In the circuit shown in the figure, in steady state:
The current drawn from the battery is

Options

A. `2 A`
B. `3 A`
C. `4 A`
D. `1 A`

Answer

To find the current drawn from the battery in the given circuit in steady state, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Circuit Configuration**:
- In steady state, capacitors behave as open circuits. Therefore, no current flows through the capacitors.

2. **Identify the Resistors in the Circuit**:
- The resistors in the circuit are connected in series. Let's denote their resistances as follows:
- R1 = 4 ohms
- R2 = 2 ohms
- R3 = 2 ohms
- R4 = 2 ohms

3. **Calculate the Equivalent Resistance (R_eq)**:
- Since the resistors are in series, the equivalent resistance can be calculated by simply summing up all the resistances:
\[ R_{eq} = R1 + R2 + R3 + R4 = 4 + 2 + 2 + 2 = 10 \text{ ohms} \]

4. **Apply Ohm's Law**:
- Ohm's Law states that \( V = I \times R \), where:
- \( V \) is the voltage across the circuit,
- \( I \) is the current flowing through the circuit,
- \( R \) is the equivalent resistance.
- Rearranging the formula to find the current gives us:
\[ I = \frac{V}{R_{eq}} \]

5. **Substitute the Values**:
- Assuming the voltage \( V \) from the battery is 10 volts (as inferred from the context), we substitute the values into the equation:
\[ I = \frac{10 \text{ volts}}{10 \text{ ohms}} = 1 \text{ ampere} \]

6. **Conclusion**:
- The current drawn from the battery in steady state is 1 ampere.

### Final Answer:
The current drawn from the battery is **1 ampere**.

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