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

To do: To find the number of $176\ \Omega$ resistors that are required to carry $5\ A$ on a $220\ V$ line$(in\ parallel)$.

Solution:

Let $x$ be the numbers of the resistors of $176\ \Omega$ in parallel.

Here given, Current $I=5\ A$

Voltage $V=220\ V$

We know that, $\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}+......+\frac{1}{R_x}$

Or $\frac{1}{R}=x\times \frac{1}{176\ \Omega}$

Or $R=\frac{176\ \Omega}{x}$

We know that $\frac{V}{I}=R$

Or $\frac{220\ V}{5\ A}=\frac{176\ \Omega}{x}$

Or $x=\frac{176\ \Omega\times 5\ A}{220\ V}$

Or $x=4$

Therefore, 4 resistors of $176\ \Omega$ in parallel are required to draw the current of $5\ A$ on a $220\ V$ line.

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Updated on: 10-Oct-2022

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