Question

A Delta-connected network with its Wye-equivalent is shown in the figure. The values of resistances \(R_1\), \(R_2\) and \(R_3\) (in \(\Omega\)) are respectively.

• A. \(1.5\Omega,\ 3\Omega \text{ and } 9\Omega\)
• B. \(3\Omega,\ 9\Omega \text{ and } 1.5\Omega\)
• C. \(9\Omega,\ 3\Omega \text{ and } 1.5\Omega\)
• D. \(3\Omega,\ 1.5\Omega \text{ and } 9\Omega\)

Hint:

For Delta-to-Wye conversion: \[ R_{\text{Wye arm}}= \frac{\text{Product of adjacent Delta resistors}} {\text{Sum of all Delta resistors}} \] Always use the resistor pair connected to the corresponding terminal.

Answer 1

The Correct Option is D

Concept: Delta-to-Wye transformation is used to simplify resistor networks. For a delta network: \[ R_{ab},\ R_{bc},\ R_{ca} \] the equivalent wye resistances are: \[ R_a=\frac{R_{ab}R_{ca}}{R_{ab}+R_{bc}+R_{ca}} \] \[ R_b=\frac{R_{ab}R_{bc}}{R_{ab}+R_{bc}+R_{ca}} \] \[ R_c=\frac{R_{bc}R_{ca}}{R_{ab}+R_{bc}+R_{ca}} \]

Step 1: Identify delta resistances from the figure. From the given delta network: \[ R_{ab}=5\Omega \] \[ R_{bc}=15\Omega \] \[ R_{ca}=30\Omega \] Total sum: \[ S=5+15+30 \] \[ S=50\Omega \]

Step 2: Calculate \(R_1\). \(R_1\) is connected to terminal \(a\). Thus: \[ R_1=\frac{R_{ab}R_{ca}}{S} \] Substituting: \[ R_1=\frac{5\times30}{50} \] \[ R_1=\frac{150}{50} \] \[ R_1=3\Omega \]

Step 3: Calculate \(R_2\). \(R_2\) is connected to terminal \(b\). Therefore: \[ R_2=\frac{R_{ab}R_{bc}}{S} \] Substituting: \[ R_2=\frac{5\times15}{50} \] \[ R_2=\frac{75}{50} \] \[ R_2=1.5\Omega \]

Step 4: Calculate \(R_3\). \(R_3\) is connected to terminal \(c\). Thus: \[ R_3=\frac{R_{bc}R_{ca}}{S} \] Substituting: \[ R_3=\frac{15\times30}{50} \] \[ R_3=\frac{450}{50} \] \[ R_3=9\Omega \]

Step 5: Write the final answer. Therefore: \[ R_1=3\Omega \] \[ R_2=1.5\Omega \] \[ R_3=9\Omega \] Hence, the correct option is: \[ \boxed{(D)\ 3\Omega,\ 1.5\Omega \text{ and } 9\Omega} \]