Network Elements and the Concept of Circuit
- The Laplace transforms of the functions tu(t) and u (t) sin t are respectively—
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L {t u (t)} = 1 s2 L {sin t} = 1 s2+ 1 Correct Option: C
L {t u (t)} = 1 s2 L {sin t} = 1 s2+ 1
- The Thevenin equivalent of the following network between A and B is—
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Calculation for Rth to calculate the Rth. Let us assume an imaginary source say 10 V is connected across the open terminal and current I is flowing due to this imaginary source than
Rth = 1 0 I
applying KVL, we get 10 – 6I – 10 (I – 0.8I1) = 0 .....(i)
KlL at node A, gives I = – I1 ....(ii)
from (i) and (ii)
10 = 6 I + 10 (I + 0.8 I) (I – 0.8 I1) 6Ω
10 = 6 I + 18 I
10 = 24 II = 10 24 Rth = V = 10 = 24 I 10/24
Calculation for Vth: As, I1 = 0 dependant source will be acts as open circuit and the circuit becomes as shown.
So, the Thevenin's equivalent circuit between terminal A and B is shown below.
Correct Option: A
Calculation for Rth to calculate the Rth. Let us assume an imaginary source say 10 V is connected across the open terminal and current I is flowing due to this imaginary source than
Rth = 1 0 I
applying KVL, we get 10 – 6I – 10 (I – 0.8I1) = 0 .....(i)
KlL at node A, gives I = – I1 ....(ii)
from (i) and (ii)
10 = 6 I + 10 (I + 0.8 I) (I – 0.8 I1) 6Ω
10 = 6 I + 18 I
10 = 24 II = 10 24 Rth = V = 10 = 24 I 10/24
Calculation for Vth: As, I1 = 0 dependant source will be acts as open circuit and the circuit becomes as shown.
So, the Thevenin's equivalent circuit between terminal A and B is shown below.
- The voltmeter in the shown circuit is ideal. The transformer has identical windings with perfect coupling. The voltmeter reads—
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As the transformer has identical winding i.e.
N1 = N2 (1: 1)
So, V1 = V2, It means the reading of the voltmeter will be zero.Correct Option: D
As the transformer has identical winding i.e.
N1 = N2 (1: 1)
So, V1 = V2, It means the reading of the voltmeter will be zero.
- The capacitor C1 in the circuit shown has a voltage of 20 V across it before the switch S is closed at time t = 0. C1 = 2 µF and C2 = 3 µF. The voltage across C2 after S is closed is—
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Initial charge on C1 = 2 × 10– 6 × 20 = 40 × 10– 6 C Since charge remain the same so,
C = C1 + C2 = 5 µFV= Q + 10 = 40 × 10–6 = 8 V C 5 × 10–6
Alternate method
q0 = C1 V0
q0 = q1 + q2V= C V0 = 2 × 20 = 8 V C1 + C2 5 Correct Option: A
Initial charge on C1 = 2 × 10– 6 × 20 = 40 × 10– 6 C Since charge remain the same so,
C = C1 + C2 = 5 µFV= Q + 10 = 40 × 10–6 = 8 V C 5 × 10–6
Alternate method
q0 = C1 V0
q0 = q1 + q2V= C V0 = 2 × 20 = 8 V C1 + C2 5
- A long uniform coil of inductance 2L and associated resistance 2R ohms is physically cut into two exact halves which are rewound in parallel. The resistance and inductance of the combination are—
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Given inductance of the coil is 2L and cut into two exact equal halves.
L eq = L || L = L 2 R eq = R || R = R 2 Correct Option: C
Given inductance of the coil is 2L and cut into two exact equal halves.
L eq = L || L = L 2 R eq = R || R = R 2