Question

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

(b) Heat dissipated by the bulb 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 $8\Omega$

Thus, the net resistance of the bulb $R_B=R-r=8-2.5=5.5\Omega$

Step 3- To find out the Heat dissipated by the bulb -

$H=I^2{R}_{B}t$

$$\begin{gathered} \Rightarrow H={\left(0.5\right)}^2\times 5.5\times 600 \\ \Rightarrow H=820J \end{gathered}$$

Thus, the energy provided by the battery is $820J$