In an electric circuit shown in the figure. The external resistance `R` is variable. It is found that same power is developed `R` is variable. Its is found that same power is developed in external resistance `R` if its value is either `5 Omega` or `20 Omega`. within this information find the internal resistance `(r)` of the voltage source.

Options

A. `20 Omega`
B. `8 Omega`
C. `0.3 Omega`
D. `10 Omega`

Answer

D

Explanation

(d) `I = (E)/(R + r), P = I^(2) = (E^(2) R)/((R + r)^(2))`
Now `P_(1) = P_(2) implies (E^(2) 5)/((5 + r)^(2)) = (E^(2) 20)/((20 + r)^(2)) implies r = 10 Omega``