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

The current is 0.248 A

Step-by-step explanation:

Given that,

Inductor
`L= 45*10^(3)\ H`

Resistance
`R= 60\Omega`

Voltage = 15 volt

Time
`t =7.0*10^(-3)\ sec`

We need to calculate the current

Using formula of current

`I=(V)/(R)(1-e^{(-R)/(L)}t)`

Where, V = voltage

R = resistance

L = inductance

T = time

Put the value into the formula

`I=(15)/(60)(1-e^{(-60)/(45*10^(3))}*7*10^(-3))`

`I=0.248\ A`

Hence, The current is 0.248 A.