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The switch of below circuit was open for long and at t = 0 it is closed. What is the final steady state voltage across the capacitor and the time constant of the circuit?
  • a)
    0 V and 0.1 sec
  • b)
    20 V and 0.2 sec
  • c)
    10 V and 0.2 sec
  • d)
    10 V and 0.1 sec
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
The switch of below circuit wasopen for long and at t = 0 it isclosed....
Final value = 20/2 = 10 Volt
Time constant τ= 5 × 103 × 20 × 10-6
= 0.1 sec
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Most Upvoted Answer
The switch of below circuit wasopen for long and at t = 0 it isclosed....
Final value = 20/2 = 10 Volt
Time constant τ= 5 × 103 × 20 × 10-6
= 0.1 sec
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The switch of below circuit wasopen for long and at t = 0 it isclosed....
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