The figure shows three circuits with identical batteries, inductors, and resisto

Question

The figure shows three circuits with identical batteries, inductors, and resistors. Rank the circuits according to the current through the battery (a) just after the switch is closed and (b) a long time later, greatest first. A charged capacitor and an inductor are connected in series at time t = 0. In terms of the period T of the resulting oscillations, determine how much later the following reach their maximum value: (a) the charge on the capacitor; (b) the voltage across the capacitor, with its original polarity; (c) the energy stored in the electric field; and (d) the current.

Explanation / Answer

Problem 1.

(a)

Just after the switch is closed, the current through inductor will be zero as initially the current was zero and the current through inductor cannot change abruptly;

So, in circuit (1), just after switch is closed, the current through inductor will be zero. But, the resistors and the inductor are in series so current through the circuit will be zero.

in circuit (2),  just after switch is closed, the current through inductor will be zero. So the current will flow through the left resistor as there will be no current through the resistor connected in series with the inductor. So the battery will provide current I = V/R, where V is the voltage of the battery and R is the resistance.

in circuit (2),  just after switch is closed, the current through inductor will be zero. So the current will flow through the two resistors which will be in series as no current is flowing through the inductor which makes the same current go through both the resistors making them in series. So, the battery provides current I = V/(R+R) = V/2R

From above data the current flow ranking will be, battery 2 > battery 3 > battery 1, so maximum current is through battery 2 in circuit 2.

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(b)

A long time after the switch closed, the inductor is short circuited in all the three circuits.

In circuit 1, Battery provides current I = V/2R as both resistors are in series.

In circuit 2, equivalent resistance seen by the battery is Req = R.R/(R +R) = R/2, as the two resistor are now in parallel. So, Battery provides current I = V/(R/2) = 2V/R

In circuit 3, the resistor having common terminal with the inductor is short circuited as the inductor is short circuited and the current simply flows through the top resistor and inductor without leaving the other resistor.

So  Battery provides current I = V/(R)

From above data the ranking of current provided by the three circuit batteries is;

battery 2 > battery 3 > battery 1

So ranking remains the same as before.

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This concludes the answer. Check the answer and let me know if it's correct. If you need any more clarification or correction, feel free to ask.....

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