Question

Three resistors of resistances $3\Omega ,4\Omega$ and $5\Omega$ are combined in parallel. This combination is connected to a battery of emf $12$V and negligible internal resistance, current through each resistor in ampere is?

• A. $4,3,2.4$
• B. $8,7,3.4$
• C. $2,5,1.8$
• D. $5,5,8.2$

Answer 1

The correct option is A $4,3,2.4$

Given three resistances

$R_1=3\Omega$

$R_2=4\Omega$

$R_3=5\Omega$

Let $I_1,I_2$ and $I_3$ be the current through resistor 1,2 and 3

In parallel combination, potenial difference will be equal to the em.f of the battery and will be same.

E.m.f of battery, $V=12V$

Hence,

$I_1=\frac{V}{R_1}=\frac{12}{3}=4A$

$I_2=\frac{V}{R_2}=\frac{12}{4}=3A$

$I_3=\frac{V}{R_3}=\frac{12}{5}=2.4A$

Hence, option $\left(A\right)$ is correct.