Question

A capacitor C1 =1F is connected to a 1V battery using a wire with a total resistance R = 1Ohm

Suppose after the charging is complete the capacitor C1 is connected to another capacitor C2 = 2 F using a wire with a total resistance of R =0.3 Ohm. Now the first capacitor discharges while the second one charges.

2.8 [1pt] How much charge was transferred through the resistor R during the discharge?

ANSWER

2.9 [1pt] How much energy was dissipated in the resistor R during the discharge

ANSWER

Let’s derive the differential equation describing the discharge in this 2-capacitor circuit. Let’s label the charge on the capacitor C1 as q1(t) and the charge on the capacitor C2 as q2(t).

2.10 [1pt] Sketch the circuit diagram, label the charges and their signs at the capacitors plate, and link q1(t) and q2(t).

Hint, what is q1(t=0) and q2(t=0) (t=0 is the instant the connection was made)?

ANSWER

2.11 [1pt] Express the current through the resistor in terms of the charge derivative and write down the voltages across both capacitors and the resistor add up to zero. From the resulting equation, deduce the characteristic charging time without solving it.

ANSWER

Answer 1

Initially C₁ = 1 F is charged using battery 1 V, Hence charge Q_o on C₁ = C₁ 1 V = 1F 1 V = 1 C

During discharging time, charge on capacitor C₁ keep on decreasing and

charge on capacitor C₂ keep on increasing

Charge on discharging capacitor C₁ as a function of time, q₁(t) = [Q_o - Q₁]e^-t/RC₂ + Q₁

where Q₁ is the final charge on capacitor C₁ after completion of charging second capacitor

Voltage on discharging capacitor C₁ as a function of time,

q₁(t)/C₁ = v₁(t) = [ ( Q_o - Q₁]) /C₁ ] e^-t/RC₂ + Q₁/C₁

Charge on charging capacitor C₂ as a function of time, q₂(t) = Q₂[ 1 - e^-t/RC₂ ]

where Q₂ is the charge on capacitor after long time

Voltage on charging capacitor C₂ as a function of time, q₂(t)/C₂ = v₂(t) = ( Q₂/C₂ ) [ 1 - e^-t/RC₂ ]

In the circuit shown in figure, if we apply Kirchoff's voltage law, we get

v₁(t) = i(t) R + v₂(t)

[ ( Q_o - Q₁]) /C₁ ] e^-t/RC₂ + Q₁/C₁ = i(t) R + ( Q₂/C₂ ) [ 1 - e^-t/RC₂ ]

After long time , charging of C₂ is complete, then charging current i(t) is zero.

when t = , we get , Q₁/C₁ = Q₂/C₂

By substituting C₁ and C₂ in the above eqn., 2Q₁ = Q₂

But by charge conservation, Q₁ + Q₂ = Q₀ = 1 C

Hence Q₁ = (1/3) C   and Q₂ = (2/3) C

Answer to Qn.(2.8) :- charge transferred through R during discharge = Q₂ = (2/3) C

Answer to Qn.(2.9)

Energy dissipated in Resistor = Initial stored energy in capacitor C₁ - ( Final stored energy in both capacitors)

Initial energy stored in capacitor C₁ = (1/2) Q₀² / C₁ = (1/2) J

Final energy stored in both capacitors = [ (1/2) Q₁² / C₁ ] + [ (1/2) Q₂² / C₂ ] = (1/2) (1/9) + (1/2)(4/9)(1/2) = (1/6) J

Energy dissipated in Resistor = [ (1/2) - (1/6) ] J = (1/3) J