Network Elements and the Concept of Circuit Easy Questions and Answers | Page - 58

Network Elements and the Concept of Circuit

Which of the following theorems can be applied to any network-linear or non-linear, active or passive, time variant or time-invariant?

1. Thevenin theorem
1. Norton theorem
1. Tellegen theorem
1. Superposition theorem

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Tellegen theorem can be applied to any network i.e.
• Linear or non-linear
• Active or passive
• Time variant or time-invariant

Correct Option: C

Tellegen theorem can be applied to any network i.e.
• Linear or non-linear
• Active or passive
• Time variant or time-invariant

The circuit shown below is under steady-state condition with the switch closed. The switch is opened at t = 0. What is the time constant of the circuit?

1. 0.1 s
1. 0.2 s
1. 5 s
1. 10 s

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Time constant
τ = L = 2H
R 10Ω

or τ = 0.2s.

Correct Option: B

Time constant
τ = L = 2H
R 10Ω

or τ = 0.2s.

In the circuit shown below, the switch is moved from position A to B at time t = 0. The current i through the inductor satisfies the following conditions—

1. i(0) = – 8A
2. di (t = 0) = 3A/s
dt

3. i(∞) = 4A The value of R is—

1. 0.5 ohm
1. 2.0 ohm
1. 4.0 ohm
1. 12 ohm

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When the switch in the position B, as shown below.

Writing voltage equation.
Ri + L di = E₂ ....(A)
dt

Taking Laplace transform
RI(s) + 2[sI(s) – I(0)] = E₂
s

or (R + 2) I(s) = E₂ – 16 (since, I(0) = – 8A)
s

or I(s) = E₂ – 16
s(R + 2s) (R + 2s)

or I(s) = E₂ 1 – 2 – 16
R s R + 2s (R + 2s)

(By using partial fraction)
Taking inverse Laplace transform
i(t) = E₂ R 1 – e^{[–(R/2) t]} – 8e^{–(R/2) t}....(B)
Given: i(t) = E₂ = 4 ....(i)
R

Again di(0) = 3 = E₂ . R + 4R ....(ii)
dt R 2

Solving equations (i) and (ii), we get
R = 0.5 ohm

Correct Option: A

When the switch in the position B, as shown below.

Writing voltage equation.
Ri + L di = E₂ ....(A)
dt

Taking Laplace transform
RI(s) + 2[sI(s) – I(0)] = E₂
s

or (R + 2) I(s) = E₂ – 16 (since, I(0) = – 8A)
s

or I(s) = E₂ – 16
s(R + 2s) (R + 2s)

or I(s) = E₂ 1 – 2 – 16
R s R + 2s (R + 2s)

(By using partial fraction)
Taking inverse Laplace transform
i(t) = E₂ R 1 – e^{[–(R/2) t]} – 8e^{–(R/2) t}....(B)
Given: i(t) = E₂ = 4 ....(i)
R

Again di(0) = 3 = E₂ . R + 4R ....(ii)
dt R 2

Solving equations (i) and (ii), we get
R = 0.5 ohm

Direction: Figure given below shows four light bulbs connected across an 85V battery. The bulbs are said to be connected in parallel.

Which one of the following is the
ratio V₂₄
V₁₃

of the network shown in the given figure—

1. 1
  3
1. 2
  3
1. 3
  4
1. 4
  3

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V₁₃ = 1 (I₁ – I) ...........(i)
V₂₄ = 1 × 1 .........(ii)

(I₁ – I) 1 = I × 1 + I × 1 + I × 1
or I₁ = 4I .......(iii)
and V₁₃ = 4I – I = 3I
V₂₄ = 1 = 1
V₁₃ 3I 3

Correct Option: A

V₁₃ = 1 (I₁ – I) ...........(i)
V₂₄ = 1 × 1 .........(ii)

(I₁ – I) 1 = I × 1 + I × 1 + I × 1
or I₁ = 4I .......(iii)
and V₁₃ = 4I – I = 3I
V₂₄ = 1 = 1
V₁₃ 3I 3

Kirchoff’s laws fail in case of—

1. Non-linear networks
1. Linear networks
1. Dual networks
1. Distributed parameter networks

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Kirchhoff’s laws fail in case of non-linear networks.

Correct Option: A

Kirchhoff’s laws fail in case of non-linear networks.