CapacitanceHard
Question
A capacitor of capacitance 10 μF is connected to a battery of emf 2V.It is found that it takes 50 ms for the charge on the capacitor to become 12.6 μC. Then the resistance of the circuit is : (Take 1/e = 0.37)
Options
A.4 kΩ
B.5 kΩ
C.6 kΩ
D.7 kΩ
Solution
Charge at steady state q0 = 20 μC
q = q0 (1 − e-t/τ)
q = 12.6 = 20 (1 − e-50×10-3/τ)
and τ = RC = 50 × 10-3 ⇒ R = 5 KΩ
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