Question

Wire of resistance 6.2 Ω is connected to a battery whose emf ε is 4.0 V and whose internal resistance is 0.38 Ω. In 2.6 min, how much energy is

(a) transferred from chemical to electrical form in the battery,
(b) dissipated as thermal energy in the wire
(c) dissipated as thermal energy in the battery?

Answer 1

a) given resistance R= 6.2Ω

Internal resistance r= 0.38Ω

Electromotive force E.m.f = 4v

Time = 2.6 × 60 = 156 seconds

E =[ (emf)^2 ÷ (R + r)] × t

=>[ (16÷ 6.58) × 156]

= 2.431 × 156

Energy transfered = 379.24J

b) current I = emf ÷ (R+r)

I= 4÷ (6.58)

I= 0.607 A

Therefore thermal energy=

I^2 × R ×t = 0.607^2 × 6.2 × 156

Thermal energy dissipated in wire

E= 356.36J

C). Thermal energy dissipated in battery

I^2 × r × t = 0.607^2 × 0.38× 156

E= 21.84J