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:
<u>Solution :</u>
(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
