If the resistance of a circuit having 12 V source is increased by `4Omega` the current drops by 0.5 A. What is the original resistance of circuit

Options

A. `4Omega`
B. `8Omega`
C. `16Omega`
D. `(1)/(16)Omega`

Explanation

To solve the problem, we need to find the original resistance of the circuit given the voltage and the change in current when the resistance is increased.

### Step-by-Step Solution:

1. **Understanding the Problem**: We have a circuit with a voltage source of 12 V. When the resistance is increased by 4 Ω, the current decreases by 0.5 A. We need to find the original resistance \( R \) of the circuit.

2. **Using Ohm's Law**: According to Ohm's Law, the relationship between voltage (V), current (I), and resistance (R) is given by:
\[ V = I \times R \]
For our circuit, we can write:
\[ 12 = I \times R \quad \text{(Equation 1)} \]

3. **Setting Up the New Condition**: When the resistance is increased by 4 Ω, the new resistance becomes \( R + 4 \) and the new current becomes \( I - 0.5 \). Applying Ohm's Law again:
\[ 12 = (I - 0.5)(R + 4) \quad \text{(Equation 2)} \]

4. **Expanding Equation 2**:
\[ 12 = (I - 0.5)(R + 4) = I \times R + 4I - 0.5R - 2 \]
Rearranging gives:
\[ 12 = IR + 4I - 0.5R - 2 \]
Adding 2 to both sides:
\[ 14 = IR + 4I - 0.5R \quad \text{(Equation 3)} \]

5. **Substituting Equation 1 into Equation 3**: From Equation 1, we know \( IR = 12 \). Substitute \( IR \) in Equation 3:
\[ 14 = 12 + 4I - 0.5R \]
Simplifying gives:
\[ 2 = 4I - 0.5R \]
Rearranging gives:
\[ 0.5R = 4I - 2 \]
Therefore,
\[ R = 8I - 4 \quad \text{(Equation 4)} \]

6. **Substituting Equation 4 into Equation 1**: Now we substitute \( R \) from Equation 4 back into Equation 1:
\[ 12 = I(8I - 4) \]
Expanding gives:
\[ 12 = 8I^2 - 4I \]
Rearranging gives:
\[ 8I^2 - 4I - 12 = 0 \]
Dividing the entire equation by 4:
\[ 2I^2 - I - 3 = 0 \]

7. **Solving the Quadratic Equation**: We can solve this quadratic equation using the quadratic formula:
\[ I = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 2, b = -1, c = -3 \):
\[ I = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \times 2 \times (-3)}}{2 \times 2} \]
\[ I = \frac{1 \pm \sqrt{1 + 24}}{4} = \frac{1 \pm \sqrt{25}}{4} = \frac{1 \pm 5}{4} \]
This gives us two possible values for \( I \):
\[ I = \frac{6}{4} = 1.5 \quad \text{or} \quad I = \frac{-4}{4} = -1 \]
Since current cannot be negative, we take \( I = 1.5 \) A.

8. **Finding the Original Resistance**: Now substituting \( I \) back into Equation 4:
\[ R = 8(1.5) - 4 = 12 - 4 = 8 \, \Omega \]

### Final Answer:
The original resistance of the circuit is \( R = 8 \, \Omega \).