In the circuit shown in the figure, in steady state:
The charge on `5 muF` capacitor is

Options

A. `30 mu C`
B. `40 mu C`
C. `50 mu C`
D. `60 mu C`

Answer

To find the charge on the 5 µF capacitor in the given circuit at steady state, we can follow these steps:

### Step 1: Understand the Steady State Condition
In a steady state, the current flowing through the branches containing capacitors is zero. This means that the capacitors are fully charged and there is no further current flowing through them.

**Hint:** Remember that capacitors block DC current in steady state.

### Step 2: Identify the Circuit Elements
We need to identify the resistances in the circuit and how they are connected. For this problem, we assume there are resistors connected in series.

**Hint:** Look for resistors that are connected in series and add their resistances.

### Step 3: Calculate the Equivalent Resistance
If the resistors are connected in series, the total or equivalent resistance (R_eq) can be calculated by simply adding the individual resistances together. For example, if we have resistors of 4Ω, 2Ω, 2Ω, and 2Ω, the equivalent resistance is:

\[ R_{eq} = 4 + 2 + 2 + 2 = 10 \, \Omega \]

**Hint:** Use the formula for resistors in series: \( R_{eq} = R_1 + R_2 + R_3 + ... \)

### Step 4: Calculate the Current in the Circuit
Using Ohm's Law (V = IR), we can find the current (I) flowing through the circuit. If we assume a total voltage (V_total) is applied across the circuit, we can rearrange the formula to find I:

\[ I = \frac{V_{total}}{R_{eq}} \]

Assuming a total voltage of 6V, we have:

\[ I = \frac{6V}{10Ω} = 0.6 \, A \]

**Hint:** Make sure to use the total voltage provided in the circuit.

### Step 5: Determine the Voltage Across the Capacitor
To find the voltage (V) across the 5 µF capacitor, we need to consider the voltage drop across the resistors in series. If the voltage drop across the 4Ω resistor is:

\[ V_1 = I \times R_1 = 0.6 \, A \times 4 \, Ω = 2.4 \, V \]

And for the 2Ω resistor:

\[ V_2 = I \times R_2 = 0.6 \, A \times 2 \, Ω = 1.2 \, V \]

The total voltage across the capacitor is the sum of the voltage drops:

\[ V = V_1 + V_2 = 2.4V + 1.2V = 3.6V \]

**Hint:** Remember to sum the voltage drops across the resistors leading to the capacitor.

### Step 6: Calculate the Charge on the Capacitor
Now that we have the voltage across the capacitor, we can use the formula for charge (Q) on a capacitor:

\[ Q = C \times V \]

Where:
- C = 5 µF = \( 5 \times 10^{-6} \, F \)
- V = 3.6 V

Substituting the values:

\[ Q = 5 \times 10^{-6} \, F \times 3.6 \, V = 18 \times 10^{-6} \, C = 18 \, µC \]

**Hint:** Use the formula \( Q = C \times V \) and ensure the units are consistent.

### Final Answer
The charge on the 5 µF capacitor is **18 µC**.