two heading coils of resistances `10Omega and 20 Omega` are connected in parallel and connected to a battery of emf 12 V and internal resistance `1Omega` Thele power consumed by the n are in the ratio

Options

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

Answer

To solve the problem, we need to find the power consumed by two heating coils with resistances of 10Ω and 20Ω connected in parallel to a battery with an EMF of 12V and an internal resistance of 1Ω. We will then determine the ratio of the power consumed by each coil.

### Step-by-Step Solution:

1. **Calculate the total resistance of the parallel combination of the coils:**
The formula for the equivalent resistance \( R_{eq} \) of two resistors \( R_1 \) and \( R_2 \) in parallel is given by:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \]
Here, \( R_1 = 10Ω \) and \( R_2 = 20Ω \).
\[ \frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{20} = \frac{2}{20} + \frac{1}{20} = \frac{3}{20} \]
Therefore,
\[ R_{eq} = \frac{20}{3}Ω \approx 6.67Ω \]

2. **Calculate the total resistance in the circuit including the internal resistance of the battery:**
The total resistance \( R_{total} \) is the sum of the equivalent resistance of the coils and the internal resistance of the battery:
\[ R_{total} = R_{eq} + R_{internal} = \frac{20}{3} + 1 = \frac{20}{3} + \frac{3}{3} = \frac{23}{3}Ω \approx 7.67Ω \]

3. **Calculate the total current flowing through the circuit using Ohm's Law:**
The total current \( I \) can be calculated using the formula:
\[ I = \frac{V}{R_{total}} = \frac{12V}{\frac{23}{3}Ω} = \frac{12 \times 3}{23} = \frac{36}{23}A \]

4. **Determine the voltage across each coil:**
The voltage drop across the internal resistance can be calculated as:
\[ V_{internal} = I \times R_{internal} = \frac{36}{23} \times 1 = \frac{36}{23}V \]
The voltage across the parallel combination of the coils is:
\[ V_{parallel} = V - V_{internal} = 12V - \frac{36}{23}V = \frac{276 - 36}{23} = \frac{240}{23}V \]

5. **Calculate the power consumed by each coil:**
The power consumed by each coil can be calculated using the formula:
\[ P = \frac{V^2}{R} \]
- For the 10Ω coil:
\[ P_1 = \frac{V_{parallel}^2}{10} = \frac{\left(\frac{240}{23}\right)^2}{10} = \frac{57600}{5290} = \frac{57600}{5290}W \]
- For the 20Ω coil:
\[ P_2 = \frac{V_{parallel}^2}{20} = \frac{\left(\frac{240}{23}\right)^2}{20} = \frac{57600}{10580} = \frac{57600}{10580}W \]

6. **Calculate the ratio of power consumed:**
The ratio of power consumed by the two coils is:
\[ \frac{P_1}{P_2} = \frac{\frac{57600}{10}}{\frac{57600}{20}} = \frac{20}{10} = 2:1 \]

### Final Answer:
The ratio of power consumed by the two heating coils is \( 2:1 \).