Question

A cell sends a current through a resistance $R$ for time $t$. Now the same cell sends current through another resistance $r$ for the same time. If same amount of heat is developed in both the resistance, then the internal resistance of cell is

• A. $\frac{(R+r)}{2}$
• B. $\frac{(R-r)}{2}$
• C. $\frac{\left(Rr\right)}{2}$
• D. $\sqrt{\left(Rr\right)}$

Answer 1

Answer: (D)

$\sqrt{\left(Rr\right)}$

$\text{Explanation:-}$

Suppose the internal resistance of the cell is x and E.M.F is E

So, while current flow through R ohm, total resistance is $(R+x)$

So, current(I) through R ohm is $\frac{E}{R+x}$

For the second case,

Similarly, Current(I) flowing is $\frac{E}{r+x}$

So if heat generated is same for time t

then, We have,

$I_1^{2}Rt=I_2^{2}rt$

$\Rightarrow \frac{E^{2}R}{(R+x{)}^2}=\frac{E^{2}r}{(r+x{)}^2}$

$\Rightarrow R(r+x{)}^2=r(R+x{)}^2$

$\Rightarrow R{r}^2+R{x}^2=r{R}^2+r{x}^2$

$\Rightarrow Rr(r-R)=x^2(r-R)$

$\Rightarrow x=\sqrt{Rr}[Forr\ne R]$

$\text{Hence the correct option is D}$