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When the resistances are connected with each other such that one end of each resistance is joined to a common point and the other end of each resistance is joined to another common point so that the number paths for the current flow is equal to the number of resistances, it is called a *parallel circuit*.

Consider three resistors R_{1}, R_{2} and R_{3} connected across a source of voltage V as shown in the circuit diagram. The total current (I) divides in three parts – I1 flowing through R_{1, I2 flowing through R2 and I3 flowing through R3. As, it can be seen that the voltage across each resistance is the same.}

By ** Ohm’s law**, current through each resistor is given by,

$$\mathrm{\mathit{I}_{1}=\frac{\mathit{V}}{\mathit{R}_{1}};\:\:\:\mathit{I}_{2}=\frac{\mathit{V}}{\mathit{R}_{2}}\:\:\:and\:\:\:I_{3}=\frac{\mathit{V}}{\mathit{R}_{3}};}$$

Referring the circuit, the total current is,

$$\mathrm{\mathit{I}=\mathit{I}_{1}\:+\:\mathit{I}_{2}\:+\:\mathit{I}_{3}}$$

$$\mathrm{\Rightarrow\:\mathit{I}=\frac{\mathit{V}}{\mathit{R}_{1}}+\frac{\mathit{V}}{\mathit{R}_{2}}+\frac{\mathit{V}}{\mathit{R}_{3}}=\mathit{V}(\frac{1}{\mathit{R}_{1}}+\frac{1}{\mathit{R}_{2}}+\frac{1}{\mathit{R}_{3}})}$$

$$\mathrm{\Rightarrow\:\frac{1}{\mathit{V}}=(\frac{1}{\mathit{R}_{1}}+\frac{1}{\mathit{R}_{2}}+\frac{1}{\mathit{R}_{3}})}$$

$$\mathrm{∵\:\frac{1}{\mathit{V}}=\frac{1}{\mathit{R}_{p}}}$$

$$\mathrm{∵\:\frac{1}{\mathit{R}_{p}}=(\frac{1}{\mathit{R}_{1}}+\frac{1}{\mathit{R}_{2}}+\frac{1}{\mathit{R}_{3}})}$$

Hence, when a number of resistance are connected in parallel, the reciprocal of total resistance is equal to the sum of the reciprocals of the individual resistances.

Also,

$$\mathrm{∵\:\frac{1}{\mathit{R}}=\mathit{G}(Conductance)}$$

$$\mathrm{∵\:\mathit{G}_{p}=\mathit{G}_{1}+\mathit{G}_{2}+\mathit{G}_{3}}$$

Thus, *total conductance G _{P}* of resistors which are connected in parallel is equal to the sum of their individual conductances.

The total *power dissipated* in the circuit is equal to sum of powers dissipated in the individual resistances. Thus,

$$\mathrm{\frac{1}{\mathit{R}_{p}}=(\frac{1}{\mathit{R}_{1}}+\frac{1}{\mathit{R}_{2}}+\frac{1}{\mathit{R}_{3}})}$$

$$\mathrm{\Rightarrow\frac{\mathit{V}^{2}}{\mathit{R}_{p}}=\frac{\mathit{V}^{2}}{\mathit{R}_{1}}+\frac{\mathit{V}^{2}}{\mathit{R}_{2}}+\frac{\mathit{V}^{2}}{\mathit{R}_{3}}}$$

$$\mathrm{\Rightarrow\:\mathit{P}_{p}=\mathit{P}_{1}+\mathit{P}_{2}+\mathit{P}_{3}}$$

The voltage across each branch is same.

As the number of parallel branches is increased, the total resistance of the circuit is decreased.

The total resistance of the circuit is less than the smallest of the resistances.

The total conductance is equal to sum of their individual conductances.

The total power dissipated in the circuit is equal to sum of powers dissipated in the individual resistances.

What is the potential difference between points A and B, also find the three branch currents in the circuit shown below?

Solution

The equivalent resistance (R_{P}) of the three parallel connected resistors is

$$\mathrm{\frac{1}{\mathit{R}_{p}}=\frac{1}{2}+\frac{1}{4}+\frac{1}{5}=\frac{19}{20}}$$

$$\mathrm{\Rightarrow\:{\mathit{R}_{p}}=1.053 \:Ω}$$

Therefore, the voltage V across the terminals A and B is

$$\mathrm{\mathit{V}=I\mathit{R}_{p}=24×1.053=25.27\:Volts}$$

Now, the branch currents are

$$\mathrm{Current\:\mathit{I}_{1}=\frac{\mathit{V}}{\mathit{R}_{1}}=\frac{25.27}{2}=12.64\:A}$$

$$\mathrm{Current\:\mathit{I}_{2}=\frac{\mathit{V}}{\mathit{R}_{2}}=\frac{25.27}{4}=6.32\:A}$$

$$\mathrm{Current\:\mathit{I}_{3}=\frac{\mathit{V}}{\mathit{R}_{3}}=\frac{25.27}{5}=5.05\:A}$$

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