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# Two batteries of different $emf$ and internal resistances connected in series with each other and with an external load resistor. The current is $3A$ . When the polarity of one battery is reversed, the current becomes $1A$ . The ratio of the $emf$ of the two batteries is:(A) $2.5:1$(B) $2:1$(C) $3:2$(D) $1:1$

Last updated date: 20th Jun 2024
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Hint: We will use Kirchhoff's rule to find a relation of the resistances and the $emf$ of the two batteries. Then, we will equate them with each other.

Step By Step Solution

Here,
${V_1}$ is the $emf$ of the first battery and ${V_2}$ is that of the second one. ${r_1}$ is the internal resistance of the first battery and ${r_2}$ is that of the second. $R$ in the load resistance.
Now,
For the first situation when current is $3A$ .
By Kirchhoff’s Law,
$\frac{{\mathop V\nolimits_1 \mathop { + V}\nolimits_2 }}{{\mathop {R + r}\nolimits_1 \mathop { + r}\nolimits_2 }} = \mathop 3\nolimits_{}$
Thus, we can say
$\mathop {(R + r}\nolimits_1 \mathop { + r}\nolimits_2 ) = \frac{{\mathop V\nolimits_1 \mathop { + V}\nolimits_2 }}{3} \cdot \cdot \cdot \cdot (1)$
Similarly for the second case when current is $1A$ ,
$\mathop V\nolimits_1 \mathop { - V}\nolimits_2 = \mathop {R + r}\nolimits_1 \mathop { + r}\nolimits_2$
Now,
Putting in equation $(1)$, we get
$\mathop {3V}\nolimits_1 \mathop { - 3V}\nolimits_2 = \mathop V\nolimits_1 \mathop { + V}\nolimits_2$
After further evaluation, we get
$\mathop {2V}\nolimits_1 = \mathop {4V}\nolimits_2$
In the question, it is asked for $\frac{{\mathop V\nolimits_1 }}{{\mathop V\nolimits_2 }}$
Thus, we get
$\frac{{\mathop V\nolimits_1 }}{{\mathop V\nolimits_2 }} = \frac{2}{1}$

Additional Information: The Kirchhoff’s rules are handy to use in the cases for internal resistance, multiple $emf$ and in the cases indulging potentiometer. These rules are simple and very intuitive. Just that they were placed in a standardized manner by Kirchhoff.