Two resistance wires of the same material and of equal lengths and equal diameters are connected in series in a simple electric circuit consisting of a voltage source of voltage V. Now, the same wires are connected in parallel in the same circuit for the same time but voltage source is replaced by another voltage source of voltage 2V. The ratio of heat produced in the two cases is

Options

A. `1:1`
B. `1:4`
C. `1:8`
D. `1:16`

Answer

To solve the problem, we need to analyze the heat produced in both cases: when the two resistance wires are connected in series and when they are connected in parallel.

### Step-by-Step Solution:

1. **Identify the Resistance of Each Wire:**
Since both wires are made of the same material, have equal lengths, and equal diameters, their resistances (R) will be equal. The resistance of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area.

2. **Case 1: Wires in Series:**
When the two wires are connected in series, the total resistance \( R_{eq1} \) is:
\[ R_{eq1} = R + R = 2R \]
The heat produced \( H_1 \) in a resistor is given by:
\[ H = \frac{V^2}{R} \cdot t \]
For the series connection, the heat produced \( H_1 \) is:
\[ H_1 = \frac{V^2}{R_{eq1}} \cdot t = \frac{V^2}{2R} \cdot t \]

3. **Case 2: Wires in Parallel:**
When the two wires are connected in parallel, the total resistance \( R_{eq2} \) is:
\[ \frac{1}{R_{eq2}} = \frac{1}{R} + \frac{1}{R} \Rightarrow R_{eq2} = \frac{R}{2} \]
The new voltage source is \( 2V \). The heat produced \( H_2 \) in this case is:
\[ H_2 = \frac{(2V)^2}{R_{eq2}} \cdot t = \frac{4V^2}{R/2} \cdot t = \frac{8V^2}{R} \cdot t \]

4. **Calculate the Ratio of Heat Produced:**
Now we need to find the ratio \( \frac{H_1}{H_2} \):
\[ \frac{H_1}{H_2} = \frac{\frac{V^2}{2R} \cdot t}{\frac{8V^2}{R} \cdot t} \]
The \( t \) and \( V^2 \) cancel out:
\[ \frac{H_1}{H_2} = \frac{1}{2} \cdot \frac{R}{8} = \frac{1}{16} \]

5. **Final Result:**
Thus, the ratio of heat produced in the two cases is:
\[ H_1 : H_2 = 1 : 16 \]