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Electrostatic Potential And Capacitance

Question

An ideal coil of 10H is connected in series with a resistance of 5 Ω and a battery of 5 V. 2 second after the connection is made, what will be the current flowing in the circuit?

Solution

Given,
Inductance of the coil = 10 H
Resistance = 5 $Ω$
Voltage of battery = 5 V
Time , t = 2 sec
Maximum current , I = $\frac{E}{R} = \frac{5}{5} = 1 A$

During growth of current ,
$I = I_{o} (1 - e^{- \frac{R}{L} t})$
$∴$ $I = 1 (1 - e^{- \frac{5}{10} \times 2}) = (1 - e^{- 1})$

which is the required amount of current flowing through the circuit.

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Electrostatic Potential And Capacitance

An ideal coil of 10H is connected in series with a resistance of 5 Ω and a battery of 5 V. 2 second after the connection is made, what will be the current flowing in the circuit?

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