Question

an electric generator has an emf of 240V and an internal resistance of 1 ohm. if the current supplied by the generator is 20A when the terminal volatge is 220V. what would be the ratio of the power supplied to the external load to the power dissipated in the generator's internal resistance

Answer 1

Answer:

11:1

Explanation:

P=I^2/R

P=20×20/1=400......eq i

P=VI=220×20=4400W

P1/P2=4400/400

11:1

Answer 2

Answer:

The ratio of the power supplied to the external load to the power dissipated in the generator's internal resistance is 11:1.

Explanation:

The power dissipated in the generator's internal resistance is given as,

$P=I^2R$     (1)

Where,

P=power dissipated in the generator's internal resistance

I=current through the resistance

R=internal resistance

From the question we have,

EMF(V)=240V

Internal resistance(R)=1Ω

Current supplied by the generator(I)=20 A

Terminal voltage(V')=220V

By substituting the value of I and R in equation (1) we get;

$P=(20)^2\times 1=400 W$      (2)

Power supplied to the external load is given as,

$P'=V'\times I$           (3)

By substituting the values in equation (3) we get;

$P'=220\times 20=4400W$    (4)

The ratio of equations (4) and (2) is given as,

$\frac{P'}{P}=\frac{4400}{400}$

$\frac{P'}{P}=\frac{11}{1}$

Hence, the ratio of the power supplied to the external load to the power dissipated in the generator's internal resistance is 11:1.