### Resistances in Parallel

• Calculate the combined resistance of resistors connected in parallel.

• Perform calculations on current divider circuits;

• Define power as VI and apply the formula to calculate power dissipation in parallel circuits.

The components sit side by side and connect to more than one other component.

The voltage is the same across all parallel components.

$\dfrac{1}{R_{\mathrm{parallel}}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} + ......................$

The sum of the currents in each branch of the circuit equals the current leaving the power supply.

## Parallel Direct Current Circuit Rules:

1. The same voltage exists across each branch of a parallel circuit and is equal to the source voltage.

2. The current through a branch of a parallel network is inversely proportional to the amount of resistance of the branch.

3. The total current of a parallel circuit is equal to the sum of the currents of the individual branches of the circuit.

4. The total resistance of a parallel circuit is equal to the reciprocal of the sum of the reciprocals of the individual resistances of the circuit.

5. The total power dissipated in a parallel circuit is equal to the sum of the individual power dissipation.

### Solved Example:

#### 19-4-01

Effective resistance of parallel combination of two resistors is $\dfrac{6}{4} \Omega$. If one of resistance is broken, then the resultant resistance becomes 2 $\Omega$. Then other resistance is:

Solution:
$\dfrac{1}{2} + \dfrac{1}{R} = \dfrac{1}{\dfrac{6}{4}}, \quad \dfrac{1}{R} = \dfrac{1}{6}, \quad R = 6 \ \Omega$

### Solved Example:

#### 19-4-02

The currents into a junction flow along two paths. One current is 4 A and the other is 3 A. The total current out of the junction is:

### Solved Example:

#### 19-4-03

When an additional resistor is connected across an existing parallel circuit, the total resistance:

### Solved Example:

#### 19-4-04

When a 1.6 $k\Omega$ resistor and a 120 $\Omega$ resistor are connected in parallel, the total resistance is:

### Solved Example:

#### 19-4-05

The power dissipation in each of four parallel branches is 1.2 W. The total power dissipation is:

Solution:
Power dissipated in parallel circuit overall is sum of power dissipated in individual branches.

### Solved Example:

#### 19-4-06

What is the total resistance, if a series-parallel circuit using $R_1$ =30 $\Omega$ in series with a parallel combination of $R_2$ = 30 $\Omega$ and $R_3$ = 30 $\Omega$ ?

### Solved Example:

#### 19-4-07

Find the current I$_x$ using superposition theorem.

Solution:
In Principle of Superposition, only one source is active at a time. Case I:

36 V is active, 12 V is short circuited, 6 A is open circuited.

5 $\Omega$ is parallel to 10 $\Omega$ = 3.33 $\Omega$

3.33 $\Omega$ is in series with 5 $\Omega$ = 8.33 $\Omega$

$I_x = \dfrac{36}{8.33} = 4.32 A$ Case II:

12 V is active, 36 V is short circuited, 6 A is open circuited.

5 $\Omega$ is parallel to 10 $\Omega$ = 3.33 $\Omega$

3.33 $\Omega$ is in series with 5 $\Omega$ = 8.33 $\Omega$

Total current = $\dfrac{12}{8.33}$ = 1.44 A

Current entering 5 $\Omega$ = $\dfrac{2}{3}$ $\times$ 1.44 A = 0.96 A

I$_x$ = -0.96 A

Case III:

6 A is active, 12 V is short circuited, 36 V is short circuited.

5 $\Omega$ is parallel to 10 $\Omega$ = 3.33 $\Omega$

3.33 $\Omega$ is in parallel with 5 $\Omega$

Current entering 5 $\Omega$ = 6 $\times$ $\dfrac{3.33}{3.33 + 5}$

I$_x$ = -2.40 A

Total I$_x$ = 4.32 - 0.96 - 2.40 = 0.96 A (From left to Right)