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Electric Circuits Test 2

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Electric Circuits Test 2

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Question 1
2 / -0.33

In the circuit of fig. v_∞ = 0 for t > 0. The initial condition are v(0) = 6V and dv(0) /dt = -3000 V s. The v(t) for t > 0 is

A
-2e^-100t + 8e ^-400t V

B
6e^-100t + 8e ^-400t V

C
6e^-100t - 8e ^-400t V

D
None of the above

Solution
Question 2
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The circuit shown in fig. P1.6.5 has been open for a long time before closing at t = 0. The initial condition is v(0) = 2V. The v(t) for t > is

A
5e^-t - 7e ^-3t V

B
7e^-t - 5e ^-3t V

C
-e^-t + 3e ^-3t V

D
3e^-t - e ^-3t V

Solution
Question 3
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Circuit is shown in fig. Initial conditions are i₁(0) = i₂(0) =11A

i₁ (1s) = ?

A
0.78 A

B
1.46 A

C
2.56 A

D
3.62 A

Solution
Question 4
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Circuit is shown in fig. P.1.6. Initial conditions are (0)i₁ = i₂(0) = 11A

i₂ (1 s)= ?

A
0.78 A

B
1.46 A

C
2.56 A

D
3.62 A

Solution
Question 5
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v_c (t ) =? for t > 0

A

B

C

D

Solution
Question 6
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The switch of the circuit shown in fig. is opened at t = 0 after long time. The v(t) , for t > 0 is

A

B

C

D

Solution
Question 7
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In the circuit of fig.the switch is opened at t = 0 after long time. The current iL(t) for t > 0 is

A

B

C

D

Solution
Question 8
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In the circuit shown in fig.all initial condition are zero.

If i_s (t) = 1 A, then the inductor current i_L(t) is

A
1 A

B
t A

C
t + 1 A

D
0 A

Solution
Question 9
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In the circuit shown in fig. all initial condition are zero

If i_s(t) = 0.5t A, then i_L(t) is

A

B

C

D
2t + 3250 A

Solution
Question 10
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In the circuit shown in fig. a steady state has been established before switch closed. The i(t) for t > 0 is

A

B

C

D

Solution
Question 11
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Two 2-port networks with following parameters are connected in series.
\(Z - parameters\;of\;network - 1 = \left[ {\begin{array}{*{20}{c}} 5&6\\ 9&2 \end{array}} \right]\)
\(Y - parameters\;of\;network - 2 = \left[ {\begin{array}{*{20}{c}} 9&5\\ 7&4 \end{array}} \right]\)
The z-parameters of series combination network is

A
\(\left[ {\begin{array}{*{20}{c}} {14}&{11}\\ {16}&6 \end{array}} \right]\)

B
\(\left[ {\begin{array}{*{20}{c}} 9&1\\ 2&{11} \end{array}} \right]\)

C
\(\left[ {\begin{array}{*{20}{c}} 1&{11}\\ {16}&{ - 7} \end{array}} \right]\)

D
\(\left[ {\begin{array}{*{20}{c}} { - 22}&{29}\\ {22}&{ - 27} \end{array}} \right]\)

Solution

\(Z - parameters\;of\;network - 1 = \left[ {\begin{array}{*{20}{c}} 5&6\\ 9&2 \end{array}} \right]\)
\(Y - parameters\;of\;network - 2 = \left[ {\begin{array}{*{20}{c}} 9&5\\ 7&4 \end{array}} \right]\)
\(Z - parameter\;of\;network - 2 = {y^{ - 1}} = {\left[ {\begin{array}{*{20}{c}} 9&5\\ 7&4 \end{array}} \right]^{ - 1}}\)
\( = \frac{1}{{\left( {36 - 35} \right)}}\left[ {\begin{array}{*{20}{c}} 4&{ - 5}\\ { - 7}&9 \end{array}} \right]\)
\(= \left[ {\begin{array}{*{20}{c}} 4&{ - 5}\\ { - 7}&9 \end{array}} \right]\)
In a series combination, the z-parameters of two network will get add.
Z-parameters of series combination \( = \left[ {\begin{array}{*{20}{c}} 5&6\\ 9&2 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 4&{ - 5}\\ { - 7}&9 \end{array}} \right]\)
\(= \left[ {\begin{array}{*{20}{c}} 9&1\\ 2&{11} \end{array}} \right]\)
Question 12
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The switch is closed after long time in the circuit of fig. The v(t) for t > 0 is

