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# Two batteries of different EMFs and different internal resistance are connected as shown. The voltage across AB in volts is:A. 4B. 5C. 6D. 7

Last updated date: 21st Jun 2024
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Hint: When we talk about small microcircuits, minute change in the internal resistance makes a huge difference in the quantities of the circuit. When cells are connected to each other in series the current flows from the higher potential to the low potential and it becomes necessary to calculate the average potential between any two points. This can be calculated with the help of a potentiometer which works as an external variable resistance applied to the circuit.
As per the given data
EMF of one cell (${{E}_{1}}$) is $6V$
The internal resistance of ${{E}_{1}}$ (${{r}_{1}}$) is $1\Omega$
EMF of another cell (${{E}_{2}}$) is $3V$
The internal resistance of ${{E}_{2}}$ (${{r}_{1}}$) is $2\Omega$

Formula used:
${{V}_{AB}}=\dfrac{\Sigma \dfrac{E}{r}}{\Sigma \dfrac{1}{r}}$

${{V}_{AB}}=\dfrac{\Sigma \dfrac{E}{r}}{\Sigma \dfrac{1}{r}}$
\begin{align} & {{V}_{AB}}=\dfrac{6+\dfrac{3}{2}}{1+\dfrac{1}{2}} \\ & \\ & \Rightarrow {{V}_{AB}}=\dfrac{\dfrac{15}{2}}{\dfrac{3}{2}} \\ & \\ & \Rightarrow {{V}_{AB}}=\dfrac{15}{2}\times \dfrac{2}{3} \\ & \\ & \Rightarrow {{V}_{AB}}=5V \\ \end{align}
Thus the voltage across points A and B is ${{V}_{AB}}=5V$.