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 e^–1 = 0.368 I₀.

But I₀ = V₀/R = 6/5 = 1.2 A.

Therefore I = 0.368×1.2 = 0.4416 A.

Free Response Practice Question on Nuclear Physics for AP Physics B

Occasionally you may get a free response question involving nuclear physics. Here is a question meant for exercise. This question carries 15 points and you can take 17 minutes for answering it. Try to answer the question yourself before going through the solution given below the question.

One proposal for the production of power is the fusion of deuterium to produce tritium in accordance with the nuclear reaction

₁D² + ₁D² → ₁T³ + ₁H¹ + Q

where the symbols D and T represent respectively deuterium and tritium. Mass of ₁D² = 2.0141 u; Mass of ₁T³ = 3.0161 u; Mass of ₁H¹ = 1.0078 u and 1 u = 1.66×10^–27 kg = 931 MeV/c²

Now answer the following:

(a) Calculate the mass defect involved in a single fusion reaction.

(b) Determine the energy Q released in a single fusion reaction.

(c) Calculate the number of fusions required per second to produce a power of 2000 mega watt.

(d) If the efficiency of the process is assumed to be 60%, determine the mass of deuterium required per day.

(a) Mass defect per fusion ∆m = (2×Mass of ₁D²) – (Mass of ₁T³ + Mass of ₁H¹)

= (2×2.0141 u) – (3.0161 u + 1.0078) = 0.0043 u

(b) Energy released in a single fusion reaction = 0.0043×931 MeV = 4.0033 MeV

(c) Since 1 MeV = 1.6×10^–13 J, the number (n) of fusions required per second to produce a power of 2000 mega watt (which is 2000×10⁶ joule per second) is given by

n = (2000×10⁶)/( 4.0033×1.6×10^–13) = 3.1224×10²¹

(d) Number of fusions required per day = 3.1224×10²¹×(24×60×60) = 2.6977×10²⁶.

Since two deuterons are required per fusion, the number of deuterons required per day (at 100% efficiency) is 2×2.6977×10²⁶ = 5.3954×10²⁶.

Since the efficiency of the process is 60% only, the required number of deuterons is 5.3954×10²⁶ ×(100/60) = 8.9923×10²⁶.

Since 1 u = 1.66×10^–27 kg, the mass of the above number of deuterium atoms = 8.9923×10²⁶×2.0141×1.66×10^–27 kg = 3.0065 kg.

Now, suppose the part (d) in the above question is replaced by the following:

(d) If the radius of the deuteron is assumed to be nearly 1.5×10^–15 m, calculate (i) the kinetic energy required by a deuteron to overcome the coulomb repulsion and (ii) the temperature to which the deuterium gas is to be heated to initiate the fusion reaction. Boltzmann’s constant, k = 1.38×10^–23 J/K

When the deuterons are in contact, the separation (r) between their centres is 2×1.5×10^–15 m (= 3×10^–15 m) and the electrostatic potential energy is (1/4πε₀)q²/r where ε₀ is the permittivity of free space and q is the charge on the deuteron (which is equal to 1.6×10^–19 coulomb). The kinetic energy required to overcome the coulomb repulsion must be at least equal to the above potential energy.

Therefore, kinetic energy = (1/4πε₀)q²/r = 9×10⁹×(1.6×10^–19)² /3×10^–15 = 7.68×10^–14 J. Since the deuteron has 3 degrees of freedom its kinetic energy is (3/2)kT where k is Boltzmann’s constant and T is the absolute (Kelvin) temperature.

Therefore (3/2)kT = 7.68×10^–14 so that T = 2×7.68×10^–14/(3×1.38×10^–23) = 3.7×10⁹ K.