Question

A bulb is connected to a battery of p.d 4V and internal resistance 2.5ohm . A steady current of 0.5A flows through the circuit calculate
1 the total energy supplied by the battery in 10 minutes
2 the resistance of the bulb
3 the energy dissipated in the bulb in 10 minutes

Answer 1

i.)heat energy supplied= VIt
=4×0.5×10×60
=1200 J
ii.)let Resistance of bulb be R.
internal resistance(r) =2.5 ohm
we know that by ohm's law,
V= IR
where V=potential difference(p.d)=4 V
I= current =0.5 A
R=Total resistance i.e. R +r
= R+2.5
so
4= 0.5(R+2.5)
on solving this equation
we get
R=5.5 ohm.
iii.) energy dissipated= I^2 Rt
=0.5×0.5×5.5×600
=825 J