Question

A battery has an open circuit potential difference of 6V between its terminals. When a load resistance of 60 Omega is connected across the battery,the total power supplied by the bttery is 0.4 W. What should be the load resistance R, so that maximum power will be dissipated in R. Calculated this power. What is the total power supplied by the battery when such a load is connected?

Answer 1

Thus the total power supplied by the battery is 0.3 W

Explanation:

When the circuit is open V = E

E = 6 V

Let r be the internal resistance of the battery

Power supplied by the battery in this case is

P = E^2 / R+r

Substituting the value, we have 0.4= (6)^2 / 60+r

Solving this we get r = 30Ω

Maximum power is dissipated in the circuit when net external resistance is equal to net internal resistance or

R = r

R = 30 Ω

further, total power supplied by the battery under this condition is

P(total) = E^2 R + r = (6)^2 / 30+30  = 0.6 W

Of this 0.6 W half of the power is dissipated in R and half in r.

Therefore, maximum power dissipated in R would be

0.62 = 0.3 W

Thus the total power supplied by the battery is 0.3 W