This set of Network Theory Question Paper focuses on “Advanced Problems on Two Port Network – 2”.
1. A network contains linear resistors and ideal voltage source S. If all the resistors are made twice their initial value, then voltage across each resistor is __________
a) Halved
b) Doubled
c) Increases by 2 times
d) Remains same
View Answer
Explanation: The voltage/resistance ratio is a constant (say K). If K is doubled then, electric current will become half. So voltage across each resistor remains same as was initially.
2. A voltage waveform V(t) = 12t2 is applied across a 1 H inductor for t ≥ 0, with initial electric current through it being zero. The electric current through the inductor for t ≥ 0 is given by __________
a) 12 t
b) 24 t
c) 12 t3
d) 4 t3
View Answer
Explanation: We know that, I = \( \frac{1}{L} \int_0^t V \,dt\)
= \(1\int_0^t 12 t^2 \,dt\)
= 4 t3.
3. The linear circuit element among the following is ___________
a) Capacitor
b) Inductor
c) Resistor
d) Inductor & Capacitor
View Answer
Explanation: A linear circuit element does not change their value with voltage or current. The resistance is only one among the others does not change its value with voltage or current.
4. Consider a circuit having resistance 10 kΩ, excited by voltage 5 V and an ideal switch S. If the switch is repeatedly closed for 2 ms and opened for 2 ms, the average value of i(t) is ____________
a) 0.25 mA
b) 0.35 mA
c) 0.125 mA
d) 1 mA
View Answer
Explanation: Since i = \(\frac{5}{10 × 2 X 10^{-3}}\) = 0.25 × 10-3 = 0.25 mA.
As the switch is repeatedly close, then i (t) will be a square wave.
So average value of electric current is (\(\frac{0.25}{2}\)) = 0.125 mA.
5. In the circuit given below the value of resistance Req is _____________
a) 10 Ω
b) 11.86 Ω
c) 11.18 Ω
d) 25 Ω
View Answer
Explanation: The circuit is as shown in figure below.
Req = 5 + \(\frac{10(R_{eq}+5)}{10 + 5 + R_{eq}}\)
Or, \(R_{eq}^2\) + 15Req = 5Req + 75 + 10Req + 50
Or, \(R_{eq} = \sqrt{125}\) = 11.18 Ω.
6. A particular electric current is made up of two components a 10 A, a sine wave of peak value 14.14 A. The average value of electric current is __________
a) 0
b) 24.14 A
c) 10 A
d) 14.14 A
View Answer
Explanation: Average dc electric current = 10 A.
Average ac electric current = 0 A since it is alternating in nature.
Average electric current = 10 + 0 = 10 A.
7. Given that, R1 = 36 Ω and R2 = 75 Ω, each having tolerance of ±5% are connected in series. The value of resultant resistance is ___________
a) 111 ± 0 Ω
b) 111 ± 2.77 Ω
c) 111 ± 5.55 Ω
d) 111 ± 7.23 Ω
View Answer
Explanation: R1 = 36 ± 5% = 36 ± 1.8 Ω
R2 = 75 ± 5% = 75 ± 3.75 Ω
∴ R1 + R2 = 111 ± 5.55 Ω.
8. Consider a circuit having a charge of 600 C, which is delivered to 100 V source in a 1 minute. The value of Voltage source V is ___________
a) 30 V
b) 60 V
c) 120 V
d) 240 V
View Answer
Explanation: In order for 600 C charges to be delivered to 100 V source, the electric current must be in reverse clockwise direction.
Now, I = \(\frac{dQ}{dt}\)
= \(\frac{600}{60}\) = 10 A
Applying KVL we get
V1 + 60 – 100 = 10 × 20 ⇒ V1 = 240 V.
9. The energy required to charge a 10 μF capacitor to 100 V is ____________
a) 0.01 J
b) 0.05 J
c) 5 X 10-9 J
d) 10 X 10-9 J
View Answer
Explanation: E = \(\frac{1}{2}\) CV2
= 5 X 10-6 X 1002
= 0.05 J.
