Question

A wire of length 2m area of cross section 0.5m msquare and resistivity 1.5 10to the power of minus 6 ohm m is connected in series with a cell of emf 4v if the current through the wire in 0.5A calculate a)the internal resistance of the cell and b)the rate of energy dissipated by the wire

Answer 1

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.