Question

A cell has emf of 4 volt and internal resistance 0.6. The maximum power which it can deliver to any external resistor is

Answer 1

Answer:

Explanation:  9.6 watt of power

Answer 2

Given:

Emf = 4 V

Internal resistance r = 0.6 ohm.

To Find:

Maximum power

Solution:

Let the cell having emf E and internal resistance r be connected with the series resistor R.

Therefore,

i = E / (r + R)

Output power,  

P= (i^2)R = (E^2)R / (r+R)

Output power is maximum when r = R.

Pmax = (E^2. r) / (r + r)^2

Pmax = (E^2. r) / 4r^2

Pmax = (E^2) /4r

Pmax = (4×4) / (4× 0.6)

Pmax = 4 / 0.6 = 6.67 W.

The maximum power which it can deliver to any external resistor is 6.67 W.