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Test: Problems on Initial Conditions - Electrical Engineering (EE) MCQ


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10 Questions MCQ Test - Test: Problems on Initial Conditions

Test: Problems on Initial Conditions for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Test: Problems on Initial Conditions questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Problems on Initial Conditions MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Problems on Initial Conditions below.
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Test: Problems on Initial Conditions - Question 1

In the circuit shown below, steady state was reached when the switch ‘s’ was open. The switch was closed at t = 0. Then initial value of the current through the capacitor 2C is?

Detailed Solution for Test: Problems on Initial Conditions - Question 1

At steady-state conditions, the capacitor acts as an open circuit, and the inductor acts as a short circuit.

Now considering t = 0-,

Now voltage across 4 Ω resistor is 12 V.

Considering t = 0+

By KCL, 

i2c (0+) = 2 A

The initial value of the current through the capacitor 2 C is 2 A.

Test: Problems on Initial Conditions - Question 2

In the circuit shown in the figure, the switch S is closed at time t = 0. The supply current at t = 0+ and the capacitor voltage at t → ∞ are, respectively

Detailed Solution for Test: Problems on Initial Conditions - Question 2

The switch s is closed at time, t = 0

At, t = 0+, the capacitor is short circuited and inductor is open circuited. Now circuit will become

At, t = ∞, the capacitor is open circuited and inductor is short circuited. Now the circuit will become

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*Answer can only contain numeric values
Test: Problems on Initial Conditions - Question 3

If initial voltage retained by the capacitor C1 is 100 V and C2 is zero, the switch k is closed at t = 0, the voltage drop across capacitor C1  in steady state is ________ V


Detailed Solution for Test: Problems on Initial Conditions - Question 3

The circuit after the switch is closed is given in the s-domain in the below figure

The voltage drop across capacitor C1 is

= 33.33 V

*Answer can only contain numeric values
Test: Problems on Initial Conditions - Question 4

The circuit shown in figure was at steady state for t < 0 with switch at position ‘A’. The switch is thrown to position ‘B’ at time t = 0. The voltage across the 10 ohm resistor at time t = 0+ is ________V.


Detailed Solution for Test: Problems on Initial Conditions - Question 4

At t < 0 the circuit was in steady state and switch at ‘A’

The current drawn by circuit i

Current through the inductor I(0-)

Immediately after t > 0 the circuit will be the current through Inductor I(0-) = I(0+)

at t= 0+

V = I(0+) × 10Ω

= -3 × 10 = -30V

*Answer can only contain numeric values
Test: Problems on Initial Conditions - Question 5

When current changes from 4A to 0A in 0.1s in an inductor the induced emf is found to be 100V, then find the value of self-inductance is :


Detailed Solution for Test: Problems on Initial Conditions - Question 5

 

*Answer can only contain numeric values
Test: Problems on Initial Conditions - Question 6

For the circuit shown in the figure, the value of  is _____ (in A/s)


Detailed Solution for Test: Problems on Initial Conditions - Question 6

At t = 0-, the circuit is as shown below.

i(0) = 12 / 6 = 2A

V(0-) = 12 V

The inductor doesn’t allow the sudden change in current.

I(0+) = i(0-) = 2 A

Capacitor doesn’t allow the sudden change in voltage.

V(0-) = V(0+) = 12 V

At t = 0+, the circuit becomes,

By applying KVL,

-12 + 4(2) + 6(2) + VL(0+) = 0

VL(0+) = -8 V

Since  

 

Test: Problems on Initial Conditions - Question 7

In the circuit shown in the figure, the switch s open for a long time. It is closed at t=0 For t>0 the current flowing through the inductor will be given by

Detailed Solution for Test: Problems on Initial Conditions - Question 7

For t < 0,S is open and network is in steady state so the inductor acts like a short circuit.

Test: Problems on Initial Conditions - Question 8

The network shown in the figure given below reaches a steady state with the switch K in position a. At t = 0, the switch is moved from a to b by a make-before-break mechanism. Assume the initial current in 2 H inductors as zero. What is the current in 1 H inductor at t = 0+and t = ∞, respectively?

Detailed Solution for Test: Problems on Initial Conditions - Question 8

Before t = 0 the circuit is

I = 5/2 = 2.5 A

As t → ∞ the current tends to zero, since no power source is present in the circuit, the power stored inductor is dissipated through the resistor.

Test: Problems on Initial Conditions - Question 9

Find  i(0+) and V (2 m sec) for the circuit shown below.

Detailed Solution for Test: Problems on Initial Conditions - Question 9

V(0) = 50 V = V(0+)

For t > 0, RC = 1.6 m sec.

Test: Problems on Initial Conditions - Question 10

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

Detailed Solution for Test: Problems on Initial Conditions - Question 10

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.

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