Question

15. A bulb is connected to a battery of p.d. 4 V and 20.
internal resistance 2.5 S2. A steady current of 0.5 A
flows through the circuit. Calculate :
21.
(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,

Potential difference, V=4V

Internal resistance, r=2.5Ω

Steady current flowing, I=0.5 A

time t=10min=10×60=600sec

Resistance of bulb R,

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

(ii) V=I(R+r)

⇒R=

I

V

−r=

0.5

4

−2.5=5.5Ω

Resistance of bulb = 5.5Ω

(iii) Energy Dissipate in bulb in 10 minutes =I

2

Rt=0.5

2

×5.5×600=825 J

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