Question

Short-circuit test on a single-phase transformer at 1 kHz gives: wattmeter \(P=8\) W, primary current \(I=2\) A, primary voltage \(V=6\) V. Assume negligible no-load loss/current and linear core. With the secondary shorted, the primary is now fed from a \(2\) V\(_{\mathrm{RMS}}\), 1 kHz source in series with a capacitor of value \( \dfrac{1}{2\pi\sqrt{5}} \) mF. The primary RMS current is _____A (rounded off to two decimals).

Hint:

From SC test: \(R=P/I^2\), \(|Z|=V/I\), \(X=\sqrt{|Z|^2-R^2}\). Choosing \(X_C=X_L\) at the same frequency cancels reactance, leaving a purely resistive path.

Answer 1

Correct Answer: 0.95

Step 1: Extract series parameters from the SC test
\(|Z_{\text{eq}}|=V/I=6/2=3~\Omega\). \(R_{\text{eq}}=P/I^2=8/4=2~\Omega\).
\(X_L=\sqrt{|Z|^2-R^2}=\sqrt{9-4}=\sqrt{5}=2.236~\Omega\) at \(1\) kHz.
Step 2: Series capacitor at 1 kHz
Given \(C=\dfrac{1}{2\pi\sqrt{5}}\,\text{mF}\Rightarrow X_C=\dfrac{1}{\omega C}=\sqrt{5}~\Omega\).
Thus the net reactance with the transformer (inductive) is \(X_L-X_C=\sqrt{5}-\sqrt{5}=0\) \(\Rightarrow\) purely resistive.
Step 3: Current with \(2\) V source
Total series resistance \(=R_{\text{eq}}=2~\Omega\). Hence \(I=\dfrac{V}{R}= \dfrac{2}{2}=1.00\ \text{A}\).