Question

A bulb is connected to a battery of p.d. 4V and internal resistance 2.5ohm. A steady current f 0.5A flows through the circuit. Calculate the : 1.total energy supplied by thr battery in 10 mins
2. Resistance of the bulb

Answer 1

Dear Student ,
Here in this question ,the potential difference of the battery 4 V and steady current is 0.5 A .
Now the resistance of the bulb is , R = $\frac{V}{I}=\frac{4}{0·5}=8\Omega$
Now the total resistance of the circuit is = resistance of the bulb + internal resistance = 8 + 2.5 = 10.5 ohm
So the energy supplied by the battery in 10 mins = E = $\frac{V^2}{R}t=\frac{{\left(4\right)}^2}{10·5}\times 10\times 60=914·28J$
Regards