PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365152
The peak voltage in the output of a half-wave diode rectifier fed with a sinusoidal signal without filter is 10\(V\). The \(d\).\(c\). component of the output voltage is
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365153
The process of converting alternating current into direct current is known as
1 Modulation
2 Amplification
3 Detection
4 Rectification
Explanation:
Conceptual Question
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365154
In the circuit shown in figure the maximum output voltage \({V_0}\) is
1 \(5V\)
2 \(0\,V\)
3 \(\frac{5}{{\sqrt 2 }}V\)
4 \(10\,V\)
Explanation:
For the positive half cycle of input the resulting network is shown below \( \Rightarrow {\left( {{V_0}} \right)_{\max }} = \frac{1}{2}{\left( {{V_i}} \right)_{\max }} = \frac{1}{2} \times 10 = 5\,V\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365155
Assertion : The half-wave rectifier work only for positive half cycle of ac Reason : In half-wave rectifier only one diode is used.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
In the context of a half-wave rectifier, a single diode becomes biased only during the positive half of the alternating current (AC) cycle. During the negative half of the AC cycle, the diode becomes reverse-biased, resulting in no corresponding output. Hence, the designationas a "half-wave rectifier. So correct option is (1).
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365152
The peak voltage in the output of a half-wave diode rectifier fed with a sinusoidal signal without filter is 10\(V\). The \(d\).\(c\). component of the output voltage is
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365153
The process of converting alternating current into direct current is known as
1 Modulation
2 Amplification
3 Detection
4 Rectification
Explanation:
Conceptual Question
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365154
In the circuit shown in figure the maximum output voltage \({V_0}\) is
1 \(5V\)
2 \(0\,V\)
3 \(\frac{5}{{\sqrt 2 }}V\)
4 \(10\,V\)
Explanation:
For the positive half cycle of input the resulting network is shown below \( \Rightarrow {\left( {{V_0}} \right)_{\max }} = \frac{1}{2}{\left( {{V_i}} \right)_{\max }} = \frac{1}{2} \times 10 = 5\,V\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365155
Assertion : The half-wave rectifier work only for positive half cycle of ac Reason : In half-wave rectifier only one diode is used.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
In the context of a half-wave rectifier, a single diode becomes biased only during the positive half of the alternating current (AC) cycle. During the negative half of the AC cycle, the diode becomes reverse-biased, resulting in no corresponding output. Hence, the designationas a "half-wave rectifier. So correct option is (1).
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365152
The peak voltage in the output of a half-wave diode rectifier fed with a sinusoidal signal without filter is 10\(V\). The \(d\).\(c\). component of the output voltage is
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365153
The process of converting alternating current into direct current is known as
1 Modulation
2 Amplification
3 Detection
4 Rectification
Explanation:
Conceptual Question
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365154
In the circuit shown in figure the maximum output voltage \({V_0}\) is
1 \(5V\)
2 \(0\,V\)
3 \(\frac{5}{{\sqrt 2 }}V\)
4 \(10\,V\)
Explanation:
For the positive half cycle of input the resulting network is shown below \( \Rightarrow {\left( {{V_0}} \right)_{\max }} = \frac{1}{2}{\left( {{V_i}} \right)_{\max }} = \frac{1}{2} \times 10 = 5\,V\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365155
Assertion : The half-wave rectifier work only for positive half cycle of ac Reason : In half-wave rectifier only one diode is used.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
In the context of a half-wave rectifier, a single diode becomes biased only during the positive half of the alternating current (AC) cycle. During the negative half of the AC cycle, the diode becomes reverse-biased, resulting in no corresponding output. Hence, the designationas a "half-wave rectifier. So correct option is (1).
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365152
The peak voltage in the output of a half-wave diode rectifier fed with a sinusoidal signal without filter is 10\(V\). The \(d\).\(c\). component of the output voltage is
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365153
The process of converting alternating current into direct current is known as
1 Modulation
2 Amplification
3 Detection
4 Rectification
Explanation:
Conceptual Question
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365154
In the circuit shown in figure the maximum output voltage \({V_0}\) is
1 \(5V\)
2 \(0\,V\)
3 \(\frac{5}{{\sqrt 2 }}V\)
4 \(10\,V\)
Explanation:
For the positive half cycle of input the resulting network is shown below \( \Rightarrow {\left( {{V_0}} \right)_{\max }} = \frac{1}{2}{\left( {{V_i}} \right)_{\max }} = \frac{1}{2} \times 10 = 5\,V\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365155
Assertion : The half-wave rectifier work only for positive half cycle of ac Reason : In half-wave rectifier only one diode is used.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
In the context of a half-wave rectifier, a single diode becomes biased only during the positive half of the alternating current (AC) cycle. During the negative half of the AC cycle, the diode becomes reverse-biased, resulting in no corresponding output. Hence, the designationas a "half-wave rectifier. So correct option is (1).
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365152
The peak voltage in the output of a half-wave diode rectifier fed with a sinusoidal signal without filter is 10\(V\). The \(d\).\(c\). component of the output voltage is
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365153
The process of converting alternating current into direct current is known as
1 Modulation
2 Amplification
3 Detection
4 Rectification
Explanation:
Conceptual Question
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365154
In the circuit shown in figure the maximum output voltage \({V_0}\) is
1 \(5V\)
2 \(0\,V\)
3 \(\frac{5}{{\sqrt 2 }}V\)
4 \(10\,V\)
Explanation:
For the positive half cycle of input the resulting network is shown below \( \Rightarrow {\left( {{V_0}} \right)_{\max }} = \frac{1}{2}{\left( {{V_i}} \right)_{\max }} = \frac{1}{2} \times 10 = 5\,V\)
PHXII14:SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS
365155
Assertion : The half-wave rectifier work only for positive half cycle of ac Reason : In half-wave rectifier only one diode is used.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
In the context of a half-wave rectifier, a single diode becomes biased only during the positive half of the alternating current (AC) cycle. During the negative half of the AC cycle, the diode becomes reverse-biased, resulting in no corresponding output. Hence, the designationas a "half-wave rectifier. So correct option is (1).
