# Startodeltaconversionsolvedproblemspdf40

## Star Delta Conversion Solved Problems PDF 40

Star delta conversion is a method of simplifying a circuit that contains resistors connected in a star or delta configuration. Star delta conversion can be used to find the equivalent resistance of a complex network, or to analyze the current and voltage in each branch of the circuit. In this article, we will explain the basic principles of star delta conversion, and provide some solved problems with detailed solutions.

## What is Star Delta Conversion?

A star network is a network of three resistors that are connected to a common point, called the neutral point. A star network can also be called a wye or Y network. A delta network is a network of three resistors that are connected in a closed loop, forming a triangle. A delta network can also be called a mesh or Î” network.

## startodeltaconversionsolvedproblemspdf40

Star delta conversion is the process of transforming a star network into an equivalent delta network, or vice versa. The equivalent network has the same total resistance and the same behavior when connected to an external voltage or current source. Star delta conversion can be useful for simplifying complex circuits that contain both star and delta networks, or for finding the current and voltage in each branch of the circuit.

## How to Perform Star Delta Conversion?

To perform star delta conversion, we need to use some formulas that relate the resistances of the original network and the equivalent network. The formulas are different depending on whether we are converting from star to delta, or from delta to star.

### Star to Delta Conversion

To convert a star network into an equivalent delta network, we use the following formulas:

R_AB = (R_A R_B)/(R_A + R_B + R_C) R_BC = (R_B R_C)/(R_A + R_B + R_C) R_CA = (R_C R_A)/(R_A + R_B + R_C)

Where R_AB, R_BC, and R_CA are the resistances of the delta network, and R_A, R_B, and R_C are the resistances of the star network.

### Delta to Star Conversion

To convert a delta network into an equivalent star network, we use the following formulas:

R_A = (R_AB R_CA)/(R_AB + R_BC + R_CA) R_B = (R_AB R_BC)/(R_AB + R_BC + R_CA) R_C = (R_BC R_CA)/(R_AB + R_BC + R_CA)

Where R_A, R_B, and R_C are the resistances of the star network, and R_AB, R_BC, and R_CA are the resistances of the delta network.

## Solved Problems

In this section, we will provide some solved problems on star delta conversion. We will show how to apply the formulas and how to find the equivalent resistance, current, and voltage in each case.

### Problem 1

Find the equivalent resistance between terminals A and B of the circuit shown below.

Solution:

We can simplify this circuit by converting the star network into an equivalent delta network. Using the formulas for star to delta conversion, we get:

R_AB = (6 x 6)/(6 + 6 + 6) = 12/3 = 4 ohms R_BC = (6 x 6)/(6 + 6 + 6) = 12/3 = 4 ohms R_CA = (6 x 6)/(6 + 6 + 6) = 12/3 = 4 ohms

The equivalent circuit is shown below.

The equivalent resistance between terminals A and B is simply the parallel combination of R_AB and R_CA, which are both equal to 4 ohms. Using the formula for parallel resistance, we get:

R_eq = (R_AB x R_CA)/(R_AB + R_CA) = (4 x 4)/(4 + 4) = 16/8 = 2 ohms

Therefore, the equivalent resistance between terminals A and B is 2 ohms.

### Problem 2

Find the current in each branch of the circuit shown below.

Solution:

We can simplify this circuit by converting the star network into an equivalent delta network. Using the formulas for star to delta conversion, we get:

R_AB = (2 x 2)/(2 + 2 + 2) = 4/3 ohms R_BC = (2 x 2)/(2 + 2 + 2) = 4/3 ohms R_CA = (2 x 2)/(2 + 2 + 2) = 4/3 ohms

The equivalent circuit is shown below.

