Question

In the circuit shown, the \(n:1\) step-down transformer and the diodes are ideal. The diodes have no voltage drop in forward-biased condition. If the input voltage (in Volts) is \(V_s(t) = 10\sin\omega t\) and the average value of load voltage \(V_L(t)\) (in Volts) is \(2.5/\pi\), the value of \(n\) is \(\_\_\_\_\). \begin{center} \includegraphics[width=5cm]{23.png} \end{center}

• A. 4
• B. 12
• C. 16
• D.

Hint:

For rectifier circuits involving transformers, calculate the turns ratio by analyzing the relationship between the input and rectified output voltages, and remember the average voltage formula for rectified signals.

Answer 1

The Correct Option is A

Step 1: Analyze the circuit functionality.
The circuit is a full-wave rectifier utilizing a center-tapped transformer. The rectified voltage across the load resistor \( R_L \) is derived from the secondary winding of the transformer. Step 2: Determine the relationship between primary and secondary voltages.
The primary voltage is given as: \[ V_s(t) = 10 \sin\omega t. \] The secondary voltage \( V_{sec}(t) \) is scaled by the transformer turns ratio \( n:1 \), such that: \[ V_{sec}(t) = \frac{10}{n} \sin\omega t. \] Step 3: Rectified output voltage.
In a full-wave rectifier, the output is the absolute value of the secondary voltage: \[ V_L(t) = \left| \frac{10}{n} \sin\omega t \right|. \] Step 4: Compute the average output voltage.
For a full-wave rectified sine wave, the average voltage is: \[ V_{{avg}} = \frac{2V_{{peak}}}{\pi}. \] Substituting \( V_{{peak}} = \frac{10}{n} \), we get: \[ V_{{avg}} = \frac{2}{\pi} \cdot \frac{10}{n}. \] Step 5: Solve for \(n\).
Given that \( V_{{avg}} = \frac{2.5}{\pi} \), equate and solve for \(n\): \[ \frac{2}{\pi} \cdot \frac{10}{n} = \frac{2.5}{\pi}. \] Simplify: \[ \frac{10}{n} = 2.5. \] \[n = \frac{10}{2.5} = 4.\] The transformer turns ratio \(n\) is therefore 4, which matches option (1). Final Answer: \[\boxed{4}\]