Question

A coil of self inductance 50 mH and resistance 1 is connected through a switch to a battery of internal resistance 9.The switch is closed till it attain saturation state. Now the battery is short circuited. The time taken for current to attain 80% of saturation value is------secL

Answer 1

Explanation:

Given A coil of self inductance 50 mH and resistance 1 is connected through a switch to a battery of internal resistance 9.The switch is closed till it attain saturation state. Now the battery is short circuited. The time taken for current to attain 80% of saturation value

• Given Self inductance L = 50 m h
• Resistance = 1 ohm
• Internal resistance = 9 ohm
• We need to find the time taken for current to attain 80% saturation value.
• So i = io(1 – e^-t / T)
•   80 / 100 io = io (1 – e^-t / T)
•    0.8 = 1 – e^-t / T
•  Or e^- t/T = 0.2
• Now 1/T = log (1/5)
• Or t/T = - log 5
• Or t = T. log 5
• Or t = L / R x log 5
•        = 50 x 10^-3 / (1 + 9)  x 0.698
• Or t = 3.49 x 10^-3 secs

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