To find the current in each branch, we can use Kirchhoff's current law (KCL) and Kirchhoff's voltage law (KVL). KCL states that the sum of the currents entering a node is equal to the sum of the currents leaving that node. KVL states that the sum of the voltages around a closed loop is equal to zero. Applying KCL at node A, we get:

I_1 - I_2 - I_3 = 0

Applying KVL around loop ABCA, we get:

-12 + I_1 x 3 + I_2 x (3 + 4/3) = 0

Applying KVL around loop BCAB, we get:

-12 + I_1 x (3 + 4/3) + I_3 x 3 = 0

Solving this system of equations, we get:

I_1 = -1.5 A I_2 = -0.75 A I_3 = -0.75 A

The negative sign indicates that the current is flowing in the opposite direction of the assumed direction. Therefore, the current in each branch is as follows:

### I_AB = -1.5 A (from B to A) I_BC = -0.75 A (from C to B) I_CA = -0.75 A (from A to C) Problem 3

Find the voltage across each resistor of the circuit shown below.

Solution:

We can simplify this circuit by converting the delta network into an equivalent star network. Using the formulas for delta to star conversion, we get:

R_A = (10 x 10)/(10 + 10 + 10) = 100/3 ohms R_B = (10 x 10)/(10 + 10 + 10) = 100/3 ohms R_C = (10 x 10)/(10 + 10 + 10) = 100/3 ohms

The equivalent circuit is shown below.

To find the voltage across each resistor, we can use the voltage divider rule. The voltage divider rule states that the voltage across a resistor in a series circuit is proportional to its resistance and the total voltage across the circuit. Applying the voltage divider rule to the series combination of R_A and R_1, we get:

V_1 = (R_1 x V_s)/(R_1 + R_A) = (5 x 15)/(5 + 100/3) = 75/23 V

Where V_s is the source voltage of 15 V. Similarly, applying the voltage divider rule to the series combination of R_B and R_2, we get:

V_2 = (R_2 x V_s)/(R_2 + R_B) = (5 x 15)/(5 + 100/3) = 75/23 V

And applying the voltage divider rule to the series combination of R_C and R_3, we get:

V_3 = (R_3 x V_s)/(R_3 + R_C) = (5 x 15)/(5 + 100/3) = 75/23 V

Therefore, the voltage across each resistor in the original star network is equal to 75/23 V. To find the voltage across each resistor in the original delta network, we can use Kirchhoff's voltage law (KVL). KVL states that the sum of the voltages around a closed loop is equal to zero. Applying KVL around loop ABCA, we get:

-V_s + V_AB + V_BC + V_CA = 0

Where V_AB, V_BC, and V_CA are the voltages across the delta network. Substituting the values of V_s and V_CA, we get:

-15 + V_AB + V_BC + (V_1 - V_2) = 0

Substituting the values of V_1 and V_2, we get:

-15 + V_AB + V_BC = 0

This implies that V_AB and V_BC are equal and opposite in sign. Therefore, we can write:

V_AB = -V_BC

To find the value of V_AB, we can use Ohm's law. Ohm's law states that the voltage across a resistor is equal to the product of its resistance and the current flowing through it. Using Ohm's law for resistor R_AB, we get:

V_AB = I_AB x R_AB

Where I_AB is the current flowing through R_AB. To find I_AB, we can use Kirchhoff's current law (KCL). KCL states that the sum of the currents entering a node is equal to the sum of the currents leaving that node. Applying KCL at node A, we get:

I_s - I_A - I_AB = 0

Where I_s is the source current and I_A is the current flowing through R_A. Substituting the values of I_s and I_A, we get:

(V_s/(R_A + R_1)) - (V_1/R_A) - I_AB = 0

Solving for I_AB, we get:

I_AB = (V_s x R_1 - V_1 x (R_A + R_1))/(R_A x (R_A + R_1))

Substituting the values of V_s, V_1, R_A, and R_1, we get:

I_AB = (15 x 5 - 75/23 x (100/3 + 5))/(100/3 x (100/3 + 5)) = -25/161 A

The negative sign indicates that the current is flowing in the opposite direction of the assumed direction. Therefore, the voltage across R_AB is:

V_AB = I_AB x R_AB = (-25/161) x 10 = -25/16.1 V

Therefore, the voltage across each resistor in the original delta network is as follows:

### V_AB = -25/16.1 V (from B to A) V_BC = 25/16.1 V (from C to B) V_CA = 75/23 V (from A to C) Problem 4

Find the power dissipated by each resistor of the circuit shown below.

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