This set of Network Theory Questions & Answers for Exams focuses on “Advanced Problems on Reciprocity Theorem”.

1. In Reciprocity Theorem, which of the following ratios is considered?

a) Voltage to current

b) Current to current

c) Voltage to voltage

d) No ratio is considered

View Answer

Explanation: The Reciprocity Theorem states that if an Emf E in one branch produces a current I in a second branch, then if the same emf E is moved from the first to the second branch, it will produce the same current in the first branch, when the Emf E in the first branch is replaced with a short circuit. Therefore the ratio of Voltage to Current is considered in case of Reciprocity Theorem.

2. The Reciprocity Theorem is valid for ___________

a) Non-Linear Time Invariant circuits

b) Linear Time Invariant circuits

c) Non-Linear Time Variant circuits

d) Linear Time Variant circuits

View Answer

Explanation: A reciprocal network comprises of linear time-invariant bilateral elements. It is applicable to resistors, capacitors, inductors (including coupled inductors) and transformers. However, both dependent and independent sources ate not permissible.

3. In the circuit given below, the current in the 4-ohm resistor is __________

a) 3.5 A

b) 2.5 A

c) 1.5 A

d) 0.5 A

View Answer

Explanation: R

_{th}= [(2+4) || 6] + 12 = 15 Ω

I

_{S}= \(\frac{45}{15}\) = 3 A

Now, by current division rule, we get, I = \(\frac{3 × 6}{12} = \frac{18}{12}\) = 1.5 A.

4. The Reciprocity Theorem is applicable for __________

a) Single-source networks

b) Multi-source networks

c) Both Single and Multi-source networks

d) Neither Single nor Multi-source networks

View Answer

Explanation: According to Reciprocity Theorem, the voltage source and the resulting current source may be interchanged without a change in current. Therefore the theorem is applicable only to single-source networks. It therefore cannot be employed in multi-source networks.

5. In the circuit given below, the current through the 12 Ω resistance is _________

a) 1.5 A

b) 2.5 A

c) 3.5 A

d) 4.5 A

View Answer

Explanation: Equivalent resistance, R

_{EQ}= [(12 || 6) + 2 + 4] = 10 Ω

I

_{S}= \(\frac{45}{10}\) = 4.5 A

Now, by using Current division rule, we get, I = \(\frac{4.5 × 6}{12+6} = \frac{27}{18}\) = 1.5 A.

6. A circuit is given in the figure below. We can infer that ________

a) The circuit follows Reciprocity Theorem

b) The circuit follows Millman’s Theorem

c) The circuit follows Superposition Theorem

d) The circuit follows Tellegen Theorem

View Answer

Explanation: Let us consider this circuit,

R

_{th}= [(2+4) || 6] + 12 = 15 Ω

I

_{S}= \(\frac{45}{15}\) = 3 A

Now, by current division rule, we get, I

_{1}= \(\frac{3 × 6}{12} = \frac{18}{12}\) = 1.5 A.

Again, let us consider this circuit,

Equivalent resistance,R

_{EQ}= [(12 || 6) + 2 + 4] = 10 Ω

I

_{S}= \(\frac{45}{10}\) = 4.5 A

Now, by using Current division rule, we get, I

_{2}= \(\frac{4.5 × 6}{12+6} = \frac{27}{18}\) = 1.5 A.

Since I

_{1}= I

_{2}, the circuit follows Reciprocity Theorem.

7. In the circuit given below, the current in the resistance 20 Ω(far end) is _________

a) 8.43 A

b) 5.67 A

c) 1.43 A

d) 2.47 A

View Answer

Explanation: Equivalent Resistance R

_{EQ}= 20 + [30 || (20 + (20||20))]

= 20 + [30 || (20 + \(\frac{20×20}{20+20}\))]

= 20 + [30 || (20+10)]

= 20 + [30 || 30]

= 20 + \(\frac{30 × 30}{30+30}\)

= 20 + 15 = 35 Ω

The current drawn by the circuit = \(\frac{200}{35}\) = 5.71 A

Now, by using current division rule, we get, I

_{2Ω}= 1.43 A.

8. In the circuit given below, the value of I is __________

a) 2.47 A

b) 5.67 A

c) 8.43 A

d) 1.43 A

View Answer

Explanation: Equivalent Resistance, R

_{EQ}= [[((30 || 20) + 20) || 20] + 20]

= \(\Big[\Big[\left(\left(\frac{30 × 20}{30+20}\right) + 20\right) || 20\Big] + 20\Big]\)

= [[(12 + 20) || 20] + 20]

= [[32 || 20] + 20]

= \(\Big[\left(\frac{32 × 20}{32+20}\right) + 20\Big]\)

= [12.31 + 20] = 32.31 Ω

The current drawn by the circuit = \(\frac{200}{32.31}\) = 6.19 A

Now, by using current division rule, we get, I

_{2Ω}= 1.43 A.

