This circuit problem involves a battery with two resistors and a capacitor, where one resistor has a variable resistance $R_1$, the other resistor has a fixed resistance $R_2$, and the capacitor has a capacitance $C$. The task consists of analyzing the behavior of the charge on the capacitor in steady-state conditions when the resistance $R_1$ changes.

Let's break it down:

(a) Analysis without deriving equations:

As the value of $R_1$ increases, the total resistance in the circuit increases. The time constant $\tau$ for an RC circuit is given by: $\tau = (R_1 + R_2) C$ Since $R_1$ is increased, the time constant increases, meaning the capacitor will charge more slowly. However, once the capacitor is fully charged, it will reach the same final voltage (and charge) as if $R_1$ had been at any other value. This is because the charge on the capacitor at steady-state depends on the emf of the battery, the total resistance, and the capacitance.

At steady state, the current is zero (as the capacitor is fully charged), and the charge on the capacitor, $Q_0$, is given by: $Q_0 = C \cdot \epsilon$ where $\epsilon$ is the emf of the battery. Therefore, the charge on the capacitor does not depend on $R_1$ in the steady state, and the new charge on the capacitor is equal to $Q_0$.

Thus, increasing $R_1$ does not change the charge on the capacitor in steady-state conditions.

(b) Derivation for the steady-state charge $Q$:

To derive the equation for the steady-state charge on the capacitor, we need to consider the voltage division between the resistors and the capacitor in the steady-state condition.

1. The total resistance in the circuit is $R_1 + R_2$.
2. The current through the circuit will be the same for both resistors because they are in series, and the capacitor will act like an open circuit at steady state (no current through the capacitor).
3. The voltage across the capacitor will be equal to the voltage drop across the resistors.

Using Kirchhoff's loop rule: $\epsilon = I(R_1 + R_2)$ where $I$ is the steady-state current. At steady state, the current through the circuit is zero because the capacitor is fully charged. Thus, the voltage across the capacitor equals the battery's emf.

Therefore, the charge on the capacitor is: $Q = C \cdot \epsilon$

Summary:

• The charge on the capacitor is independent of $R_1$ in steady state.
• The steady-state charge is given by $Q = C \cdot \epsilon$.

Would you like further details on any part of this solution, or would you like to explore more about RC circuits?

Here are some follow-up questions you might find interesting:

1. How does the time constant of an RC circuit change with different resistor values?
2. What is the role of the capacitor in a series RC circuit?
3. How would the charge on the capacitor change if the emf $\epsilon$ were increased?
4. How does the behavior of current in the circuit differ between transient and steady-state conditions?
5. Can you explain the energy stored in a capacitor at steady state?

Tip: The time constant $\tau = (R_1 + R_2)C$ describes the rate at which the capacitor charges or discharges. The larger the resistance, the slower the charge/discharge process.