How many resistors of `88Omega` are connected in parallel to carry 10 A current on a 220 V line ?

Options

A. 2 resistors
B. 1 resistors
C. 3 resistors
D. 4 resistors

Explanation

To solve the problem of how many resistors of 88 ohms are connected in parallel to carry a current of 10 A on a 220 V line, we can follow these steps:

### Step 1: Calculate the total resistance required for the circuit.
Using Ohm's Law, we know that:
\[ V = I \times R \]
Where:
- \( V \) = Voltage (220 V)
- \( I \) = Current (10 A)
- \( R \) = Total resistance

Rearranging the formula to find \( R \):
\[ R = \frac{V}{I} \]
Substituting the values:
\[ R = \frac{220 \, \text{V}}{10 \, \text{A}} = 22 \, \Omega \]

### Step 2: Use the formula for total resistance in parallel.
When resistors are connected in parallel, the total resistance \( R_t \) can be calculated using the formula:
\[ \frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots \]
For \( n \) identical resistors of resistance \( R \):
\[ \frac{1}{R_t} = \frac{n}{R} \]
Thus:
\[ R_t = \frac{R}{n} \]

### Step 3: Set the total resistance equal to the required resistance.
We already calculated \( R_t = 22 \, \Omega \) and each resistor has a resistance of \( 88 \, \Omega \):
\[ 22 = \frac{88}{n} \]

### Step 4: Solve for \( n \).
Rearranging the equation gives:
\[ n = \frac{88}{22} \]
Calculating this:
\[ n = 4 \]

### Conclusion:
Therefore, **4 resistors of 88 ohms** are required to carry a current of 10 A on a 220 V line.

---