Current Electricity and Electrical InstrumentEasy
Question
An inductor (L = 0.03 H) and a resistor (R = 0.15 kΩ) are connected in series to a battery of 15V EMF in a circuit shown below. The key K1 has been kept closed for a long time. Then at t = 0, K1 is opened and key K2 is closed simultaneously. At t = 1ms, the current in the circuit will be (e5 ≅ 150)
Options
A.6.7 mA
B.6.7 mA
C.0.67 mA
D.67 mA
Solution
Decay of current
I = I0e-tR/L
= 6.66 × 10-4
= 0.666 × 10-3
= 0.67 mA
I = I0e-tR/L
= 6.66 × 10-4
= 0.666 × 10-3
= 0.67 mA
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