Question

A 40-mH ideal inductor is connected in series with a 50 Ω resistor through an ideal 15-V DC power supply and an open switch. If the switch is closed at time t = 0 s, what is the current 7.0 ms later? O 650 mA O 850 mA O 300 mA O 280 mA O 550 mA

Answer 1

Answer:i=300 mA

Step-by-step explanation:

Given

inductance(L)=40 mH

Resistor(R)=
`50 \Omega`

Voltage(V)=15 V

Time constant(
`\tau`)=
`(L)/(R)`

`\tau =(40* 10^(-3))/(50)=8* 10^(-4)`

current
`i_0=(V)/(R)`

`i_0=(15)/(50)=0.3 A`

Current as a function of time is given by

`i=i_0\left ( 1-e^{-(t)/(\tau )}\right )`

`i=0.3* 0.9998`

i= 299.95 mA