The current in a simple series circuit is 5.0 amp . When an additional resistance of 2.0 ohms is inserted, the current drops to 4.0 amp . The original resistance of the circuit in ohms was

Options

A. 1.25
B. 8
C. 10
D. 20

Answer

To find the original resistance of the circuit, we can use Ohm's law and the relationship between current, voltage, and resistance in series circuits.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial current (I1) = 5.0 A
- Current after adding resistance (I2) = 4.0 A
- Additional resistance (R_add) = 2.0 Ω

2. **Use Ohm's Law:**
- Ohm's law states that V = I × R, where V is voltage, I is current, and R is resistance.
- For the initial circuit, we have:
\[ V = I_1 \times R_1 \]
- For the circuit after adding the resistance, we have:
\[ V = I_2 \times R_2 \]
- Where \( R_2 = R_1 + R_{add} \) (since resistances in series add up).

3. **Set Up the Equations:**
- From the first case:
\[ V = 5.0 \times R_1 \quad (1) \]
- From the second case:
\[ V = 4.0 \times (R_1 + 2.0) \quad (2) \]

4. **Equate the Two Expressions for Voltage:**
- Set equations (1) and (2) equal to each other:
\[ 5.0 \times R_1 = 4.0 \times (R_1 + 2.0) \]

5. **Expand and Rearrange the Equation:**
- Expanding the right side:
\[ 5.0 R_1 = 4.0 R_1 + 8.0 \]
- Rearranging gives:
\[ 5.0 R_1 - 4.0 R_1 = 8.0 \]
\[ R_1 = 8.0 \, \text{Ω} \]

6. **Conclusion:**
- The original resistance of the circuit, \( R_1 \), is 8.0 Ω.