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A 45-mH ideal inductor is connected in series with a 60-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?

User Emilsen
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1 Answer

5 votes

Answer:

0.249978A

Step-by-step explanation:

V= IR+ Ve^(-tR/L)

I= V/R(1-e^(-tR/L))

R/L= 60/0.045= 1333.3

e^(-tR/L)= e^(-0.007*1333.3)= 0.0000887

I=15/60(1-0.0000887)

I=0.249978A

User Mikkel Larsen
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