Question

A bulb is connected to a battery of p.d 4 V and internal resistance 2.5 Ω. A steady current of 0.5 A flows through the circuit. Calculate:

1. the total energy supplied by the battery in 10 minutes
2. the resistance of the bulb and
3. the energy dissipated in the bulb in 10 minutes.

Answer 1

Given, V = 4V, I = 0.5 A,

t = (60 × 10) sec

1. Total energy supplied by battery W = VI t
   W = 4 × `1/2` × 600
   = 1200 J
2. Let R be the resistance of the bulb. int. Resistance = r = 2.5 Ω
   I = `E/("R" + r)`
   `1/2 = 4/("R" +2.5)`
   R + 2.5 = 8
   ∴ R = 8 - 2.5
   = 5.5 Ω
3. Energy dissipated in bulb in 10 min.
   W = I² R t
   W = `1/2 xx 1/2 xx 55/10 xx 600`
   = 825 J