Question

Find out potential difference between point A and B.

Answer 1

In the above circuit, each 2Ω resistor is connected in series with a 1Ω resistor. There are two batteries of 3V and 2V corresponding to these resistors. To find the potential difference between point A and B, we have to find the equivalent potential of the whole circuit. We will find the equivalent resistances near the given source which will be useful in finding the required potential difference.

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Formula used:

If two dc sources ${\varepsilon _1}$ and ${\varepsilon _2}$ are connected with internal resistances $r_1$ and $r_2$ respectively, then their equivalent resistance is given by the formula

${\varepsilon _{eq}} = \dfrac{{{\varepsilon _1}{r_2} + {\varepsilon _2}{r_1}}}{{{r_1} + {r_2}}}$

${\varepsilon _{eq}} =$ Equivalent battery potential (In this case it is the potential difference between point A & B)

${\varepsilon _1}\& {\varepsilon _2} =$ Potentials of batteries 1 & 2 respectively.

${r_1}\& {r_2} =$ Internal resistances of batteries (In this case these are the equivalent resistances on the left and right)

Complete step by step solution:

The given circuit is

Let ${\varepsilon _1} = 2V$ and ${\varepsilon _2} = 3V$

To find the equivalent potential, we need the $r_1$ and $r_2$.

To find ${r_1}$, we add resistances on the left, because they are in series. So

${r_1} =1 + 1= 2$ Ω.

Again we find ${r_2}$, we add resistances on the right. So

$${r_2}$$$= 2+2 = 4$ Ω.

Using the formula

${\varepsilon _{eq}} = \dfrac{{{\varepsilon _1}{r_2} + {\varepsilon _2}{r_1}}}{{{r_1} + {r_2}}}$

Substituting the corresponding values in the above equation,

${\varepsilon _{eq}} = \dfrac{{2 \times 4 + 3 \times 2}}{{4 + 2}} = \dfrac{{15}}{6}$

On simplification,

${\varepsilon _{eq}}= 2.5V$

Therefore, the equivalent potential between A and B is $2.5V$.

Note: This problem can also be solved by using Kirchhoff's loop law. With loop law we would have to consider two loops, first we would have taken the entire circuit to find the value of current flowing and then the either half of the circuit (top half or bottom half) to apply loop law for the second time and this would have given us the required potential difference.