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A cell of emf $10 \mathrm{~V}$ and internal resistance $3 \Omega$ is connected in parallel with another cell of emf $7 \mathrm{~V}$ and internal resistance $\frac{3}{5} \Omega$, such that their positive terminals are joined together and so their are negative terminals. Their positive terminals are joined with the negative terminal and their negative terminal is joined with the positive terminal of a third cell of emf $20 \mathrm{~V}$ with internal resistance $2 \Omega$. The combination can be replaced by a battery of emf $E$ and internal resistance $r$, then the values of $E$ and $R$ are respectively
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Verified Answer
The correct answer is:
$E=2 V, r=0.4 \Omega$
$$
\text { Circuit according to the question will be }
$$
All three cells are connected in parallel, but $B_3$ connected wrongly, so equivalent emf and equivalent internal resistance is given by
$$
\begin{aligned}
& \frac{E_{\mathrm{eq}}}{r_{\mathrm{eq}}}=\frac{B_1}{r_1}+\frac{B_2}{r_2}-\frac{B_3}{r_3} \\
& \frac{E_{\mathrm{eq}}}{r_{\mathrm{eq}}}=\frac{10}{3}+\frac{7}{3 / 5}-\frac{20}{2}=\frac{10}{3}+\frac{35}{3}-10 \\
& \frac{E_{\mathrm{eq}}}{r_{\mathrm{eq}}}=5
\end{aligned}
$$
Also, $\frac{1}{r_{\mathrm{eq}}}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3} \Rightarrow \frac{1}{r_{\mathrm{eq}}}=\frac{1}{3}+\frac{5}{3}+\frac{1}{2}$
$$
\Rightarrow \quad r_{\text {eq }}=0.4 \Omega
$$
Putting the value req in Eq. (i), we get
$$
\Rightarrow \quad \frac{E_{\text {eq }}}{0.4}=5 \Rightarrow B_{\text {eq }}=2 \mathrm{~V}
$$
\text { Circuit according to the question will be }
$$
All three cells are connected in parallel, but $B_3$ connected wrongly, so equivalent emf and equivalent internal resistance is given by
$$
\begin{aligned}
& \frac{E_{\mathrm{eq}}}{r_{\mathrm{eq}}}=\frac{B_1}{r_1}+\frac{B_2}{r_2}-\frac{B_3}{r_3} \\
& \frac{E_{\mathrm{eq}}}{r_{\mathrm{eq}}}=\frac{10}{3}+\frac{7}{3 / 5}-\frac{20}{2}=\frac{10}{3}+\frac{35}{3}-10 \\
& \frac{E_{\mathrm{eq}}}{r_{\mathrm{eq}}}=5
\end{aligned}
$$
Also, $\frac{1}{r_{\mathrm{eq}}}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3} \Rightarrow \frac{1}{r_{\mathrm{eq}}}=\frac{1}{3}+\frac{5}{3}+\frac{1}{2}$
$$
\Rightarrow \quad r_{\text {eq }}=0.4 \Omega
$$
Putting the value req in Eq. (i), we get
$$
\Rightarrow \quad \frac{E_{\text {eq }}}{0.4}=5 \Rightarrow B_{\text {eq }}=2 \mathrm{~V}
$$
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