Network Elements and the Concept of Circuit Easy Questions and Answers | Page - 20

Network Elements and the Concept of Circuit

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

1. 6 Ω
  5
1. 5 Ω
  6
1. 2 Ω
1. 5 Ω

View Hint View Answer Discuss in Forum

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

View Hint View Answer Discuss in Forum

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

View Hint View Answer Discuss in Forum

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}

Find the Thevenin voltage and resistance for the network shown below across the terminal A.B—

1. 27 V, 3 Ω
1. 18 V, 3 Ω
1. 27 V, 1 Ω
1. 18 V, 8 Ω

View Hint View Answer Discuss in Forum

Calculation for R_th
To calculate R_th when dependent source are taking into account for this, assume a imaginary source say 5 V is connected across terminal A and B and current produced by that current source is I then R_th is given by the relation.
R_th = 5/1
the equivalent circuit for calculating R_th is shown below:
5 = 2 V_x + I. 1 ....(i)
V_x = 1. I
5 = 2 I + I
I = 5 Ω
3

so, R_th = 5 = 3 Ω
5 / 3

Calculation for Vth:
Here the 3 A current source will drop across 1 Ω resistance

V_x = 3 × 1 = 3 V
or V_th = 2 V_x + V_x + 18
or V_th =3 V_x + 18 = 3 × 3 + 18 = 27 V

Correct Option: A

Calculation for R_th
To calculate R_th when dependent source are taking into account for this, assume a imaginary source say 5 V is connected across terminal A and B and current produced by that current source is I then R_th is given by the relation.
R_th = 5/1
the equivalent circuit for calculating R_th is shown below:
5 = 2 V_x + I. 1 ....(i)
V_x = 1. I
5 = 2 I + I
I = 5 Ω
3

so, R_th = 5 = 3 Ω
5 / 3

Calculation for Vth:
Here the 3 A current source will drop across 1 Ω resistance

V_x = 3 × 1 = 3 V
or V_th = 2 V_x + V_x + 18
or V_th =3 V_x + 18 = 3 × 3 + 18 = 27 V

The Thevenin equivalent of the given at terminal a – b will be—

1. None of these

View Hint View Answer Discuss in Forum

Because we cannot connect a resistance in series with the current source.

Correct Option: D

Because we cannot connect a resistance in series with the current source.