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

(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.

(i) Given,

Voltage (V) = 4 V

Internal resistance = 2.5 Ω

Current (I) = 0.5 A

time (t) = 10 min = 600 sec

Energy supplied by the battery (E) = V x I x t

Substituting the values in the formula, we get,

E = 4 x 0.5 x 600 = 1200 J

Hence, total energy supplied = 1200 J

(ii) resistance of the bulb = ?

From relation V = I (R + r)

We get,

4 = 0.5 (R + 2.5)
⇒ 4 = (0.5 R) + (0.5 x 2.5)
⇒ 4 = (0.5 R) + 1.25
⇒ 0.5 R = 4 - 1.25
⇒ R = 2.75 / 0.5 = 5.5 Ω

Hence, resistance of the bulb = 5.5 Ω

(iii) Energy dissipated in the bulb in 10 min = ?

From relation

E = I²Rt

Substituting the values in the formula above, we get,

E = 0.5 x 0.5 x 5.5 x 600 = 825 J

Hence, energy dissipated = 825 J