Question

a wire of length 2m area of cross-section 0.5mm2 and resistivity 1.5 x 10^-6 ohm meter is connected in series with a cell of emf 4 v if the circuit through the wire is o.5 A calculate INTERNAL RESISTANCE OF THE CELL AND RATE F ENERGY DISPERSED

plz send ans in pic formate

Answer 1

Given info : a wire of length 2m area of cross-section 0.5mm² and resistivity 1.5 x 10¯⁶ ohm meter is connected in series with a cell of emf 4volts if the circuit through the wire is 0.5 A

To find : the internal resistance of the cell and the rate of energy dispersed are..

solution : first find resistance of wire,

using formula, R = ρl/A

= (1.5 × 10¯⁶ ohm m) × (2m)/(0.5 × 10¯⁶ m²)

= 6 ohm

let internal resistance of the cell is r.

current through the circuit is given by, i = E/(r + R)

where E is emf of cell,

here E = 4 volts, R = 6 ohm, i = 0.5 A

so, 0.5 = 4/(r + 6)

⇒8 = r + 6

⇒r = 2 ohm

now rate of energy dispersed by the wire = power = i²R

= (0.5)² × 6

= 0.25 × 6

= 1.5 watt

Therefore internal resistance of the cell is 1.5 watt and rate of energy dispersed is 1.5 watt.

Answer 2

Answer:

using formula,

R = ρl/A

= (1.5 × 10¯⁶ ohm m) × (2m)/(0.5 × 10¯⁶ m²)

= 6 ohm

let internal resistance of the cell is r.

current through the circuit is given by, i = E/(r + R)

where E is emf of cell E = 4 volts, R = 6 ohm, i = 0.5 A

so, 0.5 = 4/(r + 6)

=>8 = r + 6

=>r = 2 ohm

now rate of energy dispersed by the wire = power = i²R

= (0.5)² × 6

= 0.25 × 6

= 1.5 watt

Therefore internal resistance of the cell is 1.5 watt and rate of energy dispersed is 1.5 watt.