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 \(\_\_\_\_\).
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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.
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:
\[\boxed{4}\]