9. A circuit is given in the figure below. We can infer that ________

a) The circuit follows Reciprocity Theorem

b) The circuit follows Millman’s Theorem

c) The circuit follows Superposition Theorem

d) The circuit follows Tellegen Theorem

View Answer

Explanation: Let us consider this circuit,

Equivalent Resistance R

_{EQ}= 20 + [30 || (20 + (20||20))] = 20 + [30 || (20 + \(\frac{20×20}{20+20}\))]

= 20 + [30 || (20+10)]

= 20 + [30 || 30]

= 20 + \(\frac{30 × 30}{30+30}\)

= 20 + 15 = 35 Ω

The current drawn by the circuit = \(\frac{200}{35}\) = 5.71 A

Now, by using current division rule, we get, I

_{1}= 1.43 A

Again, let us consider this circuit,

Equivalent Resistance, R

_{EQ}= [[((30 || 20) + 20) || 20] + 20]

= \(\Big[\Big[\left(\left(\frac{30 × 20}{30+20}\right) + 20\right) || 20\Big] + 20\Big]\)

= [[(12 + 20) || 20] + 20]

= [[32 || 20] + 20]

= \(\Big[\left(\frac{32 × 20}{32+20}\right) + 20\Big]\)

= [12.31 + 20] = 32.31 Ω

The current drawn by the circuit = \(\frac{200}{32.31}\) = 6.19 A

Now, by using current division rule, we get, I

_{2}= 1.43 A.

Since I

_{1}= I

_{2}, the circuit follows Reciprocity Theorem.

10. In the circuit given below, the current in the 30 Ω resistor is _________

a) 1 A

b) 2 A

c) 3 A

d) 4 A

View Answer

Explanation: Equivalent Resistance, R

_{EQ}= 20 + [60 || 30]

= 20 + \(\frac{60 × 30}{60+30}\)

= 20 + 20 = 40 Ω

Total current from the source, I = \(\frac{120}{40}\) = 3A

Now, by using current division rule, I

_{3Ω}= \(\frac{3 × 60}{30+60}\) = 2 A.

11. In the circuit given below, the current in the 20 Ω resistor is _________

a) 5 A

b) 1 A

c) 1.5 A

d) 2 A

View Answer

Explanation: Equivalent resistance, R

_{EQ}= [[20 || 60] + 30]

= \(\Big[\frac{20 × 60}{20+60} + 30\Big]\)

= [15 + 30] = 45 Ω

Total current = \(\frac{120}{45}\) = 2.67 A

Current through the 20Ω resistor is, I

_{20Ω}= \(\frac{2.67 × 60}{60+20}\) = 2 A.

12. A circuit is given in the figure below. We can infer that ________

a) The circuit follows Reciprocity Theorem

b) The circuit follows Millman’s Theorem

c) The circuit follows Superposition Theorem

d) The circuit follows Tellegen Theorem

View Answer

Explanation: Let us consider this circuit,

Equivalent Resistance, Equivalent Resistance, R

_{EQ}= 20 + [60 || 30]

= 20 + \(\frac{60 × 30}{60+30}\)

= 20 + 20 = 40 Ω

Total current from the source, I = \(\frac{120}{40}\) = 3A

Now, by using current division rule, I

_{1}= \(\frac{3 × 60}{30+60}\) = 2 A.

Again, let us consider this circuit,

Equivalent resistance, R

_{EQ}= [[20 || 60] + 30]

= \(\Big[\frac{20 × 60}{20+60} + 30\Big]\)

= [15 + 30] = 45 Ω

Total current = \(\frac{120}{45}\) = 2.67 A

Current through the 20Ω resistor is, I

_{2}= \(\frac{2.67 × 60}{60+20}\) = 2 A

Since I

_{1}= I

_{2}, the circuit follows Reciprocity Theorem.

13. In the circuit given below, the value of load R_{L}, for which maximum power is transferred through it is ___________

a) 2 Ω

b) 3 Ω

c) 1 Ω

d) 6 Ω

View Answer

Explanation: I + 0.9 = 10 I

Or, I = 0.1 A

V

_{OC}= 3 × 10 I = 30 I

Or, V

_{OC}= 3 V

Now, I

_{SC}= 10 I = 1 A

R

_{th}= 3/1 = 3 Ω.

14. In the circuit given below, the maximum power absorbed by the load resistance R_{L} is ___________

a) 2200 W

b) 1250 W

c) 1000 W

d) 621 W

View Answer

Explanation: R

_{L}= \(\sqrt{R_{TH}^2+X_{TH}^2}\)

= \(\sqrt{3^2+4^2}\) = 5

Now, 110 = (6 + j8 + 5) I

_{1}+ 5I

_{2}

And 90 = (6 + j8 + 5) I

_{2}= 5I

_{1}

∴ I

_{1}= 5.5 – 2.75j and I

_{2}= 4.5 – 2.2j

Total current in R

_{L}= I

_{1}+ I

_{2}= (10 – 4.95j) A = 11.15 A

∴ Power absorbed by R

_{L}= I

^{2}R

= 11.15

^{2}× 5 = 621 W.

15. In the circuit given below, the maximum power delivered to the load is ___________

a) 3 W

b) 5.2 W

c) 3.2 W

d) 4.2 W

View Answer

Explanation: Equivalent resistance of the circuit is = [{(3 + 2) || 5} + 10] = (2.5 + 10) = 12.5 Ω

Total current drawn by the circuit is I

_{T}= \(\frac{50}{12.5}\) = 4 A

Current in 3 Ω resistor is I

_{3}= I

_{T}× \(\frac{5}{5+5} = \frac{4 × 5}{10}\) = 2 A

V

_{TH}= V

_{3}= 3 × 2 = 6V

R

_{TH}= R

_{AB}= [(2 + 5) || 3] = 2.1 Ω

For maximum power transfer R

_{L}= R

_{TH}= 2.1 Ω

∴ Current drawn by R

_{L}is I

_{L}= \(\frac{6}{2.1+2.1} = \frac{6}{4.2}\) = 1.42 A

∴ Power delivered to the load = \(I_L^2 R_L\)

= (1.42)

^{2}(2.1) = 4.2 W.

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