Question

A bulbs is connected to a battery of p.d. 4 V and internal resistance $2.5\Omega$. A stedy current of 0.5 A
(i) the total energy supplied by the battery in 10 minutes,
(ii) the resistance of the bulb,and
(iii) the energy displated in the bulb in 10 minutes

Answer 1

Given,

Potential difference, $V=4V$

Internal resistance, $r=2.5\Omega$

Steady current flowing, $I=0.5A$

time $t=10min=10\times 60=600\sec$

Resistance of bulb $R$,

(i)Total energy supplied in 10 minutes $=I\times V\times t=0.5\times 4\times 600=1.2kJ$

(ii) $V=I(R+r)$

$\Rightarrow R=\frac{V}{I}-r=\frac{4}{0.5}-2.5=5.5\Omega$

Resistance of bulb $=5.5\Omega$

(iii) Energy Dissipate in bulb in 10 minutes $=I^{2}Rt={0.5}^2\times 5.5\times 600=825J$