Network Elements and the Concept of Circuit
- In the circuit shown in figure, the switch S is closed at time t = 0. The voltage across the inductance at t = 0+, is—
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The given circuit
at t = 0+ capacitor is replaced by short circuit and inductor is replaced by open circuit. The given circuit can be redrawn asNow, VL (0+) = 2 × 10 = 4V 2 + 3
Hence alternative (B) is the correct choice.Correct Option: B
The given circuit
at t = 0+ capacitor is replaced by short circuit and inductor is replaced by open circuit. The given circuit can be redrawn asNow, VL (0+) = 2 × 10 = 4V 2 + 3
Hence alternative (B) is the correct choice.
- The initial current in the circuit shown below when the switch is opened for t > 0—
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So in steady state when switch closed current 0 Amp.
∵ at initially capacitor is uncharged and there is no current flow due to open switch.Correct Option: B
So in steady state when switch closed current 0 Amp.
∵ at initially capacitor is uncharged and there is no current flow due to open switch.
- The initial current in the circuit shown below when the switch is opened for t > 0 is—
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Applying KVL in the loop
20 – 10I – 2I = 0
at t < 0 20 = 12I
or I = 1·67 ACorrect Option: A
Applying KVL in the loop
20 – 10I – 2I = 0
at t < 0 20 = 12I
or I = 1·67 A
- 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—
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As we know that τ = L = 2 Req 10
or τ = 0·2 SCorrect Option: B
As we know that τ = L = 2 Req 10
or τ = 0·2 S
- For the network shown below, the switch S is closed at t = 0 with the capacitor uncharged. The value of di(t) / dt at t = 0+ will be—
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When capacitor opened
at t = 0+IC(0–) = V = .1000 R 100
= 0.1 Amp
at t = o+
⇒ 100 = 1000 IC + 1 / C ∫ICdt
After differentiating, we get0 = 1000 dIC + 1 lc dt C
Capacitor in short circuited-1000 dIC = Ic = .1 dt C 1 × 10 -6 dI(+) = - .1 × 10 +6 dt 103
= – 100 Ampere.
Hence alternative (B) is the correct choice.Correct Option: B
When capacitor opened
at t = 0+IC(0–) = V = .1000 R 100
= 0.1 Amp
at t = o+
⇒ 100 = 1000 IC + 1 / C ∫ICdt
After differentiating, we get0 = 1000 dIC + 1 lc dt C
Capacitor in short circuited-1000 dIC = Ic = .1 dt C 1 × 10 -6 dI(+) = - .1 × 10 +6 dt 103
= – 100 Ampere.
Hence alternative (B) is the correct choice.