When the terminals of a cell are connected by a wire of resistance $4 \Omega$, the potential difference across the cell is 1.6 $\mathrm{V}$. If a wire of the same resistance is connected in parallel with the first, the potential difference becomes $1.33 \mathrm{~V}$. The emf and internal…

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When the terminals of a cell are connected by a wire of resistance $4 \Omega$, the potential difference across the cell is 1.6 $\mathrm{V}$. If a wire of the same resistance is connected in parallel with the first, the potential difference becomes $1.33 \mathrm{~V}$. The emf and internal resistance of the cell are respectively

Aicharya · Accepted Answer

$E=V+I r$ Given, $V=1.6 \mathrm{~V} ; R=4 \Omega$ $ I_1=\frac{1.6}{4}=0.4 \mathrm{~A} $

When a resistance of $4 \Omega$ is connected in parallel the resultant resistance $ \frac{1}{R^{\prime}}=\frac{1}{4}+\frac{1}{4}=\frac{2}{4} \Rightarrow R^{\prime}=2 \Omega $

$I_2=\frac{133}{2}=0.66 \mathrm{~A}$

From Eqs. $(i)$ and $(i i)$, we get $ \begin{aligned} & 1.6+0.4 r=1.33+0.66 r \\ & \Rightarrow \quad r=\frac{0.27}{0.26}=1 \end{aligned} $ Substituting of $r$ in Eq. (i), we get $ E=1.6+0.4 \times 1=2 \mathrm{~V} $

When the terminals of a cell are connected by a wire of resistance $4 \Omega$, the potential difference…

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