# How can i calculate voltage drop in a parallel circuit?

Mar 3, 2018

Well for that you need a circuit to show me,still I am giving you two examples with how to work out.

See the circuit below,

It is clear that, $2 \Omega$ is in parallel with the other $2 \Omega$,and their resultant is in series with the $1 O h m$ and the $3 O h m$,so net resistance of the circuit is $\frac{2 \cdot 2}{2 + 2} + 1 + 3 = 5 O h m$

So,current flowing through the circuit is $\frac{20}{5} = 4 A$

so,potential drop across each resistance can be calculated just by multiplying their resistance value with the current flowing through them.

thus ${V}_{1 O h m} = 1 \cdot 4 = 4 V$

${V}_{3 O h m} = 3 \cdot 4 = 12 V$

and, ${V}_{2 | | 2 O h m} = \left(20 - 12 - 4\right) = 4 V$

or,you could have multiplied the current flowing through them with their equivalent resistance,i.e $\frac{2 \cdot 2}{2 + 2} \cdot 4 = 1 \cdot 4 = 4 V$

In the example below, in the upper wire, $1 \Omega$ is in parallel with $1 \Omega$,their resultant is in series with $0.5 \Omega$,and the equivalent resistance is again in parallel with the equivalent resistance of $4 \Omega$ and $4 \Omega$ in parallel combination.

So,net resistance of thee upper wire is $\frac{1 \cdot 1}{1 + 1} + 0.5 = 1 \Omega$

and that of the lower wire is $\frac{4 \cdot 4}{4 + 4} = 2 \Omega$

So,net resistance of the circuit is $\frac{1 \cdot 2}{1 + 2} = \frac{2}{3} \Omega$

So,current flowing through the circuit is $\frac{14}{\frac{2}{3}} = 21 A$

now,volatge drop across the parallel combination of two $4 \Omega$ resistors is similar to the voltage drop across the battery i.e $14 V$

the same amount of voltage has been dropped across the upper wire,as both are in parallel combination.

So,we have to find the current flowing through the upper circuit in order to calculate that.

So,current flowing through the lower wire is $\frac{14}{2} = 7 A$ (as the net resistance of the lower wire is $2 \Omega$

so,current flowing through the upper circuit is $\left(21 - 7\right) = 14 A$

So,potential drop across $0.5 \Omega$ is $14 \cdot 0.5 = 7 V$

So,the rest i.e $\left(14 - 7\right) = 7 V$ must have dropped across the parallel combination of $1 \Omega$ resistors.