Network Elements and the Concept of Circuit Easy Questions and Answers | Page - 10

Network Elements and the Concept of Circuit

The Laplace transform of the given function—
f (t) = t, t ≥ 4 sec
0, Otherwise will be—

1. e^–4s
  s²
1. e^{–4(s + 1)}
  s²
1. e^–4s
  s
1. e^–4s t, t ≥ 4 sec
  4s + 1
  s²

View Hint View Answer Discuss in Forum

Given f (t) = t, t ≥ 4 sec
0, otherwise

f (t) = t u (t – 4)
or f (t)= (t – 4 + 4) u (t – 4)
or f (t)= (t – 4) u (t – 4) + 4 u (t – 4)
or f (s) = 1 e^{– 4s} +
4 e^{– 4s}
s² 5

or f (s) = e^{– 4s} 1 + 4
s² s

or f (s) = e^{– 4s} 4s + 1
s²

Correct Option: D

Given f (t) = t, t ≥ 4 sec
0, otherwise

f (t) = t u (t – 4)
or f (t)= (t – 4 + 4) u (t – 4)
or f (t)= (t – 4) u (t – 4) + 4 u (t – 4)
or f (s) = 1 e^{– 4s} +
4 e^{– 4s}
s² 5

or f (s) = e^{– 4s} 1 + 4
s² s

or f (s) = e^{– 4s} 4s + 1
s²

The initial voltage on the capacitor is V_c (0^–) = 2 V, I (s) = u (t) cos t find V (t) for t > 0—

1. V (t) = 1 cos (t – 45º) +
  3 e^{– t}
  2 2
1. V (t) = 2 cos (t – 45º) + 3 e^{+ t}
  2
1. V (t) = cos (t – 45º) + 1 e^{– 2t}
  2
1. V (t) = cos (t + 45º) + 3 e^{– 2t}
  2

View Hint View Answer Discuss in Forum

Here Z_eq = 1 .( 1 / Jω) , ω = 1 on comparing with I (s) = u (t) cos
1 + (1 / Jω)

or Z _eq = 1
1 + J

Since initial voltage on the capacitor is V(0^–) = 2 V,
so, V (t) J (s), Z_eq = cos t. 1 + A e^{– st}
1 + J

s= 1 cos (t– 45°) + A e^{– st} given that at t = 0. V (0) = 2 V
√2

so, 2 = 1 2 cos (0 – 45º) + A e^–5.0
√2

2 = 1 √2 + A
√2

or A = 3
2

S = 1 =
1 = 1
RC 1 x 1

V (t) = cos (t – 45°) + 3 e^{– t}
2

Correct Option: A

Here Z_eq = 1 .( 1 / Jω) , ω = 1 on comparing with I (s) = u (t) cos
1 + (1 / Jω)

or Z _eq = 1
1 + J

Since initial voltage on the capacitor is V(0^–) = 2 V,
so, V (t) J (s), Z_eq = cos t. 1 + A e^{– st}
1 + J

s= 1 cos (t– 45°) + A e^{– st} given that at t = 0. V (0) = 2 V
√2

so, 2 = 1 2 cos (0 – 45º) + A e^–5.0
√2

2 = 1 √2 + A
√2

or A = 3
2

S = 1 =
1 = 1
RC 1 x 1

V (t) = cos (t – 45°) + 3 e^{– t}
2

Direction: Fig. shows the statement for given question.

Current i at time t = 4 sec will be—

1. 5 amp
1. 3 amp
1. – 3 amp
1. – 5 amp

View Hint View Answer Discuss in Forum

At t = 4 sec
sources 30 u (t + 1) → exist and equals to 30 V 2 u (1 – t) → does not exist and equals to 0 amp Now, we will draw the equivalent circuit under this situation.

The current I = 30 = 3 amp.
10

Correct Option: B

At t = 4 sec
sources 30 u (t + 1) → exist and equals to 30 V 2 u (1 – t) → does not exist and equals to 0 amp Now, we will draw the equivalent circuit under this situation.

The current I = 30 = 3 amp.
10

Current i at t = 0.5 sec will be—

1. 5 amp
1. 5 amp
  3
1. 3 amp
1. 3 amp
  5

View Hint View Answer Discuss in Forum

At t = 0.5 sec
30 u (t + 1) → exist and equal to 30 V
2 u (1 – t) → exist and equal to 2 amp.

Now, we will draw the equivalent circuit under this situation we get this situation is same as the above question so,
I = 5 amp.

Correct Option: A

At t = 0.5 sec
30 u (t + 1) → exist and equal to 30 V
2 u (1 – t) → exist and equal to 2 amp.

Now, we will draw the equivalent circuit under this situation we get this situation is same as the above question so,
I = 5 amp.

The percentage of power loss in the question 115—

1. 5%
1. 2.5%
1. 10%
1. 20%

View Hint View Answer Discuss in Forum

% loss
= Power dissipated × 100 =
2.5 × 10^–3 × 100
Max. energy 0.1

= 2.5%

Correct Option: B

% loss
= Power dissipated × 100 =
2.5 × 10^–3 × 100
Max. energy 0.1

= 2.5%