Question

A wire of length 2 m, area of cross-section 0.5 mm

2

and resistivity 1.5 × 10-6 Ωm is connected in

series with a cell of emf 4 V. If the current through the wire is 0.5 A, calculate: (a) the internal

resistance of the cell and (b) the rate of energy dissipated by the wire.​

Answer 1

Answer:

a) internal resistance,r = 2ohm

b) energy dissipated by wire,P = 1.5Watt

Explanation:

resistance = (resitivity × length) ÷ Area of Cross section

resistance = 6ohm ; the the wire.

now, total resistance of the circuit,

R = V ÷ I = 4÷0.5 = 8 ohm.

now, (a) internal resistance,r = R - (resistance of wire)

= 8 - 6 = 2 ohm

now, (b) rate of energy dissipated, which is power.

Power, P = V×I

= (voltage across wire) × current

= (0.5 × 6) × 0.5 [ since, V = I × R]

= 1.5 watt.