A battery of 3.0 V is connected to a resistor dissipating 0.5 W of power. If the terminal voltage of the battery is 2.5 V , the power dissipated within the internal resistance is:

Step-by-step solution

Given, Emf of the battery, E = 3   V Potential difference across resister R is V R = 2 . 5   V Power dissipation in the resister is P = 0 . 5   W By using Kirchhoff's voltage law is given by, V r + V R = E     . . . ( 1 ) V r is voltage across internal resister. Using equation (1) and substitute the given values, ⇒ V r + 2 .5 = 3 ⇒ V r = 0 .5 Now the ratio of potential across R and r is given by, V R V r = I R I r = 2 .5 0 .5 = 5 ⇒ R r = 5 Now the power dissipation acr