R-C Circuits Problems and Solution 2

 Problem#1

In the circuit shown in Fig. 1 each capacitor initially has a charge of magnitude 3.50 nC on its plates. After the switch S is closed, what will be the current in the circuit at the instant that the capacitors have lost 80.0% of their initial stored energy?

Fig.1

Answer:

Known:
capacitor initially has a charge of magnitude q₀ = 3.50 nC = 3.50 x 10^─9 C
The stored energy is proportional to the square of the charge on the capacitor, so it will obey an exponential equation, but not the same equation as the charge.
The energy stored in the capacitor is

U = Q²/2C

and the charge on the plates is

Q = Q₀e^−t/RC

The current is

i = i₀e^−t/RC

then

U = (Q₀e^−t/RC)²/2C = (Q₀²/2C) e^−2t/RC
U = U₀ e^−2t/RC

When the capacitor has lost 80% of its stored energy, the energy is 20% of the initial energy, which is 0.2U₀ or U₀/5,

so that

U₀/5 = U₀ e^−2t/RC
1/5 = e^−2t/RC
2t/RC = ln 5
t = (RC/2) ln 5 = (25.0 Ω)(4.62 pF)(ln 5)/2
t = 92.9 ps

At this time, the current is

i = i₀e^−t/RC = (Q₀/RC) e^−t/RC, so
i = (3.5 nC)/[(25.0 Ω)(4.62 pF)] e^{–(92.9 ps)/[(25.0 Ω)(4.62 pF)]}
i =13.6 A

Problem#2
In the circuit in Fig. 2 the capacitors are all initially uncharged, the battery has no internal resistance, and the ammeter is idealized. Find the reading of the ammeter (a) just after the switch S is closed and (b) after the switch has been closed for a very long time.

Fig.2

Answer:

In both cases, simplify the complicated circuit by eliminating the appropriate circuit elements. The potential across an uncharged capacitor is initially zero, so it behaves like a short circuit. A fully charged capacitor allows no current to flow through it.

(a) Just after closing the switch, the uncharged capacitors all behave like short circuits, so any resistors in parallel with them are eliminated from the circuit.

The equivalent circuit consists of 50 Ω and 25 Ω in parallel

R_p = 50 Ω x 25 Ω/(75 Ω) = 16.7 Ω

with this combination in series with 75 Ω, 15 Ω, and the 100 V battery,

The equivalent resistance is

R_eq = 75 Ω + 15 Ω + 16.7 Ω = 106.7 Ω,

Which gives

I = (100 V)/(106.7 Ω) = 0.937 A

(b) Long after closing the switch, the capacitors are essentially charged up and behave like open circuits since no charge can flow through them. They effectively eliminate any resistors in series with them since no current can flow through these resistors.

The equivalent circuit consists of resistances of 75 Ω, 15 Ω, and three 25 Ω resistors, all in series with the 100 V battery, for a total resistance of

R_s = 75 Ω + 15 Ω + 3 x 25 Ω = 165 Ω.

Therefore

I = (100V)/(165Ω) = 0.606 A

Problem#3
Figure 3 displays two circuits with a charged capacitor that is to be discharged through a resistor when a switch is closed. In Fig. 5a, R₁= 20.0 Ω and C₁ = 5.00 µF. In Fig. 5b, R₂ = 10.0 Ω and C₂ = 8.00 µF. The ratio of the initial charges on the two capacitors is q₀₂/q₀₁ = 1.50. At time t = 0, both switches are closed. At what time t do the two capacitors have the same charge?

Fig.3

Answer:

Known:
R₁= 20.0 Ω and C₁ = 5.00 µF, then R₁C₁ = (20.0 Ω)(5.00 µF) = 1 x 10^─4 s
R₂ = 10.0 Ω and C₂ = 8.00 µF, then R₂C₂ = (10.0 Ω)(8.00 µF) = 8 x 10^─5 s

The ratio of the initial charges on the two capacitors is q₀₂/q₀₁ = 1.50

the charge in the capacitor is

q₁ = q₀₁ e^−t/R1C1 and q₂ = q₀₂ e^−t/R2C2
time t do the two capacitors have the same charge is

q₁ = q₂
q₀₁ e^−t/R1C1 = q₀₂ e^−t/R2C2
e^−t/R1C1 = 1.50 e^−t/R2C2
−t/R₁C₁ + t/R₂C₂ = ln (1.50)
−t/(1 x 10^─4 s) + t/(8 x 10^─5 s) = ln (1.50)
2500t  = ln (1.50)
t = 1.62 x 10^─4 s = 162 μs  

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