## Section36.3Resistors in Parallel

We say that two resistors $R_1$ and $R_2$ are connected in parallel if a current entering the combination splits and recombines immediately after going through each of the resistors as shown in Figure 36.3.1. The two resistors in parallel circuit are also called two branches of the circuit and points $a$ and $b$ where they split or join are called junctions or nodes.

Another property to note of resistors in parallel is that the potential drops across each resistor is same for all resitors in parallel since every resistor is connected to same junction points on the two ends.

### Subsection36.3.1Potential Drop Across Resistors in Parallel

In Figure 36.3.1, the resitors in parallel are connected to a voltage source such that we set up the same potential differece across both resistors.

$$\Delta \phi_1 = \Delta \phi_2 = \phi_a - \phi_b = V.\label{eq-over-all-potential-drop-R-in-series}\tag{36.3.1}$$

### Subsection36.3.2Currents in Resitors in Parallel

Applying Ohm's law to the two resistors separatelty gives us currents through the individual resitors immediately.

\begin{align} I_1 \amp = \dfrac{\Delta\phi_1}{R_1} = \dfrac{V}{R_1}\label{eq-resistors-in-parallel-current-I1}\tag{36.3.2}\\ I_2 \amp = \dfrac{\Delta\phi_2}{R_2} = \dfrac{V}{R_2} \label{eq-resistors-in-parallel-current-I2}\tag{36.3.3} \end{align}

These show that if $R_1 \gt R_2\text{,}$ then $I_1 \lt I_2\text{.}$ That is, higher current will flow through the branch with smaller resistance. If a branch has no resistance, then there will be no current will be left to flow through any other branch. Such as branch is called a short in the circuit.

Since a total current $I$ comes into junction $a$ and currents $I_1$ and $I_2$ leave and no charge accumates at $a\text{,}$ we can apply conservation of charge to conclude the following current conservation at each junction.

\begin{equation*} I - I_1 - I_2 = 0, \end{equation*}

that is, overall current in the circuit is

$$I = I_1 + I_2 = \dfrac{V}{R_1} + \dfrac{V}{R_2} = V \left(\dfrac{1}{R_1} + \dfrac{1}{R_2} \right) .\label{eq-resistors-in-parallel-current-conservation}\tag{36.3.4}$$

This shows that resitors in parallel divide up the total current among themselves. That is why we often refer to parallel resitors as current dividers. Solving this for $V$ and using it in Eqs. (36.3.2) and (36.3.3), we find current in each branch in terms of the total current entering or exiting the parallel resistors. After simplification we get the folloing useful formulas.

\begin{align} I_1 \amp = I\,\dfrac{R_2}{R_1+R_2},\label{eq-resistors-in-parallel-current-I1-in-I}\tag{36.3.5}\\ I_2 \amp = I\,\dfrac{R_1}{R_1+R_2}. \label{eq-resistors-in-parallel-current-I2-in-I}\tag{36.3.6} \end{align}

Note the resistor in the numerator of each formula is the resitor in the other branch. Therefore, if other branch 2 has more resistance than branch 1, more current will flow in branch 1.

### Subsection36.3.3Equivalent Resistor of Resistors in Parallel

Eq. (36.3.4) can be written in the following way, where voltage across the parallel combination is equal to the total current and a resistor, which is called equivalent resistor of the combination, $R_\text{eqiv}$ .

\begin{equation*} V = I\left( \dfrac{1}{R_1} + \dfrac{1}{R_2} \right)^{-1} \equiv I R_\text{equiv}. \end{equation*}

The total current and total voltage related this way implies that the parallel combination of resistors “act” as if there was a single resitor of an equivalent resistance. We will also denote this equivalent resistance by $R_\text{parallel}$ as illustrated in Figure 36.3.2.

$$R_\text{parallel} = \left( \dfrac{1}{R_1} + \dfrac{1}{R_2} \right)^{-1} \equiv R_\text{equiv}.\label{eq-equiv-res-parallel-resitors-resistance}\tag{36.3.7}$$

This can be written in calculationally more friendly way,

$$R_\text{parallel} = \dfrac{R_1 R_2}{R_1 + R_2}.\tag{36.3.8}$$

We will often write Eq. (36.3.7) as an inverse resistance, i.e., a conductance relation.

$$\dfrac{1}{R_\text{parallel}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} \Longleftrightarrow G_\text{parallel} = G_1 + G_2,\label{eq-equiv-res-parallel-resitors-inv-resistance}\tag{36.3.9}$$

where I have used the inverse relation between conductance $G$ and resistance $R\text{.}$

\begin{equation*} G = \dfrac{1}{R}. \end{equation*}

For $N$ resistors in parallel, you can show that

$$\frac{1}{R_\text{parallel} } = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \cdots + \frac{1}{R_{N}},\tag{36.3.10}$$

and the following for the conductances.

$$G_\text{parallel} = G_1 + G_2 + \cdots + G_N.\tag{36.3.11}$$

Hence, conductances add in a parallel circuit. Earlier, we found that resistances add up in series.

Consider the circuit given in Figure 36.3.4.

(a) Find percentage of current that passes through each resistor.

(b) Find equivalent resitance across the voltage source.

(c) Find current through the voltage source.

(d) Find current in each resistor.

Hint

Use definitions.

(a) $75\%,\ 25\%\text{,}$ (b) $75\ \Omega \text{,}$ (c) $267\text{ mA}\text{,}$ (d) $67\text{ mA},\ 200\text{ mA}\text{.}$

Solution 1 (a)

(a) In this example, current $I_1$ will be more than current $I_2$ since $I_1$ passes through the smaller resistance.

\begin{align*} I_1 \amp = \left(\frac{300\Omega}{100\Omega + 300\Omega } \right) I = \frac{3}{4} I. \\ I_2 \amp = \left(\frac{100\Omega}{100\Omega + 300\Omega } \right) I = \frac{1}{4} I. \end{align*}

Hence 75% of the total current passes through the $100\;\Omega$ resistor and $25\%$ through the $300\; \Omega$ resistor.

Solution 2 (b)

(b) The equivalent resistance across the source is the equivalent resistance of parallel resitors.

\begin{align*} R_\text{equiv} \amp = \left( \dfrac{1}{R_1} + \dfrac{1}{R_2} \right)^{-1} = \dfrac{R_1 R_2}{R_1 + R_2}\\ \amp = \dfrac{100\Omega \times 300\Omega }{100\Omega + 300\Omega } = 75\ \Omega. \end{align*}
Solution 3 (c)

(c) The equivalent resistance across the source is the equivalent resistance of parallel resitors.

\begin{align*} I \amp = \dfrac{V}{R_\text{parallel}} = \dfrac{20\text{ V}}{75\ \Omega} = 267\text{ mA} \end{align*}
Solution 4 (d)

(d) Using the results of (a) and (c) we get

\begin{align*} I_1 \amp = \frac{3}{4} I = 67\text{ mA}.\\ I_2 \amp = \frac{1}{4} I = 200\text{ mA}. \end{align*}

Find current through each resistor in the circuit given in Figure 36.3.6.

Hint

Think of potential difference across each resistor.

$\frac{1}{10}\text{ A},\ \frac{1}{20}\text{ A},\ \frac{1}{30}\text{ A}$