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

Explanation:

Given

inductance(L)=40 mH

Resistor(R)=$50 \Omega$

Voltage(V)=15 V

Time constant($\tau$)=$\frac{L}{R}$

$\tau =\frac{40\times 10^{-3}}{50}=8\times 10^{-4}$

current $i_0=\frac{V}{R}$

$i_0=\frac{15}{50}=0.3 A$

Current as a function of time is given by

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

$i=0.3\times 0.9998$

i= 299.95 mA