Question

Find an expression for the current in a circuit where the resistance is 8 Ω, the inductance is 4 H, a battery gives a constant voltage of 32 V, and the switch is turned on when t

Answer 1

Answer:

$I(t)=4[1-e^{-2t}]$

Explanation:

For a LR circuit as shown the current at any time t in the circuit is given by

$I(t)=\frac{V}{R}[1-e^{\frac{-Rt}{L}}]$

where

'V' is the voltage

'R' is resistance in the circuit

'L' is the inductance of the circuit

't' is time after circuit is turned on

Applying the given values we get

$I(t)=\frac{32}{8}[1-e^{\frac{-8t}{4}}]\\\\I(t)=4[1-e^{-2t}]$