The resistance of a heater coil is `110 Omega`. A resistance `R` is connected in parallel with it and the combination is joined in series with a resistance of `11 Omega` to a `220 V` main line. The heater operates with a power of `110 W`. The value of `R` in `Omega` is

Options

A. 12.22
B. 24.42
C. negative
D. that the given values are not correct

Answer

To solve the problem, we need to find the value of the resistance \( R \) that is connected in parallel with the heater coil. We will follow these steps:

### Step 1: Identify the given values
- Resistance of heater coil, \( R_{\text{coil}} = 110 \, \Omega \)
- Series resistance, \( R_s = 11 \, \Omega \)
- Total voltage, \( V = 220 \, V \)
- Power consumed by the heater, \( P = 110 \, W \)

### Step 2: Calculate the current through the heater coil
Using the power formula:
\[ P = I_1^2 \cdot R_{\text{coil}} \]
We can rearrange this to find \( I_1 \):
\[ I_1^2 = \frac{P}{R_{\text{coil}}} = \frac{110}{110} = 1 \]
Taking the square root:
\[ I_1 = 1 \, A \]

### Step 3: Determine the voltage across the heater coil
Since the heater operates at \( 1 \, A \):
\[ V_{\text{coil}} = I_1 \cdot R_{\text{coil}} = 1 \cdot 110 = 110 \, V \]

### Step 4: Calculate the voltage across the parallel combination
The total voltage is \( 220 \, V \) and the voltage across the series resistance \( R_s \) is:
\[ V_s = V - V_{\text{coil}} = 220 - 110 = 110 \, V \]

### Step 5: Find the total current in the circuit
Using Ohm's law, the total current \( I \) through the series circuit can be calculated as:
\[ I = \frac{V}{R_s + R_p} \]
Where \( R_p \) is the equivalent resistance of the parallel combination of \( R \) and \( R_{\text{coil}} \).

### Step 6: Set up the equation for the parallel resistance
The current through the parallel resistance \( R \) can be expressed as:
\[ I_2 = \frac{V_{\text{coil}}}{R} \]
And since \( I = I_1 + I_2 \):
\[ I = I_1 + I_2 = 1 + \frac{110}{R} \]

### Step 7: Substitute into the total current equation
We know that:
\[ I = \frac{220}{11 + R_p} \]
Where \( R_p \) can be calculated as:
\[ R_p = \frac{R \cdot R_{\text{coil}}}{R + R_{\text{coil}}} \]

### Step 8: Solve for \( R \)
Substituting \( R_p \) into the equation:
\[ I = 1 + \frac{110}{R} \]
And equating it to:
\[ \frac{220}{11 + \frac{R \cdot 110}{R + 110}} \]

After manipulating and simplifying the equations, we will arrive at a quadratic equation in terms of \( R \).

### Step 9: Calculate the value of \( R \)
After solving the quadratic equation, we find:
\[ R \approx 12.22 \, \Omega \]

### Conclusion
The value of \( R \) is \( 12.22 \, \Omega \).

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