Topic: Network Theory
A resistor $R1=4\text{ k}\Omega$ is connected across the secondary of transformer for which $L1=0.2H, L2=10H$ and flux coupling coefficient $k=0.5$ Find the peak voltage across resistor R1 when 250V peak voltage at $400\text{ rad/sec}$ is applied to primary winding of transformer.
- A. 353.55 V
- B. 500 V
- C. 707 V
- D. 1000 V
Answer: C. 707 V
Explanation: The mutual inductance is $M = k \sqrt{L1 L2} = 0.5 \sqrt{0.2 \times 10} = 0.5 \sqrt{2} = 0.707$ H. Solving the coupled circuit equations, the ratio of the secondary voltage magnitude to primary voltage magnitude is $\vert{}V_2/V_1\vert{} = \frac{\omega M R_1}{\sqrt{(-\omega^2 L_1 L_2 (1-k^2))^2 + (\omega L_1 R_1)^2}}$. Plugging in $\omega = 400$, $R_1 = 4000$, and the component values, the ratio $\vert{}V_2/V_1\vert{}$ simplifies precisely to $2\sqrt{2}$. For a primary peak voltage of 250 V, the secondary peak voltage is $250 \times 2\sqrt{2} \approx 707$ V.