Question

An electric bulb is connected to a 15V battery of negligible resistance. The resistance offered by the bulb is 5\[\Omega \]. Find the power of the bulb.
A) 3W
B) 15W
C) 75W
D) 45W

Answer 1

Hint: We need to understand the relation between the potential applied in a circuit and the resistance in the circuit with the power dissipated by the resistor in the circuit. We can easily find the solution for this problem using this relation between the parameters.

Complete step-by-step solution
We are given an electric circuit in which a bulb of resistance 5\[\Omega \] is connected to an external potential difference of 15 V. We know that the heat dissipated from a resistance is given by Joule’s law of heating. According to the law, the heat dissipated by the resistor R through which a current I pass due to the external potential V is proportional to the potential V, the current through the circuit, and the time duration ‘t’.
It is given as –
\[H=VIt\]
The power dissipation in the same circuit is given as the rate of change of the heat energy dissipated. So, it can be given as –
\[P=VI\]
Now, we can use the Ohm’s law to convert the current in to the terms of resistance and voltage drop as –
\[\begin{align}
  & V=IR \\
 & \therefore I=\dfrac{V}{R} \\
\end{align}\]
Now, the power dissipated in the circuit will become as –
\[\begin{align}
  & P=VI \\
 & \therefore P=\dfrac{{{V}^{2}}}{R} \\
\end{align}\]
The power dissipated in the circuit with the given value of resistance and voltage can be given as –
\[\begin{align}
  & P=\dfrac{{{V}^{2}}}{R} \\
 & \Rightarrow P=\dfrac{{{15}^{2}}}{5} \\
 & \therefore P=45W \\
\end{align}\]
The power dissipated in the circuit is 45W. The correct answer is option D.

Note: The power dissipated is the rate of energy expenditure in the circuit due to the involved resistance and the voltage applied in the circuit. The power rating for a device is a constant given that the device operates in the voltage as per the specifications.