In the given circuit diagram, when the current reaches steady state in the circuit, the charge on th

Q: Why doesn't current flow through the capacitor?

In a DC circuit, a capacitor charges up until its voltage equals the voltage applied across its branch. Once fully charged, it acts as an open circuit, blocking any further flow of direct current.

Q: Why is the voltage drop across $r_1$ zero?

According to Ohm's law, the voltage drop across a resistor is $V = IR$. Since the capacitor blocks the current in the steady state, $I = 0$ for that entire bottom branch. Therefore, $V = 0 \times r_1 = 0$.

Q: Why don't we include $r_1$ in the total resistance calculation?

Because no current flows through the bottom branch, it doesn't draw any current from the battery. The battery only 'sees' the top and middle branches, so only $r$ and $r_2$ contribute to the equivalent resistance for the current calculation.

Q: What if the capacitor was in the middle branch instead?

If the capacitor were in the middle branch, then in the steady state, no current would flow through the middle branch. The active loop would then be the top and bottom branches, and the current would be $I = \frac{E}{r + r_1}$.

Q: How would we find the charge before the steady state is reached?

Before the steady state, the circuit is in a transient state. You would need to write the differential equation using Kirchhoff's laws or find the Thevenin equivalent resistance across the capacitor to use the standard charging formula $Q(t) = Q_{\text{max}}(1 - e^{-t/\tau})$, where $\tau = R_{\text{eq}}C$.

Question

In the given circuit diagram, when the current reaches steady state in the circuit, the charge on the capacitor of capacitance $C$ will be

Options:

(A) $CE \frac{r_1}{(r_2+r)}$

(B) $CE \frac{r_2}{(r+r_2)}$

(C) $CE \frac{r_1}{(r_1+r)}$

(D) $CE$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at a multi-lane highway for electrons. Our circuit consists of a battery with an electromotive force $E$ and an internal resistance $r$ on the top branch. Below it, we have a middle branch with a resistor $r_2$, and a bottom branch containing a capacitor $C$ in series with another resistor $r_1$.

The question asks us to find the charge stored on the capacitor once the circuit reaches its **steady state**. This is a classic scenario in DC circuits, and understanding the behavior of the capacitor is the key to unlocking the solution.

### The Magic of the Steady State

What exactly happens when a DC circuit reaches a steady state? When you first close the switch, current rushes in to charge the capacitor. However, as the capacitor fills up with charge, it pushes back against the battery's voltage.

Eventually, after a long time ($t 	o \infty$), the capacitor becomes fully charged. At this point, its voltage perfectly opposes the flow of any more charge. It acts as an **open circuit**. Because of this, absolutely zero current flows through the bottom branch where the capacitor resides.

$$I_{	ext{bottom}} = 0$$

### The Active Current Loop

Since the bottom branch is effectively a dead end for continuous current, the electrons from the battery have only one path to take. They flow out of the battery, through the internal resistance $r$, down through the middle resistor $r_2$, and back to the battery.

This forms a simple, single active loop. We can easily calculate the steady-state current $I$ circulating in this loop using Ohm's law. The total resistance of this active path is the series combination of $r$ and $r_2$.

$$I = \frac{E}{r + r_2}$$

### Unlocking the Capacitor's Voltage

To find the charge on the capacitor, we first need to know the potential difference (voltage) across it. Let's look at the bottom branch again. It contains the capacitor $C$ and the resistor $r_1$.

Because the steady-state current through this branch is zero, the voltage drop across the resistor $r_1$ must also be zero ($V = I \cdot r_1 = 0 \cdot r_1 = 0$).

This is a crucial realization! It means that the entire potential difference across the bottom branch is dropped exclusively across the capacitor. Furthermore, since the bottom branch is in parallel with the middle branch, the voltage across the capacitor $V_C$ must be exactly equal to the voltage across the middle resistor $r_2$.

$$V_C = V_{r_2}$$

We can find $V_{r_2}$ by multiplying the steady-state current $I$ by the resistance $r_2$:

$$V_C = I \cdot r_2 = \left( \frac{E}{r + r_2} ight) r_2$$

### The Final Charge

Finally, the relationship between charge, capacitance, and voltage is given by the fundamental capacitor equation:

$$Q = C \cdot V_C$$

Substituting the voltage we just found into this equation, we arrive at our final answer:

$$Q = CE \frac{r_2}{r + r_2}$$

This elegant result shows how the internal resistance of the battery and the parallel resistor dictate the final energy state of the capacitor.

Animated Solution for Physics - Current Electricity: In the given circuit diagram, when the current reaches steady state in the circuit, the charge on the capacitor of capacitance $C$ will be

Select Answer:

Visualized Solution

Analyzing the Circuit

Steady State of a Capacitor

Active Current Loop

Calculating Steady State Current

Potential Difference Across Capacitor

Voltage Across Middle Branch

Final Charge on Capacitor

The Way Forward

The Sigma Insight: RC Circuit

Analyzing the Setup

The Magic of the Steady State

The Active Current Loop

Unlocking the Capacitor's Voltage

The Final Charge

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