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

Options

A. 8 resistors
B. 5 resistors
C. 10 resistors
D. 4 resistors

Explanation

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

### Step 1: Understand the relationship between voltage, current, and resistance
We know from Ohm's Law that:
\[ V = I \times R \]
Where:
- \( V \) is the voltage across the circuit (220 V)
- \( I \) is the total current (10 A)
- \( R \) is the equivalent resistance of the circuit

### Step 2: Calculate the equivalent resistance
Rearranging Ohm's Law to find the equivalent resistance:
\[ R = \frac{V}{I} \]
Substituting the known values:
\[ R = \frac{220 \, \text{V}}{10 \, \text{A}} = 22 \, \Omega \]

### Step 3: Determine the equivalent resistance of resistors in parallel
When resistors are connected in parallel, the equivalent resistance \( R_{eq} \) can be calculated using the formula:
\[ R_{eq} = \frac{R}{N} \]
Where:
- \( R \) is the resistance of each resistor (88 Ohms)
- \( N \) is the number of resistors connected in parallel

### Step 4: Set up the equation
We already calculated \( R_{eq} \) to be 22 Ohms. Now we can set up the equation:
\[ \frac{88}{N} = 22 \]

### Step 5: Solve for \( N \)
To find \( N \), we can rearrange the equation:
\[ N = \frac{88}{22} \]
Calculating this gives:
\[ N = 4 \]

### Conclusion
Thus, we need **4 resistors of 88 Ohms** connected in parallel to carry a total current of 10 A on a 220 V line.

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