The switch in the circuit has been open for a long time. Immediately after it is

Question

The switch in the circuit has been open for a long time. Immediately after it is closed, what is the CURRENT through THE INDUCTOR?

E/R1
0
E/(R1+R2)
E/L
E/R2
  

Immediately after the switch has been closed, what can you say about the current in R1?

It is the same as the current in R2, and they are both equal to zero.
It is zero, and not the same as the current through R2
It is greater than the current through R2
It is the same as the current in R2, but greater than zero.
  

What is the correct loop rule for a loop that goes through the battery, the inductor, and Resistor 1?

Immediately after the switch is closed, what is the VOLTAGE across the INDUCTOR?
I L
E?I1 R1
E?I2 R2
E
0

Explanation / Answer

When switch is closed then inductor initially behave as open circuit. so R1 and R2 come in series across the battery

so current I = E/(R1+R2)

Part B

As discussed above inductor become open circuit and R1 and R2 are in series so current in R1 and R2 is same.

Part C

Loop equation

E-VL - IR1=0

Part D

Voltage across inductor is equal to the voltage across resistor R2

V = IR2

V = ER2/(R1+R2)

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