Question

If each branch of a Delta circuit has impedance $\sqrt{3}\,Z$, then each branch of the equivalent Wye circuit has impedance

• A. $\dfrac{Z}{\sqrt{3}}$
• B. $3Z$
• C. $3\sqrt{3}\,Z$
• D. $\dfrac{Z}{3}$

Hint:

For balanced networks, remember: $\; Z_Y = \dfrac{Z_{\Delta}}{3}$. This formula directly helps in quick Delta–Wye conversions.

Answer 1

The Correct Option is A

Step 1: The standard delta-to-wye relation for a symmetric network is \(Z_Y = Z_\Delta/3\).
Step 2: Substitute the given branch impedance \(Z_\Delta = \sqrt{3}\,Z\): \(Z_Y = \dfrac{\sqrt{3}\,Z}{3}\).
Step 3: Simplify by rationalising: \(\dfrac{\sqrt{3}}{3} = \dfrac{1}{\sqrt{3}}\).
\[ \boxed{Z_Y = \frac{Z}{\sqrt{3}}} \]

Answer 2

Plugging in a concrete numerical value for \(Z\) makes it easy to check each option against the known \(Z_Y = Z_\Delta/3\) relation. Let \(Z = 3\,\Omega\), so \(Z_\Delta = \sqrt{3}\times 3 = 3\sqrt{3} \approx 5.196\,\Omega\). The wye equivalent should then be \(Z_Y = Z_\Delta/3 = \sqrt{3} \approx 1.732\,\Omega\).

1. \(Z/\sqrt{3}\): With \(Z=3\), this gives \(3/\sqrt{3} = \sqrt{3} \approx 1.732\,\Omega\), matching the expected wye impedance exactly.
2. \(3Z\): With \(Z=3\), this gives \(9\,\Omega\), far larger than the expected \(1.732\,\Omega\).
3. \(3\sqrt{3}\,Z\): With \(Z=3\), this gives \(9\sqrt{3}\approx 15.6\,\Omega\), also far too large.
4. \(Z/3\): With \(Z=3\), this gives \(1\,\Omega\), close in order of magnitude but not equal to the required \(1.732\,\Omega\).

Therefore, the correct answer is \(Z/\sqrt{3}\).