Question: Q. 8. Two cells of emf $E_{1}$ and $E_{2}$ have internal resistance $r_{1}$ and $r_{2}$. Deduce an expression for equivalent emf of their parallel combination. $\square$ [SQP II 2017]
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Solution:
Ans.
$$ \begin{aligned} I= & I_{1}+I_{2} \ = & \frac{E_{1}-V}{r_{1}}+\frac{E_{2}-V}{r_{2}} \ I= & \left(\frac{E_{1}}{r_{1}}+\frac{E_{2}}{r_{2}}\right)-V\left(\frac{1}{r_{1}}+\frac{1}{r_{2}}\right) \ & I=\frac{E_{1} r_{2}+E_{2} r_{1}}{r_{1} r_{2}}-V\left(\frac{r_{1}+r_{2}}{r_{1}}\right) \end{aligned} $$
or $\left.\quad V=\left(\frac{E_{1} r_{2}+E_{2} r_{1}}{r_{1} r_{2}}\right) \times \frac{r_{1} r_{2}}{r_{1}+r_{2}}\right)-I\left(\frac{r_{1} r_{2}}{r_{1}+r_{2}}\right)_{1 / 2}$
or $\quad V=\left(\frac{E_{1} r_{2}+E_{2} r_{1}}{r_{1}+r_{2}}\right)-I\left(\frac{r_{1} r_{2}}{r_{1}+r_{2}}\right)$
Comparing above with
$$ V=E_{e q}-I_{e q} \times R_{e q} $$
We get $E_{e q}=\frac{E_{1} r_{2}+E_{2} r_{1}}{r_{1}+r_{2}}$
AI Q. 9. A cell of emf $4 \mathrm{~V}$ and internal resistance $1 \Omega$ is connected to a d.c. source of $10 \mathrm{~V}$ through a resistor of $5 \Omega$. Calculate the terminal voltage across the cell during charging.
U [O.D. III 2017]
Ans. Calculation of Current
Calculation of Terminal Voltage
$10-4=I(1+5)$
$\therefore \quad I=1 \mathrm{~A}$ $\therefore$ Terminal voltage across cell
$$ \begin{align*} & =(4+1 \times 1) \mathrm{V} \ & =5 \mathrm{~V} \end{align*} $$
[CBSE Marking Scheme 2017]
