Question

A 45-mH inductor is connected in series with a 60-Ω resistor through a 15-V dc power supply and a switch. If the switch is closed at t = 0 s, what is the current after 7.0 ms?

Answer 1

Answer:

The current is 0.248 A

Explanation:

Given that,

Inductor $L= 45\times10^{3}\ H$

Resistance $R= 60\Omega$

Voltage = 15 volt

Time $t =7.0\times10^{-3}\ sec$

We need to calculate the current

Using formula of current

$I=\dfrac{V}{R}(1-e^{\dfrac{-R}{L}}t)$

Where, V = voltage

R = resistance

L = inductance

T = time

Put the value into the formula

$I=\dfrac{15}{60}(1-e^{\dfrac{-60}{45\times10^{3}}}\times7\times10^{-3})$

$I=0.248\ A$

Hence, The current is 0.248 A.