Question

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

Answer 1

Hint:In order to answer this question, first we will suppose the number of resistors as a variable and then we will apply Equivalent resistance of the combination to find the resistance. Now, we have resistance, current and voltage, so we can apply Ohm’s Law to find the value of a variable which is the number of resistors, what we have to find.

Formulas used: As we have the value of Resistance, current and the supplied voltage, so we will apply Ohm’s Law formula;
$\dfrac{V}{I} = R$

Complete step by step answer:
Let the number of resistors of resistance $176\Omega $ be $x$.
Supplied Voltage, $V = 220V$ and current, $I = 5A$.
So, applying Equivalent resistance of the combination $ = R$ , given as:
$\because \dfrac{1}{R} = x \times (\dfrac{1}{{176}}) \\
\Rightarrow R = \dfrac{{176}}{x} \\ $
Now, according to the Ohm’s Law-
$\dfrac{V}{I} = R \\
\Rightarrow \dfrac{{220}}{5} = \dfrac{{176}}{x} \\
\Rightarrow x = \dfrac{{176 \times 5}}{{220}} \\
\therefore x= 4 \\ $
Therefore the required number of resistors, $x = 4$.

Hence, 4 resistors of $176\Omega $ are required to carry $5A$ on a $220V$ line.

Note: Under a weak electric field, some materials become non-ohmic. Only a conductor at a constant temperature obeys Ohm's law. Temperature affects resistivity. Ohm's law is an empirical law that holds true for the majority of experiments but not all.