Question

A battery of emf ${}^{\prime }{E}^{\prime }$ and internal resistance ${}^{\prime }{r}^{\prime }$ is connected to a resistor of resistance '$r_1$' and $Q$ joules of heat is produced in a certain time ${}^{\prime }{t}^{\prime }$. When the same battery is connected to another resistor of resistance '$r_2$', the same quantity of heat is produced in the same time ${}^{\prime }{t}^{\prime }$. Then , the value of ${}^{\prime }{r}^{\prime }$ is:

• A. $\frac{r_2^2}{r_1}$
• B. $\frac{1}{2}(r_1+r_2)$
• C. $\sqrt{r_1{r}_2}$
• D. $\frac{r_1^2}{r_2}$

Answer 1

The correct option is C $\sqrt{r_1{r}_2}$

When the battery is connected to a resistor of resistance R, the current flowing will be $i=\frac{E}{R+r}$. Hence, power dissipated will be:

$P=i^{2}R={\left(\frac{E}{R+r}\right)}^{2}R=\frac{R{E}^2}{R+r}$

$P(R^2+r^2+2Rr)=E^{2}R$

$R^2+(2r-\frac{E^2}{P})R+r^2=0$

Since $r_1$ and $r_2$ are roots of this equation,

$r_1{r}_2=r^2$

Hence, $r=\sqrt{r_1{r}_2}$