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Consider an RL series circuit having resistance R = 3 Ω, inductance L = 3 H and is excited by 6V. The current after a long time after closing of switch is ____________
- a)1 A
- b)2 A
- c)0 A
- d)Infinity
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
Consider an RL series circuit having resistance R = 3 , inductance L =...
At t = ∞ the circuit has effectively two 6Ω resistances in parallel
Given voltage = 6 V
So, current = 2 A.
So, current = 2 A.
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Consider an RL series circuit having resistance R = 3 , inductance L =...
Solution:
Given: R = 3 Ω, L = 3 H, E = 6V
The circuit diagram is shown below:
The time constant of the circuit is
τ = L/R = 3/3 = 1 sec
At t = 0, the current through the circuit is
I0 = 0
After a long time, the inductor behaves as a short circuit. Hence, the circuit can be simplified as shown below:
The equivalent resistance of the circuit is
Req = R = 3 Ω
The current through the circuit is
I = E/Req = 6/3 = 2 A
Therefore, the current after a long time after closing of switch is 2 A.
Hence, the correct answer is option (b) 2 A.
Given: R = 3 Ω, L = 3 H, E = 6V
The circuit diagram is shown below:
The time constant of the circuit is
τ = L/R = 3/3 = 1 sec
At t = 0, the current through the circuit is
I0 = 0
After a long time, the inductor behaves as a short circuit. Hence, the circuit can be simplified as shown below:
The equivalent resistance of the circuit is
Req = R = 3 Ω
The current through the circuit is
I = E/Req = 6/3 = 2 A
Therefore, the current after a long time after closing of switch is 2 A.
Hence, the correct answer is option (b) 2 A.
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