Question

A 0.160 H inductor is connected in series with a 93 $\Omega$ resistor and an ac source. The voltage across the inductor is $v_L = -(12.5V)\sin[(490\text{rad/s})t]$.

Name: A 0.160 H inductor is connected in series with a 93 Ω resistor and an ac source. The voltage across the inductor is vL = -(12.5V)sin[(490rad/s)t].
Uploaded: 2022-03-02T18:44:10-08:00
Duration: 2 min 4 s
Channel: Kamlesh Goyal
Description: A 0.160 H inductor is connected in series with a 93 Ω resistor and an ac source. The voltage across the inductor is vL = -(12.5V)sin[(490rad/s)t].

          A 0.160 H inductor is connected in series with a 93 $\Omega$ resistor and an ac source. The voltage across the inductor is $v_L = -(12.5V)\sin[(490\text{rad/s})t]$.

Added by Tiffany A.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Kamlesh Goyal

03/02/2022

Step 1

5V)sin[(490rad/s)t]. Show more…

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A 0.160H inductor is connected in series with a 93Omega resistor and an ac source. The voltage across the inductor is v_(L)=-(12.5V)sin[(490ra(d)/(s))t]. A 0.160 H inductor is connected in series with a 93 S2 resistor and an ac source. The voltage across the inductor is vL= -(12.5V) sin[(490rad/s)t].