A

B

C

D

Solution
Question 13
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In the circuit of fig. i(0) = 1A and v(0) = 0. The current i(t) for t > 0 is

A

B

C

D

Solution
Question 14
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In the circuit of fig. a steady state has been established before switch closed. The v_o (t) for t >0 is

A

B

C

D

Solution
Question 15
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In the circuit of fig. a steady state has been established before switch closed. The i(t) for t > 0 is

A

B

C

D

Solution
Question 16
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An Inductor works as a ________ circuit for DC supply.

A
Open

B
Short

C
Polar

D
Non-polar

Solution

Induced voltage across an inductor is zero if the current flowing through it is constant, i.e. Inductor works as a short circuit for DC supply.
Question 17
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A power factor of a circuit can be improved by placing which, among the following, in a circuit?

A
Inductor

B
Capacitor

C
Resistor

D
Switch

Solution

Power factor = Real power / Apparent power = kW/kVA

By adding a capacitor in a circuit, an additional kW load can be added to the system without altering the kVA.

Hence, the power factor is improved.
Question 18
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Which among the following equations is incorrect?

A
Q = CV

B
Q = C/V

C
V = Q/C

D
C = Q/V
Solution
- Q is directly proportional to V.
- The constant of proportionality in this case is C, that is, the capacitance.
- Hence Q = CV. From the given relation we can derive all the equations except for Q = C/V.
Question 19
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What is the value of capacitance of a capacitor which has a voltage of 4V and has 16C of charge?

A
2F

B
4F

C
6F

D
8F

Solution

Q is directly proportional to V.

The constant of proportionality in this case is C, that is, the capacitance.

Hence Q = CV

From the relation,

C = Q/V = 16/4 = 4F
Question 20
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What is the total capacitance when two capacitors C1 and C2 are connected in series?

A
(C1 + C2) / C1*C2

B
(1/C1) + (1/C2)

C
C1 * C2 / (C1 + C2)

D
C1 + C2

Solution

When capacitors are connected in series, the equivalent capacitance is:
1/C_total=1/C1 + 1/C2

therefore,

C_total = C1*C2 / (C1+C2)
Question 21
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A resistance of 7 ohm is connected in series with an inductance of 31.8mH. The circuit is connected to a 100V 50Hz sinusoidal supply. Calculate the current in the circuit.

A
2.2 A

B
4.2 A

C
6.2 A

D
8.2 A

Solution

X_L= 2*π*f*L = 10 ohm

Z²= (R²+X_L²)

Therefore the total impedance

Z = 12.2 ohm

V = IZ

therefore,

I= V/Z = 100/12.2 = 8.2A
Question 22
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The equivalent circuit of the capacitor shown below is:

A

B

C

D

Solution

Due to initial condition, at t = 0 capacitor will act as a constant voltage source (at t = 0, capacitor acts as short-circuit). Hence, option (d) is correct.
Question 23
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The strength of current in 1 Henry inductor changes at a rate of 1 A/sec. The magnitude of energy stored in the inductor after 3 sec is:

A
2 Joules

B
3.8 Joules

C
4.5 Joules

D
2.2 Joules

Solution
Question 24
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The current and voltage profile of an element vs time has been shown in given figure. The element and its value are respectively:

A
A resistor with R = 2 mΩ

B
A capacitor with C = 2 μF

C
An inductor with L = 0.5 H

D
An inductor with L= 10 mH
Solution
- Since V is not proportional to R therefore, the element can’t be a resistor.
- At t = 5 ms, even if i ≠ 0, the element behaves as a short circuit therefore, the element can’t be a capacitor (since at t = 0 only capacitor behaves as short circuit).
- The current at t = 0 is zero and at t = 5 ms voltage across the element is zero therefore, the element must be an inductor (at t = 0, an inductor acts as open circuit and at t =∞ it behaves as short circuit).
- From the given voltage and current profile, we have:
Question 25
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Figure shown below exhibits the voltage-time profile of a source to charge a capacitor of 50 μF. The value of charging current in amperes is:

A
0.8

B
0.5

C
0.04

D
0.1

Solution

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