How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

Sure, let's solve this step by step.

Step 1: First, we need to find the total resistance required to carry 5A on a 220V line. We can use Ohm's law for this, which states that V = IR, where V is voltage, I is current, and R is resistance. Rearranging for R, we get R = V/I.

Step 2: Substituting the given values into the formula, we get R = 220V / 5A = 44 Ω. This is the total resistance required.

Step 3: Now, we need to find out how many 176 Ω resistors in parallel would give us this total resistance. The formula for total resistance (Rt) of resistors in parallel is 1/Rt = 1/R1 + 1/R2 + ... + 1/Rn. In this case, all resistors are the same (176 Ω), so the formula simplifies to 1/Rt = n/R, where n is the number of resistors.

Step 4: Rearranging for n, we get n = R/Rt.

Step 5: Substituting the values into the formula, we get n = 176 Ω / 44 Ω = 4.

So, 4 resistors of 176 Ω each, connected in parallel, are required to carry 5A on a 220V line.