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A $8 \Omega$ resistor is connected to a battery that has an internal resistance of $0.2 \Omega$. If the voltage across the battery (the terminal voltage) is $10 \mathrm{~V}$, then the emf of the battery is
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Verified Answer
The correct answer is:
$10.25 \mathrm{~V}$
The given situation is shown below

If $I=$ current in circuit then
Where,
$$
\begin{gathered}
E-I r=V \\
I=V / R
\end{gathered}
$$
Here, given,
$$
\begin{aligned}
& R=8 \Omega \\
& r=0.2 \Omega V=10 \mathrm{~V}
\end{aligned}
$$
Hence, circuit current
$$
I=\frac{V}{R}=\frac{10}{8}=\frac{5}{4} \mathrm{~A}
$$
Now, from eq. (i) with values we have,
$$
E=V+I r=10+\frac{5}{4} \times 0.2=10+0.25=10.25 \mathrm{~V}
$$

If $I=$ current in circuit then
Where,
$$
\begin{gathered}
E-I r=V \\
I=V / R
\end{gathered}
$$
Here, given,
$$
\begin{aligned}
& R=8 \Omega \\
& r=0.2 \Omega V=10 \mathrm{~V}
\end{aligned}
$$
Hence, circuit current
$$
I=\frac{V}{R}=\frac{10}{8}=\frac{5}{4} \mathrm{~A}
$$
Now, from eq. (i) with values we have,
$$
E=V+I r=10+\frac{5}{4} \times 0.2=10+0.25=10.25 \mathrm{~V}
$$
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