Question

A bulb is connected to a battery of potential difference 4 V and internal resistance 2.5 $\Omega$. A steady current of 0.5 A flows through the circuit. The resistance of the bulb is :

• A. $5\Omega$
• B. $5.5\Omega$
• C. $2.5\Omega$
• D. $10\Omega$

Answer 1

Answer: (B)

$5.5\Omega$

Energy ($power\times time$) is measured in Joules and by including time (t) in the power formulae, the energy dissipated by a component or circuit can be calculated.
Energy dissipated $=Pt$ or $VI\times t$ as $P=VI$.
In the question, it is given that the current $I$ is $0.5A$ and voltage $V$ is $4V$. Thus the power dissipated is calculated as $P=VI=4V\times 0.5A=2W$.

Substituting, $V=IR$ (Ohm's law) in the above formula, we get, $P=I^2\times R$.
Therefore, the total resistance in the circuit is calculated using the relation: $R=\frac{P}{I^2}=\frac{2}{{0.5}^2}=8ohms$.

The internal resistance of the cell is $2.5ohms$, hence the resistance of the bulb is $8-2.5=5.5ohms$.