An inductor of 10 mH is connected to a 20 V battery through a resistor of 10 k Ω 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 μ s is (x/100) m A. Then x is equal to. (Take e-1=0.37 )

Question

An inductor of 10 mH is connected to a 20 V battery through a resistor of 10 k Ω 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 μ s is (x/100) m A. Then x is equal to. (Take e-1=0.37 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

I max =(V/R)=(20 V /10 K Ω)=2 mA For LR - decay circuit I=I max e-R / L I=2 m Ae (-10 × 103 × 1 × 10-6/10 × 10-3) I=2 mA e-1 I=2 × 0.37 mA I=(74/100) mA x=74

Q. An inductor of 10mH is connected to a 20V battery through a resistor of 10kΩ 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 1μs is 100x​mA. Then x is equal to_______. (Take e−1=0.37 )

Imax​=RV​=10KΩ20V​=2mA For LR− decay circuit I=Imax​e−R/L I=2mAe10×10−3−10×103×1×10−6​ I=2mAe−1 I=2×0.37mA I=10074​mA x=74

$I_{m a x} = \frac{V}{R} = \frac{20 V}{10 K Ω} = 2 m A$
For $L R -$ decay circuit
$I = I_{m a x} e^{- R / L}$
$I = 2 m A e \frac{- 10 \times 1 0 ^{3} \times 1 \times 1 0 ^{- 6}}{10 \times 1 0 ^{- 3}}$
$I = 2 m A e^{- 1}$
$I = 2 \times 0.37 m A$
$I = \frac{74}{100} m A$
$x = 74$