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Answer :
Sure, let's walk through the solution for this question step-by-step.
1. Understanding the Full-Wave Bridge Rectifier:
- In a full-wave bridge rectifier, an AC voltage is converted to DC voltage.
- The output DC voltage is derived from the peak value of the input AC voltage after rectification.
2. Peak Value of the Input Voltage:
- The input given is 120 VAC (root mean square, RMS value).
- The peak value [tex]\( V_{peak} \)[/tex] of an AC voltage is calculated using the formula:
[tex]\[
V_{peak} = \sqrt{2} \times V_{AC_{RMS}}
\][/tex]
- So, for 120 VAC:
[tex]\[
V_{peak} = \sqrt{2} \times 120 \approx 169.7 \text{V}
\][/tex]
3. Diode Voltage Drop in the Bridge Rectifier:
- Each diode in the bridge rectifier has a voltage drop, typically 0.7V for silicon diodes.
- There are 2 diodes in the conduction path during each half-cycle.
- Thus, the total voltage drop due to the diodes is:
[tex]\[
2 \times 0.7 = 1.4 \text{V}
\][/tex]
4. Calculating the DC Output Voltage:
- The DC output voltage [tex]\( V_{DC} \)[/tex] is the peak voltage minus the total diode voltage drop:
[tex]\[
V_{DC} = V_{peak} - 1.4
\][/tex]
- Substituting the calculated peak voltage:
[tex]\[
V_{DC} = 169.7 - 1.4 \approx 168.3 \text{V}
\][/tex]
5. Final Answer:
- Since the question asks for an approximate value, we can round 168.3 V to the nearest whole number, which is 168 V.
So, the correct answer to the question is:
[tex]\[ \boxed{D. \text{Approximately 170 VDC}} \][/tex]
1. Understanding the Full-Wave Bridge Rectifier:
- In a full-wave bridge rectifier, an AC voltage is converted to DC voltage.
- The output DC voltage is derived from the peak value of the input AC voltage after rectification.
2. Peak Value of the Input Voltage:
- The input given is 120 VAC (root mean square, RMS value).
- The peak value [tex]\( V_{peak} \)[/tex] of an AC voltage is calculated using the formula:
[tex]\[
V_{peak} = \sqrt{2} \times V_{AC_{RMS}}
\][/tex]
- So, for 120 VAC:
[tex]\[
V_{peak} = \sqrt{2} \times 120 \approx 169.7 \text{V}
\][/tex]
3. Diode Voltage Drop in the Bridge Rectifier:
- Each diode in the bridge rectifier has a voltage drop, typically 0.7V for silicon diodes.
- There are 2 diodes in the conduction path during each half-cycle.
- Thus, the total voltage drop due to the diodes is:
[tex]\[
2 \times 0.7 = 1.4 \text{V}
\][/tex]
4. Calculating the DC Output Voltage:
- The DC output voltage [tex]\( V_{DC} \)[/tex] is the peak voltage minus the total diode voltage drop:
[tex]\[
V_{DC} = V_{peak} - 1.4
\][/tex]
- Substituting the calculated peak voltage:
[tex]\[
V_{DC} = 169.7 - 1.4 \approx 168.3 \text{V}
\][/tex]
5. Final Answer:
- Since the question asks for an approximate value, we can round 168.3 V to the nearest whole number, which is 168 V.
So, the correct answer to the question is:
[tex]\[ \boxed{D. \text{Approximately 170 VDC}} \][/tex]
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