Question

An external resistor with resistance R is connected to a battery that has an emf E and an internal resistance r. Let P be the electrical power output of the source. By conservation of energy, P is equal to the power consumed by R.
(a) What is the value of P in the limit that R is very small?

(b) What is the value of P in the limit that R is very large?

Answer 1

Answer:

a) When R is very small R << r, therefore the term R+ r will equal r and the current becomes  

b) When R is very large, R >> r, therefore the term R+ r will equal R and the current becomes

Explanation:

Solution  :

(a) We want to get the consumed power P when R is very small. The resistor in the circuit consumed the power from this battery. In this case, the current I is leaving the source at the higher-potential terminal and the energy is being delivered to the external circuit where the rate (power) of this transfer is given by equation  in the next form  

P=∈*I-I^2*r                (1)

Where the term ∈*I is the rate at which work is done by the battery and the term I^2*r is the rate at which electrical energy is dissipated in the internal resistance of the battery. The current in the circuit depends on the internal resistance r and we can apply equation to get the current by  

I=∈/R+r                     (2)

When R is very small R << r, therefore the term R+ r will equal r and the current becomes  

I= ∈/r

Now let us plug this expression of I into equation (1) to get the consumed power  

P=∈*I-I^2*r

 =I(∈-I*r)

 =0

The consumed power when R is very small is zero  

(b) When R is very large, R >> r, therefore the term R+ r will equal R and the current becomes  

I=∈/R

The dissipated power due toll could be calculated by using equation.

P=I^2*r                (3)

Now let us plug the expression of I into equation (3) to get P  

P=I^2*R=(∈/R)^2*R

 =∈^2/R