Question 56: Current Electricity

Question

Two cells with the same emf $$E$$ and different internal resistances $$r_1$$ and $$r_2$$ are connected in series to an external resistance $$R$$. If the potential difference across the first cell is zero then value of $$R$$.

Accepted Answer

Here are two batteries with emf E each and the internal resistances r$$_1$$ and r$$_2$$, respectively
We have $$I\left(R+r_1+r_2\right)=2 E$$
Current in the circuit, $$I=\frac{2 E}{R+r_1+r_2}$$ .... (i)
Now, the potential difference across the first cell would be equal to $$V=E-I r_1$$. From the question, $$V=0$$, there
$$E=I r_1=\frac{2 E r_1}{R+r_1+r_2}\quad$$ [From Eq. (i)]
where, $$R+r_1+r_2=2 r_1$$
Hence, $$R=r_1-r_2$$.

Answer

$$\sqrt{r_1 r_2}$$

Answer

$$r_1+r_2$$

Answer

$$r_1-r_2$$

Answer

$$\frac{r_1+r_2}{2}$$

Current Electricity

Two cells with the same emf $E and different internal resistances r_1 and r_2 are connected in series to an external resistance R. If the potential difference across the first cell is zero then value of R$.

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