An LR circuit is connected to a 4V DC power source. The resistance is 5ohms. Ima

Question

An LR circuit is connected to a 4V DC power source. The resistance is 5ohms. Imax=0.8A. When the power is shut off, the current drops exponentially. To measure the time constant, you should measure the time until the current drops to what value? B) If you measure this time as 0.1s, what is the value of the inductor? An LR circuit is connected to a 4V DC power source. The resistance is 5ohms. Imax=0.8A. When the power is shut off, the current drops exponentially. To measure the time constant, you should measure the time until the current drops to what value? B) If you measure this time as 0.1s, what is the value of the inductor?

Explanation / Answer

Solution:

Resistance in the LR circuit, R = 5

The equilibrium current is imax = 0.8 A

Part (A)

When the power source is removed, then the current decays exponentially according to the relation,

i = imax*e-t/

where is the time constant for the circuit.

In order to find the value of the current at the time equal to time constant after the power is shut, we put t = then

i = imax*e-/

i = imax*e-1

i = imax*(0.36788)

i = (0.8 A)* (0.36788)

i = 0.29 A

This means that after t =, the current has dropped from 0.8 A to 0.29 A.

Thus in order to measure the time constant, we must wait till the current drops to 0.29 A

Part (B)

The measured time constant is = 0.1 s, thus the value of the inductance is given by,

= L/R

L = *R

L = (0.1 s)*(5 )

L = 0.5 Henry

Thus the value of inductor 0.5 H

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