Question

in the circuit if the input voltage is 120 vp, what is vrect across the resistor.
select one:
a. 28.6 vp
b. 118.6 vp
c. 38.6 vp
d. 358.6 vp

Explanation:

Step1: Determine Transformer Output Voltage

The transformer has a turns ratio of \( 3:1 \) (primary to secondary? Wait, no—wait, the input is on the primary, and the turns ratio is \( 3:1 \), so if primary voltage \( V_p = 120 \, \text{V} \), the secondary voltage \( V_{sec} \) (peak) is calculated by the turns ratio. Wait, actually, the transformer's voltage ratio is \( \frac{V_{primary}}{V_{secondary}}=\frac{N_{primary}}{N_{secondary}} \). Wait, the turns ratio is \( 3:1 \), so if primary is 120 V (peak?), wait the input is 120 Vp (peak). Wait, the transformer: if the turns ratio is \( 3:1 \) (primary to secondary), then \( V_{secondary} = V_{primary} \times \frac{N_{secondary}}{N_{primary}} \)? Wait no, wait the transformer in the diagram: the primary is connected to 120 V, and the turns ratio is 3:1 (primary:secondary). Wait, no—wait, the transformer's voltage ratio is \( \frac{V_{pri}}{V_{sec}}=\frac{N_{pri}}{N_{sec}} \). So if \( N_{pri}:N_{sec}=3:1 \), then \( V_{sec} = V_{pri} \times \frac{N_{sec}}{N_{pri}} \)? Wait no, that would be a step - down, but the answer is 358.6, which is higher. Wait, maybe the turns ratio is secondary:primary = 3:1. So \( V_{sec}=V_{pri}\times\frac{N_{sec}}{N_{pri}} = 120\times3 = 360 \, \text{Vp} \) (approx, since 358.6 is close to 360 - 1.4, maybe considering diode drop? Wait, no, the rectifier is a full - wave bridge rectifier. Wait, no, first, the transformer: if the turns ratio is 3:1 (secondary to primary), so secondary voltage peak is \( 120\times3 = 360 \, \text{Vp} \). Then, for a full - wave bridge rectifier, the output voltage (Vrect) is approximately \( V_{sec(peak)} - 2V_d \), but if we ignore diode drops (or maybe the problem ignores them), but 358.6 is close to 360 - 1.4 (if two diodes are conducting at a time, each with 0.7 V drop, so \( 2\times0.7 = 1.4 \)). So \( 360 - 1.4 = 358.6 \, \text{Vp} \).

Step2: Analyze Rectifier Output

The circuit has a full - wave bridge rectifier. After the transformer, the secondary voltage is stepped up (since turns ratio 3:1, secondary has more turns). The full - wave bridge rectifier's output voltage (Vrect) is the peak voltage of the secondary minus the voltage drop across the diodes. If we calculate the secondary peak voltage as \( 120\times3 = 360 \, \text{Vp} \), and subtract the voltage drop across two diodes (assuming each diode has a drop of 0.7 V, \( 2\times0.7 = 1.4 \, \text{V} \)), we get \( 360 - 1.4 = 358.6 \, \text{Vp} \), which matches option d.

Answer:

d. 358.6 Vp