Question

A current I=2 amp flows through the cell of emf €=10 V . If the value of internal resistance is r=1 ohm and the external resistance is R=5 ohm then find (a) the power delivered by the battery (b) the power dissipated by the battery (c) the terminal voltage

Answer 1

Explanation ⇒ Using the formula,

Terminal Voltage = E.M.F. - Voltage drop.

∴ V = ε - ir

∴ V = 10 - 2 × 1

∴ V = 8 V.

(a). Power delivered by the Battery = ε/R

= 10/5

= 2 W.

(b). Power dissipated by the Battery = V/R

= 8/5

= 1.6 W.

(c). Terminal voltage is already calculated as 8 V.

Hope it helps.

Answer 2

Terminal Voltage = E.M.F. - Voltage drop.

∴ V = ε - ir

∴ V = 10 - 2 × 1

∴ V = 8 V.

(a). Power delivered by the Battery = ε/R

= 10/5

= 2 W.

(b). Power dissipated by the Battery = V/R

= 8/5

= 1.6 W.

(c). Terminal voltage is already calculated as 8 V.