Give step-by-step solution with explanation and final answer:3. A circuit is given as in the picture. The quantities Vs, V2, Cs, Cz and R are all known.
Initially, the system is in the steady state a

Question

Give step-by-step solution with explanation and final answer:3. A circuit is given as in the picture. The quantities Vs, V2, Cs, Cz and R are all known.
Initially, the system is in the steady state and C: contains charge O>. Recall that the
equivalent capacitance of two capacitors connected in series is C = C1C2/ (Ci+ C2).
(a) Find Oy, the amount of charge on Ci.
At 1=0 the switch is flipped from A to B. Now battery V2 is charging the capacitors.
(b) Using conservation of charge, find ¢;, the amount of charge on C; in terms of g2, the
amount of charge on C>.
(c) Using Kirchhoff’s loop rule and /=dg2/dt, find the differential equation for g2(%).
Va R C,
Cy 1
A
Va

VIPSolutions · Accepted Answer

Let's solve each part step by step:

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### (a) Find $ Q_1 $, the amount of charge on $ C_1 $:

**Initial State:**  
- The switch is at A.
- Only $ V_1 $ is connected to $ C_1 $.
- $ C_2 $ is not connected to the circuit (so it remains uncharged).

So, $ C_1 $ is simply charged by $ V_1 $:

$$
Q_1 = C_1 V_1
$$

**Final Answer (a):**
$$
\boxed{Q_1 = C_1 V_1}
$$

---

### (b) Using conservation of charge, find $ q_1 $ (charge on $ C_1 $) in terms of $ q_2 $ (charge on $ C_2 $) after the switch is flipped to B.

**After switch flips to B:**
- $ V_2 $ is connected to both $ C_1 $ and $ C_2 $ in series with resistor $ R $.
- Initial charge distribution: $ C_1 $ has $ Q_1 $, $ C_2 $ has 0.
- Now charge redistributes, but the total charge **originally on the upper plate of $ C_1 $ and lower plate of $ C_2 $ together** is conserved.

Let:
- $ q_1 $: charge on $ C_1 $ at time $ t $
- $ q_2 $: charge on $ C_2 $ at time $ t $

For two capacitors in series, the charge on each at any time is the same, but here, since the system started with $ Q_1 $ on $ C_1 $ and 0 on $ C_2 $, we have to use **conservation of charge** for the combination.

At any moment:
- The **net charge** that moved from $ C_1 $ to $ C_2 $ is $ q_2 $.
- So, the charge on $ C_1 $ is: $ q_1 = Q_1 - q_2 $

**Final Answer (b):**
$$
\boxed{q_1 = Q_1 - q_2}
$$
with $ Q_1 = C_1 V_1 $.

---

### (c) Using Kirchhoff’s loop rule, find the differential equation for $ q_2(t) $:

Now with the switch at B, write the loop equation:

$$
V_2 - IR - V_{C_1} - V_{C_2} = 0
$$
where:
- $ I = \frac{dq_2}{dt} $ (current through $ R $)
- $ V_{C_1} = \frac{q_1}{C_1} $
- $ V_{C_2} = \frac{q_2}{C_2} $
- $ q_1 = Q_1 - q_2 $

Substitute:
$$
V_2 - R \frac{dq_2}{dt} - \frac{Q_1 - q_2}{C_1} - \frac{q_2}{C_2} = 0
$$

Simplify the capacitor terms:
$$
V_2 - R \frac{dq_2}{dt} - \left(\frac{Q_1}{C_1} - \frac{q_2}{C_1} + \frac{q_2}{C_2}\right) = 0
$$
$$
V_2 - R \frac{dq_2}{dt} - \frac{Q_1}{C_1} + q_2\left(\frac{1}{C_1} + \frac{1}{C_2}\right) = 0
$$
Bring $ R \frac{dq_2}{dt} $ to one side:
$$
R \frac{dq_2}{dt} = V_2 - \frac{Q_1}{C_1} + q_2\left(\frac{1}{C_1} + \frac{1}{C_2}\right)
$$
$$
\frac{dq_2}{dt} = \frac{1}{R}\left[V_2 - \frac{Q_1}{C_1} + q_2\left(\frac{1}{C_1} + \frac{1}{C_2}\right)\right]
$$

**Final Answer (c):**
$$
\boxed{
\frac{dq_2}{dt} = \frac{1}{R} \left[ V_2 - \frac{Q_1}{C_1} + q_2 \left( \frac{1}{C_1} + \frac{1}{C_2} \right) \right]
}
$$
where $ Q_1 = C_1 V_1 $.

---

**Summary of Answers:**
- **(a)** $ Q_1 = C_1 V_1 $
- **(b)** $ q_1 = Q_1 - q_2 $
- **(c)** $ \displaystyle \frac{dq_2}{dt} = \frac{1}{R} \left[ V_2 - \frac{Q_1}{C_1} + q_2 \left( \frac{1}{C_1} + \frac{1}{C_2} \right) \right] $

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