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: Recall Delta to Wye conversion formula.
In a Delta ($\Delta$) connected circuit, if each branch has impedance $Z_{\Delta}$, then the equivalent Wye (Y) impedance $Z_Y$ is given by:
\[ Z_Y = \frac{Z_{\Delta}}{3} \]
Step 2: Substitute the given value.
It is given that each branch of the Delta circuit has impedance:
\[ Z_{\Delta} = \sqrt{3}\,Z \]
Substituting into the formula:
\[ Z_Y = \frac{\sqrt{3}\,Z}{3} \]
Step 3: Simplify the expression.
\[ Z_Y = \frac{Z}{\sqrt{3}} \]
Step 4: Match with the given options.
The obtained result matches option (A).
Step 5: Final conclusion.
Hence, each branch of the equivalent Wye circuit has impedance $\dfrac{Z}{\sqrt{3}}$.

Answer 2

Rather than quoting the standard \(Z_Y = Z_\Delta/3\) relation directly, it can be rebuilt from the requirement that the impedance measured between any two terminals must be the same whether the network is wired in delta or in an equivalent wye.

For a symmetric delta with each branch \(Z_\Delta\), the impedance between any two terminals is \(Z_\Delta \parallel (2Z_\Delta) = \dfrac{Z_\Delta \times 2Z_\Delta}{3Z_\Delta} = \dfrac{2Z_\Delta}{3}\) (one branch directly between the terminals, in parallel with the series combination of the other two). For a symmetric wye with each branch \(Z_Y\), the impedance between any two terminals is simply \(2Z_Y\) (two branches in series, through the star point). Equating the two: \[ 2Z_Y = \frac{2Z_\Delta}{3} \quad\Rightarrow\quad Z_Y = \frac{Z_\Delta}{3} \] Substituting \(Z_\Delta = \sqrt{3}\,Z\): \[ Z_Y = \frac{\sqrt{3}\,Z}{3} = \frac{Z}{\sqrt{3}} \]

1. \(Z/\sqrt{3}\): This matches exactly what the terminal-impedance matching gives.
2. \(3Z\): This would result from inverting the relationship, i.e. converting wye to delta instead of delta to wye.
3. \(3\sqrt{3}\,Z\): This is far larger than what terminal-impedance matching produces and does not correspond to any standard conversion direction.
4. \(Z/3\): This would be correct if \(Z_\Delta\) were simply \(Z\) rather than \(\sqrt{3}\,Z\); the extra factor of \(\sqrt{3}\) in the given delta impedance changes the final simplified form.

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