Question

A particular resistance wire has a resistance of 3.0 ohm per metre. The potential difference of the battery which gives a current of 2.0 A in each of the 1.5 m length when connected in parallel to the battery (assume that the resistance of battery is negligible) will be

• A. 3 V
• B. 6 V
• C. 1.5 V
• D. 9 V

Answer 1

Answer: (D)

9 V

Components connected in parallel are connected so the same voltage is applied to each component. In a parallel circuit, the voltage across each of the components is the same, and the total current is the sum of the currents through each component. In parallel circuits, each light has its own circuit, so all but one light could be burned out, and the last one will still function. A circuit composed solely of components connected in parallel is known as a parallel circuit.
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, $\frac{1}{R_{total}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}thatis,R={R_{total}}^{-1}$.

In this case, the resistance of one meter wire is $3$ ohms. So the resistance of $1.5$ meters of the same wire is given as $\frac{3\times 1.5}{1}=4.5ohms$.

When three such wires of $1.5$ meters length and resistance of $4.5$ ohms are joined in parallel the resistance is given as $\frac{1}{R_{total}}=\frac{1}{4.5}+\frac{1}{4.5}+\frac{1}{4.5}=\frac{1}{0.666}=1.5ohms$.

Therefore, the total resistance is given as $1.5$ ohms.

The total current is the sum of the currents through each component, so, total current is given as $2.0+0.2+0.2=6$ amperes.
According to ohm's law, $V=IR$.

That is, $V=6A\times 1.5\Omega =9V$.