A capacitor of 50 µF is connected in a circuit as shown in figure. The charge on the upper plate of

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 it. Once fully charged (steady state), it acts as an open circuit, blocking any further direct current.

Q: Why are the three resistors in series?

Since the capacitor branch acts as an open circuit, no current splits off into it. The exact same current from the battery must flow sequentially through all three resistors, which is the definition of a series connection.

Q: How do we know the capacitor is only across the middle resistor?

By tracing the wires! The top wire of the capacitor branch connects exactly between the top and middle resistors, and the bottom wire connects between the middle and bottom resistors.

Q: What if the capacitor was connected across all three resistors?

Then the voltage across the capacitor would be the full 6V of the battery, and the charge would be $50\mu\text{F} \times 6\text{V} = 300\mu\text{C}$.

Q: How do we determine which plate is positive?

Current flows from the positive terminal of the battery to the negative terminal. As it flows through resistors, potential drops. Therefore, the node above the middle resistor is at a higher potential than the node below it, making the upper plate positive.

Question

A capacitor of $50\mu	ext{F}$ is connected in a circuit as shown in figure. The charge on the upper plate of the capacitor is ..............$\mu	ext{C}$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are tracing the path of electricity in this fascinating circuit.

We have a $6	ext{ V}$ battery acting as our power source, connected across a vertical chain of three $2	ext{ k}\Omega$ resistors.

Right there, hugging the middle resistor, is a $50\mu	ext{F}$ capacitor. Take a moment to visualize this setup. Our main goal is to find the exact charge stored on that capacitor.

### The Steady State Secret

Here is the crucial physics concept: we are dealing with a DC circuit in a **steady state**.

What does that mean for our capacitor? Well, once it is fully charged, it acts like a roadblock—an **open circuit**.

It completely blocks any further direct current from passing through its branch. This simplifies our circuit immensely!

### The Master Equation: Ohm's Law

Because the capacitor branch is effectively a dead end for steady current, the current from the battery has only one path to take.

It must flow straight down through all three resistors. Since the same current flows through each of them sequentially, they are in a perfect **series combination**.

Let's calculate the total equivalent resistance of this path:
$$R_{eq} = 2	ext{ k}\Omega + 2	ext{ k}\Omega + 2	ext{ k}\Omega = 6	ext{ k}\Omega$$

With our equivalent resistance in hand, finding the total current is a breeze. We just call upon our good friend, Ohm's law.

By dividing the total source voltage by our total resistance, we find the steady current:
$$I = \frac{V}{R_{eq}} = \frac{6	ext{ V}}{6	ext{ k}\Omega} = 1	ext{ mA}$$

### Finding the Voltage Drop

Now, let's focus back on the capacitor. It is connected in **parallel** with that middle resistor.

In parallel circuits, the voltage is shared equally. So, the potential difference across the capacitor is exactly the same as the voltage drop across the middle resistor.

Let's calculate that specific voltage drop using Ohm's law again:
$$V_C = I 	imes R_{middle} = 1	ext{ mA} 	imes 2	ext{ k}\Omega = 2	ext{ V}$$

### Final Calculation: The Charge

We are almost there! We have the voltage across the capacitor, and we know its capacitance.

To find the stored charge, we use the master equation of capacitors:
$$Q = C 	imes V_C$$

Let's plug in our values:
$$Q = 50\mu	ext{F} 	imes 2	ext{ V} = 100\mu	ext{C}$$

### The Polarity Puzzle

Let's think about the signs. The current is flowing downwards, from the positive terminal to the negative.

This means the potential drops as we go down. Therefore, the top node of our middle resistor is at a **higher potential** than the bottom node.

Consequently, the upper plate of the capacitor must hold the **positive charge**. Our final answer is a neat $+100\mu	ext{C}$.

Animated Solution for Physics - Current Electricity: A capacitor of $50 μ F$ is connected in a circuit as shown in figure. The charge on the upper plate of the capacitor is .............. $μ C$ .

Enter Numerical Value:

Visualized Solution

$Visual Anchor$

$Steady State Concept$

$R_{e q} Calculation$

$Circuit Current$

$Voltage across Capacitor$

$Charge Calculation$

$Polarity Check$

$The Way Forward$

The Sigma Insight: RC Circuit

Analyzing the Setup

The Steady State Secret

The Master Equation

Finding the Voltage Drop

Final Calculation

The Polarity Puzzle

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visual Anchor

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Steady State Concept

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Req​ Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Circuit Current

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Voltage across Capacitor

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Charge Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Polarity Check

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward