A 189-V battery, an inductor, and a resistor are connected in series as shown in

Question

A 189-V battery, an inductor, and a resistor are connected in series as shown in the diagram below. A two-way switch makes it possible to include or exclude the battery. The switch that had been in position 1 for a long time is suddenly moved to position 2. (Enter your answers to at least two decimal places.)

There are 2 circuit diagrams, labeled (i) and (ii). They are identical to each other in all ways, except that circuit (i) has the switch in the 1 position, and circuit (ii) has the switch in the 2 position. Circuit (i) begins at a switch S at position 1. Position 1 leads to the negative pole of a battery Delta V and then the positive pole of the battery leads to a split in the wire. One wire leads to the dead end position 2. The other wire leads to an inductor L and then a resistor R before returning to the switch. Circuit (ii) begins at a switch S at position 2. Position 2 leads to a split in the wire. One wire leads to the positive pole of a battery Delta V and then the negative pole of the battery leads to the dead end position 1. The other wire leads to an inductor L and then a resistor R before returning to the switch.

(a) What is the voltage across the resistor at the end of five time constants? V

(b) At this time, what is the voltage across the inductor? V

Explain all answers in detail please.

Explanation / Answer

When the charge stored in a cap is increased, the voltage rises according to this formula V = Q/C. The concept of a 'fully charged capacitor' is therefore in principle meaningless, since any charge can be stored, depending on the voltage applied. In practice, the maximum stored charge will be that which causes the voltage to rise to a point at which the dielectric breaks down. I will assume that the applied voltage is 120kV and that the cap is not 'fully charged' - i.e. that it does not experience dielectric breakdown.

a)

At the end of the first time constant, the cap discharges down to (e^-1)*Vo or 0.3678*Vo, where Vo is the original voltage. At the end of the 2nd, it's 0.3578*the previous voltage or 0.3578*0.3578*Vo.

In table form, using Tc as time constant abreviation
Tc_____________Voltage at start of Tc____Voltage at the end Tc

0 (before start)_____ Vo___________________N/A
1_________________Vo__________________... = 0.3578*Vo
2_________________V1__________________... = (0.3578^2)*Vo
3_________________V2__________________... = (0.3578^3)*Vo
4_________________V3__________________... = (0.3578^4)*Vo

The equation for time in general is
V = e^(t/Tc) * Vo
So if t is 4X the Tc, then the e^(-t/Tc) part is e^-4

When a cap discharges through a resistor R, the time constant (tau) is R*C. The variation of voltage with time (t) is given by V = Vo*exp(-t/tau) where V is the voltage at time t and Vo is the initial voltage. In this case Vo = 120kV and t = 3*tau

V = 120*exp(-3) = 5.97kV

b) The voltage across the inductor will be the same as that across the cap, since they are joined together in parallel.

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