Question

A bulb is connected to 4 v and the internal resistance of circuit is 2.5 ohm. And 0.5 A flow through it. Find the energy dissipated in the bulb in 10 minutes and the resistance of the bulb

Answer 1

i.)heat energy supplied= VIt
the= 10×60 = 600 s
=4×0.5×600
=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)
4/.5 = R+2.5
8=R+2.5
R= 5.5 ohm.


iii.) energy dissipated= I2 Rt
=0.5²×5.5×600
=825 J