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

AcademicPhysicsNCERTClass 10

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.

raja
Updated on 10-Oct-2022 13:20:12

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