Question

A bulb is connected to a battery of potential difference $4V$ and internal resistance $2.5\Omega $ . A steady current of $0.5A$ flows through the circuit. Calculate:
(i) the total energy supplied by the battery is 10 minutes,
(ii) the resistance of the bulb, and
(iii) the energy dissipated in the bulb in 10 minutes.

Answer 1

Hint:To solve the questions, first we will write all the given information and values related with the question and then we will write the formulae of which we have to find. Energy is measured in Joule.

Complete step by step answer:
Given values: Voltage, $v = 4V$, Resistance of the battery, ${R_B} = 2.5\Omega $, Current, $I = 0.5A$
(i) Energy supplied by the battery: We have the formula to find the energy supplied when voltage, resistance and current is given,
$E = \dfrac{{{v^2}t}}{R}$
In the above equation, $E$ is the energy supplied which we have to find.
And the time for supplying energy by the battery is 10 minutes i.e.. 600 seconds.
For finding the energy supplied, we have to calculate Resistance first:
$R = \dfrac{V}{I} = \dfrac{4}{{0.5}} = 8\Omega $
Therefore, $E = \dfrac{{{{(4)}^2} \times 600}}{8} = 1200J$

So, the total energy supplied by the battery in 10 minutes is $1200J$.

(ii) Total resistance, $R = 8\Omega $
Resistance of the battery, ${R_B} = 2.5\Omega $

So, the Resistance of the bulb, ${R_b} = 8 - 2.5\Omega = 5.5\Omega $.

(iii) Energy dissipated in the bulb in 10 minutes:
$E = {I^2}Rt$
$\Rightarrow E = {(0.5)^2} \times 5.5 \times 600 \\
\therefore E= 825J$

Hence, the energy dissipated in the bulb in 10 minutes is $825J$.

Note:We refer to this conversion of potential energy into heat as dissipation. The power dissipated in a resistor is the energy dissipated per time. Instances of dissipation Energy are generally lost by warming up the environmental factors however in some cases energy is dispersed as sound. The manners by which energy is disseminated relies upon the framework: for a radio or set of speakers, the electrical work is moved into valuable sound and infrared radiation is dispersed – i.e. squandered as warmth energy.