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

Q: 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

d (d) $H_1=\frac{E^2 R}{\left(R+r_1\right)^2}$ $H_2=\frac{E^2 R}{\left(r+r_1\right)^2}$ $\frac{R}{\left(R+r_1\right)^2}=\frac{r}{\left(r+r_1\right)^2}$ $\frac{r+r_1}{R+r_1}=\frac{\sqrt{r}}{\sqrt{R}}$ $\sqrt{R} r_1+r \sqrt{R}=\sqrt{r} r_1+R \sqrt{r}$ $r_1(\sqrt{R}-\sqrt{r})=\sqrt{R r}(\sqrt{R}-\sqrt{r})$ $r_1=\sqrt{R r}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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$\frac{(R r)}{2}$
D$\sqrt{(R r)}$

Medium

STD 11 Science
NEET
STD 12 - 3. current electricity
Physics

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