Question

Two resistors of $2.0\Omega$ and $3.0\Omega$ are connected in parallel, with a battery of $6.0V$ and negligible internal resistance. The current through the battery is :

• A. 7 A
• B. 10 A
• C. 15 A
• D. 5 A

Answer 1

The correct option is D 5 A

When two resistances are connected in parallel, then to find the total resistance of all components, the reciprocals of the resistances of each component is added and the reciprocal of the sum is taken.

Total resistance will always be less than the value of the smallest resistance

That is, $\frac{1}{R_{total}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

In this case, the combined resistance across the 2 resistors of resistances 2 ohms and 3 ohms is $\frac{1}{2}+\frac{1}{3}=\frac{1}{0.083}=1.2ohm$

The emf of the cell is given as $6V$ and the internal resistance $r$ is almost $0$.
Therefore, the current across the circuit is given as follows.
The current in the circuit is calculated from the formula $I=\frac{\epsilon }{R+r}$ where,
$\epsilon$- emf of the cell
$R$- external resistance
$r$- internal resistance of the cell.
That is, $I=\frac{6}{1.2+0}=5A$.