Question

A bulb is joined to a battery of emf 4 V and internal resistance of 2.5 Ω.A steady current of 0.5 A flows through circuit. calculate-

(a) Total energy provided by battery in 10 minutes

Answer 1

Step 1- Given Data

Current $I=0.5A$

Voltage (EMF) $V=4V$

Internal resistance $r=2.5\Omega$

Time $t=10minutes=10\times 60=600seconds.$

Step 2- To find out the resistance offered by the circuit-

As per ohm's law-

$R=\frac{V}{I}$

$\Rightarrow R=\frac{4}{0.5}=8\Omega$

Thus, the resistance of the circuit is $8\Omega$

Since energy provided by the battery is asked, the internal resistance is not taken into consideration.

Step 3- To find out the energy provided by the battery-

$H=\frac{V^2}{R}t$

$$\begin{gathered} \Rightarrow H=\frac{4^2}{8}\left(600\right) \\ \Rightarrow H=1200J \end{gathered}$$

Thus, the energy provided by the battery is $1200joules.$