10. For the circuit given below, the value of the hybrid parameter h11 is ___________
a) 75 Ω
b) 80 Ω
c) 90 Ω
d) 105 Ω
View Answer
Explanation: Hybrid parameter h11 is given by, h11 = \(\frac{V_1}{I_1}\), when V2=0.
Therefore short circuiting the terminal Y-Y’, we get,
V1 = I1 ((50||50) + 50)
= I1 \(\left(\left(\frac{50×50}{50+50}\right) + 50\right)\)
= 75I1
∴ \(\frac{V_1}{I_1}\) = 75.
Hence h11 = 75 Ω.
11. For the circuit given below, the value of the hybrid parameter h21 is ___________
a) 0.6 Ω
b) 0.5 Ω
c) 0.3 Ω
d) 0.2 Ω
View Answer
Explanation: Hybrid parameter h21 is given by, h21 = \(\frac{I_2}{I_1}\), when V2 = 0.
Therefore short circuiting the terminal Y-Y’, and applying Kirchhoff’s law, we get,
-50 I2 – (I2 – I1)50 = 0
Or, -I2 = I2 – I1
Or, -2I2 = -I1
∴ \(\frac{I_2}{I_1} = \frac{1}{2}\)
Hence h21 = 0.5 Ω.
12. For the circuit given below, the value of the Inverse hybrid parameter g11 is ___________
a) 0.133 Ω
b) 0.025 Ω
c) 0.3 Ω
d) 0.25 Ω
View Answer
Explanation: Inverse Hybrid parameter g11 is given by, g11 = \(\frac{I_1}{V_1}\), when I2 = 0.
Therefore short circuiting the terminal Y-Y’, we get,
V1 = I1 ((5||5) + 5)
= I1 \(\left(\left(\frac{5×5}{5+5}\right) + 5\right)\)
= 7.5I1
∴ \(\frac{I_1}{V_1} = \frac{1}{7.5}\) = 0.133 Ω
Hence g11 = 0.133 Ω.
13. A resistor of 10 kΩ with the tolerance of 5% is connected in series with 5 kΩ resistors of 10% tolerance. What is the tolerance limit for a series network?
a) 9%
b) 12.04%
c) 8.67%
d) 6.67%
View Answer
Explanation: Error in 10 kΩ resistance = 10 × \(\frac{5}{100}\) = 0.5 kΩ
Error in 5 kΩ resistance = 5 × \(\frac{10}{100}\) = 5 kΩ
Total measurement resistance = 10 + 0.5 + 5 + 0.5 = 16 kΩ
Original resistance = 10 + 5 = 15 kΩ
Error = \(\frac{16-15}{15}\) × 100 = \(\frac{1}{15}\) × 100 = 6.67%.
14. A 200 μA ammeter has an internal resistance of 200 Ω. The range is to be extended to 500μA. The shunt required is of resistance __________
a) 20.0 Ω
b) 22.22 Ω
c) 25.0 Ω
d) 50.0 Ω
View Answer
Explanation: Ish Rsh = Im Rm
Ish = I – Im or, \(\frac{I}{I_m} – 1 = \frac{R_m}{R_{sh}}\)
Now, m = \(\frac{I}{I_m}\)
Or, m – 1 = \(\frac{R_m}{R_{sh}}\)
∴Rsh = 25 Ω.
15. A voltmeter has a sensitivity of 1000 Ω/V reads 200 V on its 300 V scale. When connected across an unknown resistor in series with a millimeter, it reads 10 mA. The error due to the loading effect of the voltmeter is
a) 3.33%
b) 6.67%
c) 13.34%
d) 13.67%
View Answer
Explanation: RT = \(\frac{V_T}{I_T}\)
VT = 200 V, IT = 10 A
So, RT = 20 kΩ
Resistance of voltmeter,
RV = 1000 × 300 = 300 kΩ
Voltmeter is in parallel with unknown resistor,
RX = \(\frac{R_T R_V}{R_T – R_V} = \frac{20 ×300}{280}\) = 21.43 kΩ
Percentage error = \(\frac{Actual-Apparent}{Actual}\) × 100
= \(\frac{21.43-20}{21.43}\) × 100 = 6.67%.
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