Question

In the given circuit, calculate the potential difference between the points P and Q.

Answer 1

Hint:The Kirchhoff’s Voltage law says that the sum of all the voltages is constant in s closed circuit. The potential difference is defined as work done in transfer of unit charge from one point A to another point B.

Formula used:The formula of the ohm's law is given by,
$ \Rightarrow V = I \times R$
Where current is I, the resistance is R and potential difference is V.

Complete step by step solution:
It is given in the problem that there is a circuit in which we need to find the potential difference between the points P and Q.
The total resistance in the circuit is equal to,
$ R = 1 + 9 + 2$
$ \therefore R = 12\Omega $.
The resultant potential difference is equal to,
$ \Rightarrow V = \left( {12 - 8} \right)V$
$ \Rightarrow V = 4V$
The formula of the ohm's law is given by,
$ \Rightarrow V = I \times R$

The potential difference is equal to 4V and the resistance is equal to$12\Omega $.
$ \Rightarrow V = I \times R$
$ \Rightarrow 4 = I \times 12$
$ \Rightarrow I = \dfrac{4}{{12}}$
$ \Rightarrow I = \dfrac{1}{3}A$

Let us apply Kirchhoff's Voltage law between the points P and Q.

The KVL equation is equal to,

$ {V_P} + \dfrac{1}{3} \times 1 + 8 - 12 + \dfrac{1}{3} \times 2 = {V_Q}$
$ \Rightarrow {V_P} + \dfrac{1}{3} - 4 + \dfrac{2}{3} = {V_Q}$
$ \Rightarrow {V_P} + \left( {\dfrac{{1 - 12 + 2}}{3}} \right) = {V_Q}$
$ \Rightarrow {V_P} - \dfrac{9}{3} = {V_Q}$
$ \Rightarrow {V_P} - 3 = {V_Q}$
$ \therefore {V_P} - {V_Q} = 3V$

The potential difference between the points P and Q is equal to 3V.

Additional information: The resistance is given in the circuit is series and the resultant is simply applied by adding the resistance simply also if the resistance is given in the parallel connection the inverse of the resistance is added and then its inverse is taken. The voltage is also added simply if it is present in the series.

Note:The potential difference is asked between the point P and point Q and therefore we have applied the Kirchhoff’s Voltage law between the points P and Q and not involved the resistance of$9\Omega $. Kirchhoff's current law says that the net current in the closed current is zero.