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Saturday, January 24, 2009

Transients in RC Circuits- Multiple Choice Practice Questions for AP Physics C

The growth of charge Q on a capacitor connected in series with a resistance R ohm and a direct voltage source of emf V0 volt is given by

Q = Q0(1– e–t/RC) where Q0 = CV0 which is the final (maximum) charge on the capacitor and e is the base of natural logarithms.

When the capacitor having charge Q0 is allowed to discharge through a resistance R, the decay of charge on the capacitor is given by

Q = Q0e–t/RC

The nature of the growth and the decay of charge are shown in the adjoining figure.

Since the voltage (V) across the capacitor is directly proportional to the charge on it, the voltage during charging and discharging can be written respectively as

V = V0(1– e–t/RC) and

V = V0e–t/RC where V0 is the charging battery voltage.

The time constant of the CR circuit is RC which is the time taken for the charge to grow from zero value to (1– 1/e) times (0.632 times) the final maximum charge. This follows from the equation for the exponential growth of charge.

The time constant RC is also equal to the time taken for the charge to decay from the initial value to 1/e times (0.368 times) the initialal value. This follows from the equation for the exponential decay of charge.

The charging current is obtained by differentiating the equation Q = Q0(1– e–t/RC) with respect to the time t. Thus the charging current (dQ/dt) is given by

I = (Q0 /RC) e–t/RC = I0 e–t/RC where I0 = Q0 /RC = V0/R where V0 is the charging voltage.

The exponential decay of the charging current is shown in the adjoining figure.

The current during the discharge of the capacitor through the resistor is obtained by differentiating the equation Q = Q0e–t/RC with respect to the time t. Thus the discharging current (dQ/dt) is given by

I = (Q0 /RC) e–t/RC = I0 e–t/RC where I0 = Q0 /RC = V0/R where V0 is the initial voltage on the capacitor.

The exponential decay of the discharging current which flows in the direction opposite to the charging current also is shown in the adjoining figure.

[The charging and discharging of a capacitor is similar to inflating and deflating a balloon. The air pressure is the analogue of voltage and the air current is the analogue of electric current].

We will now discuss a few multiple choice questions in this section:

(1) A transistor time delay relay circuit makes use of a series RC circuit. If the circuit uses a 60 μF capacitor what is the resistance required to produce a delay time of 1 minute?

(a) 60 Ω

(b) 6×10–5 Ω

(c) 1.67×104 Ω

(d) 105 Ω

(e) 106 Ω

The delay time is 60 s and hence RC = 60 so that the resistance required is given by

R = 60 s/60×10–6 F = 106 Ω

(2) A constant current source charges an initially uncharged capacitor. During the charging process, the potential difference V between the plates of the capacitor is related to the charging time t as

(a) V α t

(b) V α t0

(c) V α t–1

(d) V α (1– e–t)

(e) V α e–t

Since the current I is constant, the charge delivered to the capacitor (Q = It) is directly proportional to the time t. Since the potential difference V between the plates of the capacitor is directly proportional to the charge (V = Q/C), we have V α t.

(3) A 10 μF capacitor is connected in series with a 5 Ω resistor and a battery of negligible internal resistance and emf 6 V. After a long time, if the charging battery is removed and the capacitor is immediately connected across the 5 Ω resistor, the initial value of the discharge current is

(a) 1.2 A

(b) 0.6 A

(c) 0.3 A

(d) 0.83 A

(e) zero

The capacitor will get charged to the battery voltage (6 V) irrespective of the resistance in series with it. On connecting it across the 5 Ω resistor, the initial value of the discharge current (at t = 0) is given by

I = I0 e–t/RC where I0 = V0/R where V0 is the initial voltage on the capacitor. Let us ignore the negative sign which just indicates that the discharge current is opposite to the charging current.

Therefore, I = (6/5) e0 = 1.2 A

(4) In the above question what will be the voltage to which the capacitor will be charged if it is connected in series with the 5 Ω resistor and the 6 V battery for 50 microseconds?

(a) 1.333 V

(b) 2.208 V

(c) 3.792 V

(d) 4.5 V

(e) 6 V

The time constant of the circuit is RC =5×10×10–6 = 50 microsecond.

After one time constant, the voltage on the capacitor will be (1– 1/e) times (0.632 times) the charging battery voltage. The answer therefore is 0.632×6 V = 3.792 V.

(5) In question No.3 what will be the charging current at the instant t = 50 microseconds?

(a) 0.3680 A

(b) 0.4416 A

(c) 0.6 A

(d) 0.7326 A

(e) 1.2 A

Since the time constant of the circuit, RC =5×10×10–6 s = 50 microseconds, the charging current at the instant t = 50 microseconds will be I = I0 e–t/RC = I0 e1 = 0.368 I0.

But I0 = V0/R = 6/5 = 1.2 A.

Therefore I = 0.368×1.2 = 0.4416 A.

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