Basic Electrical Engineering Questions and Answers – Bandwidth

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This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Bandwidth”.

1. The SI unit for bandwidth is?
a) Hz
b) Watt
c) kHz
d) kW
View Answer

Answer: a
Explanation: The SI unit for bandwidth is Hz. Hertz is the SI unit because bandwidth is basically frequency and the unit for frequency is Hz.
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2. At bandwidth frequency range, the value of the current I is?
a) I=Im/2
b) I=Im2
c) I=Im
d) I=Im/√2
View Answer

Answer: d
Explanation: At the bandwidth frequency range, the value of the current is equal to the maximum value of current divided by √2.

3. At bandwidth frequency range, the value of the voltage V is?
a) V=Vm/2
b) V=Vm2
c) V=Vm
d) V=Vm/√2
View Answer

Answer: d
Explanation: At the bandwidth frequency range, the value of the voltage is equal to the maximum value of voltage divided by √2.
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4. At resonance, bandwidth includes the frequency range that allows _____ percent of the maximum current to flow.
a) 33.33
b) 66.67
c) 50
d) 70.7
View Answer

Answer: d
Explanation: At resonance, bandwidth includes the frequency range that allows 70.2 percent of the maximum current to flow. This is because, in the bandwidth frequency range, the value of the current is equal to the maximum value of current divided by √2.

5. At resonance, bandwidth includes the frequency range that allows _____ percent of the maximum voltage to flow.
a) 33.33
b) 66.67
c) 50
d) 70.7
View Answer

Answer: d
Explanation: At resonance, bandwidth includes the frequency range that allows 70.2 percent of the maximum voltage to flow. This is because, in the bandwidth frequency range, the value of the voltage is equal to the maximum value of voltage divided by √2.
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6. Find the value of current in the bandwidth range when the maximum value of current is 50A.
a) 56.65A
b) 35.36A
c) 45.34A
d) 78.76A
View Answer

Answer: b
Explanation: At the bandwidth frequency range, the value of the current is equal to the maximum value of current divided by √2. Hence I =50/√2= 35.36A.

7. Find the value of voltage in the bandwidth range when the maximum value of voltage is 100 V.
a) 56.65 V
b) 35.36 V
c) 45.34 V
d) 70.72 V
View Answer

Answer: d
Explanation: At the bandwidth frequency range, the value of the voltage is equal to the maximum value of voltage divided by √2. Hence V =100/√2= 70.72V.
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8. Bandwidth is the difference of_____________________ frequencies.
a) half power
b) full power
c) double power
d) wattless
View Answer

Answer: a
Explanation: Current for the end frequencies of bandwidth is 1/√2 times the maximum current. So, power at the end frequencies of bandwidth is half the maximum power. So, bandwidth is the difference of half-power frequencies.

9. For a sharp resonance, bandwidth is ______________
a) low
b) high
c) zero
d) infinity
View Answer

Answer: a
Explanation: For sharp resonance quality factor is high and the quality factor is inversely proportional to bandwidth so bandwidth is low for sharp resonance.
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10. Current is maximum at __________ frequency of bandwidth.
a) left end
b) middle
c) right end
d) all end
View Answer

Answer: b
Explanation: Current will be maximum at a frequency that is in the middle of bandwidth.
On both sides, it decreases and is 1/√2 times the maximum current at the ends of bandwidth.

Sanfoundry Global Education & Learning Series – Basic Electrical Engineering.

To practice all areas of Basic Electrical Engineering, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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