Question

25. A wire of length 2 m, area of cross-section 0.5 mm2 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:

from given data A=2mm

2

ρ=1.7×10

−8

Ωm.

i=1A

V=iR ( from ohms law.)

V=i×

A

ρl

E.l=i×

A

ρl

⇒E=i×

A

ρl

=1A×

2mm

2

1.7×10

−8

Ωm

=8.5×10

−3

m

V

E=8.5×10

−3

m

V

Answer 2

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.