An inductor of 10 mH is connected to a 20 V battery through a resistor of 10 kΩ and a switch. After a long time, when maximum current is set up in the circuit, the current is switched off. The current in the circuit after 1 µs is x/100 mA. Then x is equal to _______. (Take e?¹ = 0.37)
An inductor of 10 mH is connected to a 20 V battery through a resistor of 10 kΩ and a switch. After a long time, when maximum current is set up in the circuit, the current is switched off. The current in the circuit after 1 µs is x/100 mA. Then x is equal to _______. (Take e?¹ = 0.37)
Asked by Shiksha User
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1 Answer
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I = I? e? /?
= (20/10000) e^- (1×10? / 10×10? ³)
= 2 × 10? ³ e? ¹
The provided solution calculates as:
= 2 × 10? ³ e? ¹
= 2e? ¹ mA
= 2 × 0.37mA
= 0.74 mA = 74/100 mA
(Note: There seems to be a calculation discrepancy in the source image, the steps shown lead to I = 2e? ¹ mA)
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