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

To solve this problem, we first need to understand the relationship between voltage, current, and resistance, which is given by Ohm's Law: V = I * R, where V is voltage, I is current, and R is resistance.

Step 1: Calculate the total resistance needed to carry 10 A current on a 220 V line.

Using Ohm's law, we rearrange the formula to find R = V / I. Substituting the given values, we get R = 220 V / 10 A = 22 Ω. This is the total resistance needed for the circuit.

Step 2: Determine the number of 88 Ω resistors in parallel needed to achieve this total resistance.

When resistors are connected in parallel, the total resistance (Rt) of the circuit is given by the formula 1/Rt = 1/R1 + 1/R2 + 1/R3 + ... + 1/Rn. In this case, all resistors have the same resistance (88 Ω), so the formula simplifies to 1/Rt = n/R, where n is the number of resistors.

Rearranging the formula to solve for n, we get n = R / Rt. Substituting the given values, we get n = 88 Ω / 22 Ω = 4.

So, 4 resistors of 88 Ω each are needed to be connected in parallel to carry a 10 A current on a 220 V line.