

Then, we calculate the power transferred from Circuit B to A.

P = i²R + i₁ x 3

Or, P = [7 / (2 – R)]² x R + [7 / (2 – R)] x 3 —- (i)

Now, condition for transferring the maximum power is, dP / dR = 0.

So, differentiating equation (i) with respect to R, we can write:

[7 / (2 – R)]² + 98R/ (2 + R)² – 21/ (2 + R)² = 0

Or, 49 x (2 + R) – 98R – 21 x (2 + R)² = 0

Or, 98 + 42 = 49R + 21R

Or, R = 56 / 70 = 0.8 ohm

So, the R value for maximum power transfer from A to B is 0.8 ohm.

Check: Maximum Power Transfer Theorem

Circuit Theory: 9. Find out the value of the resistance for maximum power transferring. Also, find out the maximum delivered power.

• Solution: At the first step, remove the load and calculate the Thevenin’s Resistance. 

V_TH = V * R₂ / (R₁ + R₂)

Or, V_TH = 100 * 20 / (20 +30)

Or, V_TH = 4 V

The resistors are parallelly connected.

So, R_TH = R₁ || R₂

Or, R_TH = 20 || 30

Or, R_TH = 20 * 30 / (20 + 30)

Or, R_TH = 12 Ohms

Now the circuit is redrawn using the equivalent values. For maximum power transfer, R_L = R_TH = 12 ohms.

Maximum power P_MAX = V_TH² / 4 R_TH.

Or, P_MAX = 1002 / (4 × 12)

Or, P_MAX = 10000 / 48

Or, P_MAX = 208.33 Watts

So, the maximum delivered power was 208.33 watts.

Circuit Theory: 10. Calculate the load for maximum power transferring. Find out the transferred power also.

• Solution:

At the first step, remove the load, and calculate the Thevenin’s voltage now.

So, V_AB = V_A – V_B

V_A comes as: V_A = V * R₂ / (R₁ + R₂)

Or, V_A = 60 * 40 / (30 + 40)

Or, V_A = 34.28 v

V_B comes as:

V_B = V * R₄ / (R₃ + R₄)

Or, V_B = 60 * 10 / (10 + 20)

Or, V_B = 20 v

So, V_AB = V_A – V_B

Or, V_AB = 34.28 – 20 = 14.28 v

In the next step, calculation of resistance. As the rule says, remove the voltage and short circuit the connection.

R_TH = R_AB = [{R₁R₂ / (R₁ + R₂)} + {R₃R₄ / (R₃ + R₄)}]

OR, R_TH = [{30 × 40 / (30 + 40)} + {20 × 10 / (20 + 10)}]

OR, R_TH = 23.809 ohms

Now, draw the connection again with the calculated values. For maximum power transfer, R_L = R_TH = 23.809 ohms.

The load value will be = 23.809 ohms.

Maximum power is P_MAX = V_TH² / 4 R_TH.

Or, P_MAX = 14.282 / (4 × 23.809)

Or, P_MAX = 203.9184 / 95.236

Or, P_MAX = 2.14 Watts

So, the maximum delivered power was 2.14 watts.