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Question: The potential difference between \(A\) and \(B\) is- (A). \(24V\) (B). \(14V\) (C). \(56V\) ...

The potential difference between AA and BB is-
(A). 24V24V
(B). 14V14V
(C). 56V56V
(D). 42V42V

Explanation

Solution

A circuit containing resistors and batteries is shown. Due to the potential difference between A and B, a current flows from higher potential to lower potential. Using Kirchhoff’s laws, we can analyse the circuit and calculate the total potential drop in the circuit.

Formulas used:
R=VIR=\dfrac{V}{I}
VA−2×6−12−2×9−4−2×5=VB{{V}_{A}}-2\times 6-12-2\times 9-4-2\times 5={{V}_{B}}

Complete step-by-step solution:
When there is a potential difference between two points current flows between those points. It flows from higher potential to lower potential.

Current is the flow of positive charge. Therefore, by convention the flow of current is taken opposite to the flow of electrons.

Resistance is the tendency of a material to oppose the flow of current through it.
According to the Ohm’s law,
R=VIR=\dfrac{V}{I}
Here,
RR is the resistance
VV is the potential difference
II is the current
There are two laws which are used to analyse circuits, Kirchhoff’s current law and Kirchhoff’s voltage law:
Kirchhoff's current law states that the sum of all the charges passing through a junction is zero. This means the sum of incoming current is equal to the sum of outgoing current.

According to Kirchhoff's voltage law, the sum of all the voltages in a loop is zero.
In the circuit shown in the figure

Applying Kirchhoff’s voltage law in the above circuit, we get
VA−2×6−12−2×9−4−2×5=VB ⇒VA−VB=56V \begin{aligned} & {{V}_{A}}-2\times 6-12-2\times 9-4-2\times 5={{V}_{B}} \\\ & \Rightarrow {{V}_{A}}-{{V}_{B}}=56V \\\ \end{aligned}

Therefore, the total potential drop in the circuit is 56V56V.

Note:
The kirchhoff’s laws obey the law of conservation of charge. When we go through a loop, then the potential of the battery is taken according to the terminal through which we pass. If we pass through a positive terminal then potential is taken as positive. If we pass through negative terminal potential is taken as negative. By convention, when we go in the direction of current, the potential drop on appliances is taken as negative.