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 in $10$ minutes,
(ii) the resistance of the bulb, and
(iii) the energy dissipated in the bulb in $10$ minutes.

Answer 1

Given,
Voltage,v=4v
Resistance of the battery=$R_B=2.5\Omega$
Current, $I=0.5A$
i) Energy supplied by the battery, $E=\frac{v^{2}t}{R}$
$T=10\times 60=600sec$
$R=\frac{V}{I}=\frac{4}{0.5}=8\Omega$
$E=\frac{(4{)}^2\times 600}{8}=1200J$
ii) Total resistance, $R=8\Omega$
Resistance of the battery,$R_B=2.5\Omega$
Resistance of the bulb , $R_b=8-2.5\Omega =5.5\Omega$
iii) Energy dissipated in the bulb in 10 min , $E=I^{2}Rt$
$E=(0.5{)}^2\times 5.5\times 600=825J$