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The potential difference between the points A and B of adjoining figure is :
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(A) $\dfrac{2}{3}V$
(B) $\dfrac{8}{9}V$
(C) $\dfrac{4}{3}V$
(D) $2V$

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Last updated date: 19th Jun 2024
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Answer
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Hint To solve this problem, we need to first simplify the diagram and find the equivalent resistance in the circuit. Using this value of the resistance and the emf of the battery we can find the current that is flowing in the circuit. The potential difference between A and B will be the difference between the potential at A and the potential at B.

Formula Used: In this solution, we are going to use the following formula,
$\Rightarrow {R_{eq}} = {R_1} + {R_2} + {R_3} + ....$ where ${R_{eq}}$ is the equivalent resistance when the resistances are placed in series.
And $\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + ....$ where ${R_{eq}}$ is the equivalent resistance when the resistances are placed in a parallel circuit.
$\Rightarrow V = IR$ where $V$ is the voltage and $I$ is the current.

Complete step by step answer
We need to first draw a simplified form of this circuit. We can draw it as,
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From this circuit we can see that between the points A and C there are 2 wires which are parallel to each other. In the top wire, there are 3 resistances in series. So we can use the formula for calculating the equivalent resistances which are kept in series given as, ${R_{eq}} = {R_1} + {R_2} + {R_3} + ....$
We can substitute here ${R_1} = {R_2} = {R_3} = 5\Omega $
So we get the equivalent resistance as, ${R_{eq1}} = \left( {5 + 5 + 5} \right)\Omega $
On adding we have,
$\Rightarrow {R_{eq1}} = 15\Omega $
Similarly, for the bottom wire, we have three resistances in series. So again using the same formula for ${R_1} = {R_2} = {R_3} = 5\Omega $ we get,
$\Rightarrow {R_{eq2}} = \left( {5 + 5 + 5} \right)\Omega $
On adding we get
$\Rightarrow {R_{eq2}} = 15\Omega $
So in between the points A and C ${R_{eq1}}$ and ${R_{eq2}}$ are in parallel. So we can use the formula for the parallel circuits given as, $\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + ....$
Now substituting, ${R_1} = {R_{eq1}} = 15\Omega $ and ${R_2} = {R_{eq2}} = 15\Omega $
Substituting we get,
$\Rightarrow \dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{15}} + \dfrac{1}{{15}}$
On taking 15 as the LCM we have,
$\Rightarrow \dfrac{1}{{{R_{eq}}}} = \dfrac{{1 + 1}}{{15}}$
On taking reciprocal we get,
$\Rightarrow {R_{eq}} = \dfrac{{15}}{2}\Omega $
Using this resistance we can find the value of the current that is flowing in the circuit using the formula, $V = IR$. In the diagram we are given $V = 2V$. So substituting we get,
$\Rightarrow 2 = I \times \dfrac{{15}}{2}$
So we get the total current as,
$\Rightarrow I = \dfrac{4}{{15}}A$
The current in the top wire will be given by substituting $V = 2V$ and ${R_{eq1}} = 15\Omega $in the same formula,
So $2 = {I_1} \times 15$
So we get,
$\Rightarrow {I_1} = \dfrac{2}{{15}}A$
Now the potential difference between the points A and B will be the difference between the potential at point A and the potential at point B. This will be given by the potential drop of the 2 resistances between A and B. So we have the potential difference as,
$\Rightarrow {V_{AB}} = {I_1}\left( {5 + 5} \right)$
Substituting ${I_1} = \dfrac{2}{{15}}A$ we get,
$\Rightarrow {V_{AB}} = \dfrac{2}{{15}} \times 10$
So we get the potential difference between the points A and B as
$\Rightarrow {V_{AB}} = \dfrac{4}{3}V$
So the correct answer is option C.

Note
When the resistances are placed in series with one another than the current flowing is same through each of the resistors but the value of the potential drop of each depends on the value of the resistance. Here both the resistances between A and B have the same value so the potential drop between them will be equal.