The current in an RL circuits drops from 1.0 A to 10 mA in the first second following removal of the battery from the circuit. If L is 10 H, find the resistance R (in ohm) in the circuit.

Question

The current in an RL circuits drops from 1.0 A to 10 mA in the first second following removal of the battery from the circuit. If L is 10 H, find the resistance R (in ohm) in the circuit. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

we have, i = i0e-tR/L or 10-3 = 1 × e-1R/10 ⇒ R = 46Ω

Q. The current in an $R L$ circuits drops from $1.0 A$ to $10 m A$ in the first second following removal of the battery from the circuit. If $L$ is $10 H$ , find the resistance $R$ (in ohm) in the circuit.

we have, $i = i_{0} e^{- tR / L}$
or $1 0^{- 3} = 1 \times e^{- 1 R /10}$
$\Rightarrow R = 46Ω$

Q. The current in an RL circuits drops from 1.0A to 10mA in the first second following removal of the battery from the circuit. If L is 10H, find the resistance R (in ohm) in the circuit.

we have, i=i0​e−tR/L or 10−3=1×e−1R/10 ⇒R=46Ω

Q. The current in an $R L$ circuits drops from $1.0 A$ to $10 m A$ in the first second following removal of the battery from the circuit. If $L$ is $10 H$ , find the resistance $R$ (in ohm) in the circuit.

we have, $i = i_{0} e^{- tR / L}$
or $1 0^{- 3} = 1 \times e^{- 1 R /10}$
$\Rightarrow R = 46Ω$