Question

How many bulbs of resistance $8\Omega$ should be joined in parallel to draw a current of $3\text{A}$ from a battery of $3\text{V}?$

• A. $4$
• B. $8$
• C. $6$
• D. $9$

Answer 1

The correct option is B $8$

Equivalent resistance of the circuit,

$R=\frac{V}{i}$

$\Rightarrow R=\frac{3}{3}=1\Omega$

Let $n$ bulbs be joined in parallel to achieve this resistance.

Then,

$\frac{1}{R}=\frac{1}{r_1}+\frac{1}{r_2}+.....+\frac{1}{r_n}$

$\Rightarrow \frac{1}{1}=\frac{1}{8}+\frac{1}{8}+.....+\frac{1}{8}=\frac{n}{8}$

$\Rightarrow 1=\frac{n}{8}$

$\therefore n=8$