Network Elements and the Concept of Circuit Easy Questions and Answers | Page - 7

Network Elements and the Concept of Circuit

The current I in the circuit will given by—

1. 5 amp
1. 1.2 amp
1. 2.4 amp
1. 0.6 amp

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Applying the KVL
10 = 1 I + 2 (I + 2) + 2 I
10 = I + 2 I + 4 + 2 I

10 – 4 = 5I
I = 6 = 1.2 amp.
5

Correct Option: B

Applying the KVL
10 = 1 I + 2 (I + 2) + 2 I
10 = I + 2 I + 4 + 2 I

10 – 4 = 5I
I = 6 = 1.2 amp.
5

Find the Thevenin voltage and resistance for the given circuit shown below—

1. R_th = 58.82 Ω, V_th = 117.64 V
1. R_th = 117.64 Ω, V_th = 58.82 V
1. R_th = 58.82 Ω, V_th = 58.82 V
1. None of these

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Calculation for R_th
As there are many dependent sources are taking into account so in order to calculate the R_th let an imaginary current source say 2A is connected across the open terminal then R_th will be the ratio of
V_x =
V_x
I 2

On applying KCL at node N.
V_x - 1000 I_x +
V_x +
V_x = 2 ....(i)
100 100 3000

Also I_X = V_x = I₁....(ii)
3000

Solving equations (i) and (ii), we get
V_X = (58.82 × 2) V
so, R_th = V_x =
58.82 × 2 = 58.82 Ω
2 2

(∵ R_th = V_x = I₁, where I = 2 amp {imaginary source})
I

V_X = V_th = 58.82 × 2 = 117.6 V

Correct Option: A

Calculation for R_th
As there are many dependent sources are taking into account so in order to calculate the R_th let an imaginary current source say 2A is connected across the open terminal then R_th will be the ratio of
V_x =
V_x
I 2

On applying KCL at node N.
V_x - 1000 I_x +
V_x +
V_x = 2 ....(i)
100 100 3000

Also I_X = V_x = I₁....(ii)
3000

Solving equations (i) and (ii), we get
V_X = (58.82 × 2) V
so, R_th = V_x =
58.82 × 2 = 58.82 Ω
2 2

(∵ R_th = V_x = I₁, where I = 2 amp {imaginary source})
I

V_X = V_th = 58.82 × 2 = 117.6 V

Find the value of R so that V₂ = 2 volt—

1. 6 Ω
  5
1. 5 Ω
  6
1. 2 Ω
1. 5 Ω

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Given circuit

Apply KCL at node A
I ₁ + 4 = 2 I₁ + V₂
R

or 4 - 2 = I₁ (V2 = 2V)
R

4 – 2 =
8 = 5
R 5

∵ I₁ = 10 - 2 = 8
5 5

or 4R – 2 =
8
R 5

or 20 R – 10 = 8 R
or (20 – 8) R = 10
or 12 R = 10
or R = 10 =
5 Ω
12 6

Correct Option: B

Given circuit

Apply KCL at node A
I ₁ + 4 = 2 I₁ + V₂
R

or 4 - 2 = I₁ (V2 = 2V)
R

4 – 2 =
8 = 5
R 5

∵ I₁ = 10 - 2 = 8
5 5

or 4R – 2 =
8
R 5

or 20 R – 10 = 8 R
or (20 – 8) R = 10
or 12 R = 10
or R = 10 =
5 Ω
12 6

For fig. at time t ₀ after the switch K was closed, it is found that V₂ = + 5 V, determine the value of i ₂ (t ₀) and d / dt i₂ (t ₀)—

1. 2.5 amp, 5 amp/sec
1. 2.5 amp, 2.5 amp/sec
1. 5 amp, 2.5 amp/sec
1. 5 amp, 5 amp/sec

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Given V₂ = 5V
I ₁ = 10 – V₂ =
10 – 5 = 5
1 1

I ₃ = 5 = 2.5
2

I ₂ = I₁ – I₃ = 5 – 2.5 = 2.5 amp.
Voltage across 1 Ω resistance = 2.5 × 1 = 2.5 V
V_L = 5 – 2.5 = 2.5 V
Also, V_L = L di(t)
dt

2.5 = 1 / 2 di₂(t₀)
dt

or di₂(t₀) = 2.5 × 2 = 5 amp/sec
dt

Correct Option: A

Given V₂ = 5V
I ₁ = 10 – V₂ =
10 – 5 = 5
1 1

I ₃ = 5 = 2.5
2

I ₂ = I₁ – I₃ = 5 – 2.5 = 2.5 amp.
Voltage across 1 Ω resistance = 2.5 × 1 = 2.5 V
V_L = 5 – 2.5 = 2.5 V
Also, V_L = L di(t)
dt

2.5 = 1 / 2 di₂(t₀)
dt

or di₂(t₀) = 2.5 × 2 = 5 amp/sec
dt

The voltage of the source i.e. V_s, if i (t) = – 20 e^{– 2t}

1. 10 e_{– 2t}
1. 10 e_2t
1. 20 e_{– 2t}
1. 20 e_2t

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Given i (t) = – 20 e^{– 2t}
V (t) = – 20 e^{– 2t} × 1 = – 20 e^{– 2t}

i_C = C dV (t)
dt

= 2. d (– 20 e– 2t)
dt

or i_C = 80 e^{– 2t}
I = i (t) + i _C = – 20 e^{– 2t} + 80 e^{– 2t}
= 60 e^{– 2t}
V_s = V_R + V_L + V (t)
= 1 × 60 e^{– 2t} + L d i + V (t)
dt

= 1 × 60 e_{– 2t} + 1
d 60 e_{– 2t} – 20 e_{– t}
4 dt

= 60 e_{– 2t} – 15 × 2 e_{– 2t} – 20 e_{– 2t}
= 60 e_{– 2t} – 30e_{– 2t} – 20 e_{– 2t}
= 10e_{– 2t}

Correct Option: A

Given i (t) = – 20 e^{– 2t}
V (t) = – 20 e^{– 2t} × 1 = – 20 e^{– 2t}

i_C = C dV (t)
dt

= 2. d (– 20 e– 2t)
dt

or i_C = 80 e^{– 2t}
I = i (t) + i _C = – 20 e^{– 2t} + 80 e^{– 2t}
= 60 e^{– 2t}
V_s = V_R + V_L + V (t)
= 1 × 60 e^{– 2t} + L d i + V (t)
dt

= 1 × 60 e_{– 2t} + 1
d 60 e_{– 2t} – 20 e_{– t}
4 dt

= 60 e_{– 2t} – 15 × 2 e_{– 2t} – 20 e_{– 2t}
= 60 e_{– 2t} – 30e_{– 2t} – 20 e_{– 2t}
= 10e_{– 2t}