Question

An LR circuit consists of a 35-mH inductor, a resistance of 12 ohms, an 18-V battery, and a switch. What is the current 5.0 ms after the switch is closed

Answer 1

Current, I = 1.23 A

Step-by-step explanation:

Given that,

Inductance, L = 35 mH

Resistance, R = 12 ohms

Potential difference, V = 18 V

We need to find current 5 ms after the switch is closed. Current in LR circuit is given by :

`I=I_o(1-e^(-t/\tau ))` ....(1)

Here,

`I_o` is final current

`I_o=(V)/(R)\\\\I_o=(18)/(12)=1.5\ A`

`\tau` is time constant

`\tau=(L)/(R)\\\\\tau=(35* 10^(-3))/(12)\\\\\tau=0.00291\ s`

So, equation (1) becomes :

`I=1.5* (1-e^{-5* 10^(-3)/0.00291})\\\\I=1.23\ A`

So, after 5 ms the current in the circuit is 1.23 A.