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 energy dissipated in the bulb in 10 minutes is :

• A. $125J$
• B. $825J$
• C. $120J$
• D. $240J$

Answer 1

The correct option is B $825J$

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 $VI\times tasP=VI$.

In the question, it is given that the current I is 0.5 A and voltage V is 4 V. 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 2.5 ohms, hence, the resistance of the bulb is $8-2.5=5.5$ ohms.

The power dissipated by the bulb in 10 minutes is calculated from the formula $P=I^2\times R\times t$, where $I=0.5A,R=5.5$ ohms and t=10 minutes, that is 600 seconds.

Hence, $P={0.5}^2\times 5.5\times 600=825J$.

The power dissipated by the bulb in 10 minutes is 825 Joules.