It is not by sitting still at a grand distance and calling the human race larvae that men are to be helped.

– Albert Einstein

The following question on transients in LR circuit was asked in the All India Engineering/Architecture Entrance Examination (AIEEE) 2009:

An inductor of inductance L = 400 mH and resistors of resistances R_{1} = 2 Ω and R_{2} = 2 Ω are connected to a battery of emf 12 V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at t = 0. The potential drop across L as a function of time is

(1) 12 e^{–5t} V

(2) 6 e^{–5t} V

(3) (12/t) e^{–3t}

(4) 6(1– e^{–t/0.2}) V

The exponential growth of current *I* in an LR circuit is given by

*I = I*_{0} (1– e^{–Rt/L}) where *I* is the current at the instant *t*, *I*_{0} is the final maximum current equal to *V/R* where *V* is the *L* at the instant *t* is *L dI/dt*

Therefore potential drop across *L* at the instant *t* = *L** *(*I*_{0}*R/L*) e^{–Rt/L} =* I*_{0}*R* e^{–Rt/L}

In the present problem *I*_{0} = 6 A (since *V =* 12 volt and *R = R*_{2} = 2 Ω) and *L = *400 mH = 0.4 H.

Therefore, the potential drop across *L* at the instant *t* = 6×2×e^{–2t/0.4} = 12 e^{–5t} volt [Option (1)].

Let us try the following question involving transients in a CR circuits:

Suppose the inductor in the above question is replaced by a capacitor of capacitance 1000 μF. The switch S is closed at time t = 0. The charging current flowing through the capacitor as a function of time is

(1) 12 e^{–500t} A

(2) 6 e^{–500t} A

(3) 12e^{–t/1000} A

(4) 6(1– e^{–t/500}) A

The exponential growth of charge *Q *on a capacitor in a CR circuit is given by

*Q* = *Q*_{0 }(1– e^{–t/RC}) where *Q* is the charge at the instant *t*, *Q*_{0} is the final maximum charge which is equal to *CV*. The charging current *I *flowing through the capacitor is given by

*I = dQ/dt =** *(*Q*_{0}/*RC*) e^{–t/RC} = (*CV**/RC*)* *e^{–t/RC} = (*V/R*) e^{–t/RC}

Here *V = *12 volt, *C = *1000 μF = 0.001 F and *R = **R*_{2} = 2 Ω

Therefore, the charging current *I *flowing through the capacitor = 6× e^{–t/(2}^{×0.001)} = 6 e^{–500t} A.

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