Transient RL Circuit Consider the circuit shown in the figure. The resistance of

Question

Transient RL Circuit

Consider the circuit shown in the figure. The resistance of the wire used to make the inductor L is negligible compared to the resistors in the circuit. V = 11.0 V, R1 = 15.0 Ohms, R2 = 125.0 Ohms, and L = 1.70 H

For this part, assume that switch S has been closed for a long time so that steady currents exist in the circuit. Find (1) the battery current, (2) the current in resistor R2, and (3) the current through the inductor.

1. Enter the battery current here:  

2 Enter the current in resistor R2 here:  

3 Enter the current through the inductor here:  

4 Now the switch is opened. Find the initial voltage across the inductor when switch S is opened.

Explanation / Answer

here ,

after a very long time ,

inductor will behave as a shortcircuit

Equivalent resistance is Req = R1

Req = 15 ohm

for the current in battery ,

battery current = V/Req

battery current = 11/15

battery current = 0.733 A

the battery current is 0.733 A

2)

as R2 is short ,

the current in resistor 2 is 0 A

3)

current in indcutor = current in R1

current in indcutor = 0.733 A
4)

when swtich is opened ,

the current in the inductor will be same

initial voltage = 0.733 * 125

initial voltage = 91.7 V

the initial voltage across inductor is 91.7 V

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.