### 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.

### 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)