Question

plz solve question 14​

Answer 1

i.)heat energy supplied= VIt 

=40.51060 

=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= I2 Rt 

=0.50.55.5600 

=825 J

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