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

Question

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Answer 1

$Step 1 : Equivalent resistance$

Let $n$ be the number of resistors to be connected in parallel
Then, $R_{1} = R_{2} = . . . . . . = R_{n} = R = 176 Ω$
Now, $\frac{1}{R _{e q}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + . . . . . \frac{1}{R _{n}}$
$\Rightarrow \frac{1}{R _{e q}} = \frac{n}{R _{1}} = \frac{n}{1 7 6}$
$\Rightarrow R_{e q} = 176 / n$ $. . . . (1)$

$Step 2 : Applying Ohm’s law$
$V = I R_{e q}$
$\Rightarrow 220 = 5 R_{e q}$
$\Rightarrow 220 = 5 \times 176 / n$ $($ Using Equation $(1))$
$\Rightarrow n = 4$

Hence, four resistors are required.

Answer 2

Step 1 : Equivalent resistance

Let

n

be the number of resistors to be connected in parallel

Then,

R_{1} = R_{2} = . . . . . . = R_{n} = R = 176 Ω

Now,

\frac{1}{R _{e q}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + . . . . . \frac{1}{R _{n}}

\Rightarrow \frac{1}{R _{e q}} = \frac{n}{R _{1}} = \frac{n}{1 7 6}

\Rightarrow R_{e q} = 176 / n

. . . . (1)

Step 2 : Applying Ohm’s law

V = I R_{e q}

\Rightarrow 220 = 5 R_{e q}

\Rightarrow 220 = 5 \times 176 / n

(

Using Equation

(1))

\Rightarrow n = 4

Hence, four resistors are required.