Question

he circuit of Fig. 32-5 consists of a 5 T -V battery, a 20 Ω resistor key. Find the rate at which current begins change of current is 400 A/s, and the final steady current. a low-resistance 25-mH inductor, and a to rise when the key is closed, the current at the instant the rate of Ammeter

The circuit consists of a 50V battery, a 20 ohm resistor, a low-resistance 25mH inductor, and a key. Find the rate at which current begins to rise when the key is closed, the current at the instant the rate of change of current is 400A/s, and the final steady current

Answer 1

E= 50 volts

R = resistance = 20 ohm

L = inductance = 25 mH

Applying KVL to the loop given in fig.

E- L*di/dt - iR = 0 --------------------1

initially i= 0 A

E- L*di/dt - 0*R = 0

E- L*di/dt = 0

at t= 0

50 - 25*10^-3 *di/dt = 0

di/dt = 50 / 25*10^-3

di/dt = 2000 A/s

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When di/dt = 400 A/s

from eq 1:

E- L*di/dt - iR = 0   

50 - 25*10^-3 *400 - i*20 = 0

i = 2 A

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for steady current (di/dt=0), eq 1 can be written as :

50 - L(0) - 20*If = 0

50 = 20*If

If( final steady current) = 50/20 = 2.5 A