Determine the charge on the capacitor in the following circuit
(a) 2 μC
(b) 200 μC
(c) 10 μC
(d) 60

Q: Why does the capacitor act as an open circuit?

In a DC circuit, current flows into the capacitor until it is fully charged. Once the voltage across the capacitor equals the voltage of the branch it's connected to, no more charge can flow, making it behave like an open circuit (infinite resistance).

Q: Why are the $2\Omega$ and $10\Omega$ resistors in series?

Because the capacitor branch is open, any current that flows through the $2\Omega$ resistor has nowhere else to go but through the $10\Omega$ resistor. Since they share the exact same current, they are in series.

Q: Can we use nodal analysis instead of simplifying resistors?

Absolutely! You can assign the bottom wire as 0V, the top left as 72V, and write Kirchhoff's Current Law (KCL) equations for nodes P and Q. It will give you the exact same voltages, often faster if you are comfortable with equations.

Q: What if the circuit was not in steady state?

If the circuit was just turned on (transient state), the capacitor would initially act like a short circuit (0V across it), and its voltage would exponentially increase over time. You would need differential equations to find the charge at a specific time $t$.

Q: Why is the voltage across the capacitor the same as the $10\Omega$ resistor?

The capacitor and the $10\Omega$ resistor are connected in parallel. They both share node Q at the top and the ground node at the bottom. Components in parallel always share the exact same voltage.

Question

Determine the charge on the capacitor in the following circuit

Options:

(A) $2\mu	ext{C}$

(B) $200\mu	ext{C}$

(C) $10\mu	ext{C}$

(D) $60\mu	ext{C}$

Sigma Percentile · Accepted Answer

### The Steady-State Secret

When dealing with DC circuits containing capacitors, the most crucial concept to remember is the **steady state**. When a circuit is first turned on, current rushes into the capacitor to charge its plates. However, after a long time (the steady state), the capacitor becomes fully charged. At this point, the voltage across the capacitor exactly opposes the flow of any more charge, meaning it acts as an **open circuit**.

Because the capacitor branch is effectively broken, no current can flow through it. This dramatically simplifies our circuit analysis. We can temporarily ignore the capacitor and focus entirely on the resistor network to find the voltages at various nodes.

### Simplifying the Resistor Network

With the capacitor acting as an open circuit, let's trace the path of the current. The current flowing through the $2\Omega$ resistor has nowhere else to go but straight into the $10\Omega$ resistor. Because they share the exact same current, they are in **series**.

$$R_{right} = 2\Omega + 10\Omega = 12\Omega$$

Now, this entire $12\Omega$ right branch is connected in **parallel** with the $4\Omega$ resistor. We can find their equivalent resistance using the product-over-sum rule:

$$R_p = \frac{4 	imes 12}{4 + 12} = \frac{48}{16} = 3\Omega$$

We have successfully reduced the entire right side of the circuit to a single $3\Omega$ equivalent resistor. This $3\Omega$ resistor is in series with the $6\Omega$ resistor connected directly to the battery. Therefore, the total equivalent resistance of the entire circuit is:

$$R_{eq} = 6\Omega + 3\Omega = 9\Omega$$

### Tracing the Currents and Voltages

Now that we have the total resistance, we can use Ohm's Law to find the total current drawn from the $72	ext{V}$ battery:

$$I = \frac{V}{R_{eq}} = \frac{72}{9} = 8	ext{ A}$$

This $8	ext{A}$ current flows directly out of the battery and through the $6\Omega$ resistor. The voltage drop across this first resistor is simply $8	ext{A} 	imes 6\Omega = 48	ext{V}$.

To find the voltage at node P (the junction after the $6\Omega$ resistor), we subtract this drop from the battery voltage:

$$V_P = 72	ext{V} - 48	ext{V} = 24	ext{V}$$

This $24	ext{V}$ is the potential difference driving current through the right side of the circuit. Let's find the current flowing specifically through the $2\Omega - 10\Omega$ branch. Since the total resistance of that branch is $12\Omega$, the current $I_2$ is:

$$I_2 = \frac{V_P}{R_{right}} = \frac{24}{12} = 2	ext{ A}$$

This $2	ext{A}$ current flows through the $10\Omega$ resistor, creating a voltage drop across it. The voltage at node Q (which is the voltage across the $10\Omega$ resistor) is:

$$V_Q = 2	ext{A} 	imes 10\Omega = 20	ext{V}$$

### The Final Charge Calculation

We have finally found the voltage at node Q. Because the capacitor is connected in parallel with the $10\Omega$ resistor (they share the same top and bottom nodes), the voltage across the capacitor is exactly the same: $20	ext{V}$.

The fundamental equation relating charge ($Q$), capacitance ($C$), and voltage ($V$) is:

$$Q = C 	imes V$$

Substituting our known values:

$$Q = 10\mu	ext{F} 	imes 20	ext{V} = 200\mu	ext{C}$$

The charge stored on the capacitor is $200\mu	ext{C}$, making option (b) the correct answer. By breaking down the circuit logically, even complex-looking networks become highly manageable!

Animated Solution for Physics - Current Electricity: Determine the charge on the capacitor in the following circuit

Select Answer:

Visualized Solution

Circuit Analysis in Steady State

Equivalent Resistance of Right Branch

Parallel Combination

Total Equivalent Resistance

Total Circuit Current

Voltage at Node P

Current in the Right Branch

Voltage Across the Capacitor

Charge on the Capacitor

The Sigma Insight: RC Circuit

The Steady-State Secret

Simplifying the Resistor Network

Tracing the Currents and Voltages

The Final Charge Calculation

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