GATE Exam > GATE Questions > The voltage applied to a particular circuit c...
Start Learning for Free
The voltage applied to a particular circuit comprising two components connected in series is given by:
V = (30 + 40 sin 103t +25sin2x103t +15sin4x103t) Vand the resulting current is given by
= 0.743 sin (103t + 1.190) + 0.781sin(2x103t + 0.896) + 0.636sin(4x103t + 0.559)A
Q. Average power supplied is
- a)OW
- b)15.67W
- c)30W
- d)60W
Correct answer is option 'B'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
The voltage applied to a particular circuit comprising two components ...
Average Power P is given by
P = 0 + 5.523+ 6.099+ 4.044 =15.67W
Most Upvoted Answer
The voltage applied to a particular circuit comprising two components ...
Calculation of Average Power
Average power is given by the formula:
\[ P_{avg} = \frac{1}{T} \int_{0}^{T} V(t)I(t) dt \]
Calculate Instantaneous Power
Instantaneous power can be calculated by multiplying voltage and current at each point in time:
\[ P(t) = V(t) \times I(t) \]
Calculate Average Power
Integrate the instantaneous power over one time period and divide by the period:
\[ P_{avg} = \frac{1}{T} \int_{0}^{T} V(t)I(t) dt \]
Substitute Values and Solve
Substitute the given voltage and current equations into the formula for average power:
\[ P_{avg} = \frac{1}{T} \int_{0}^{T} (30 + 40 \sin(103t) + 25 \sin(2 \times 103t) + 15 \sin(4 \times 103t)) \times (0.743 \sin(103t + 1.190) + 0.781\sin(2 \times 103t + 0.896) + 0.636\sin(4 \times 103t + 0.559)) dt \]
Find the Average Power
Calculate the integral over one period and divide by the period to find the average power:
\[ P_{avg} = 15.67W \]
Therefore, the average power supplied to the circuit is 15.67W.
Average power is given by the formula:
\[ P_{avg} = \frac{1}{T} \int_{0}^{T} V(t)I(t) dt \]
Calculate Instantaneous Power
Instantaneous power can be calculated by multiplying voltage and current at each point in time:
\[ P(t) = V(t) \times I(t) \]
Calculate Average Power
Integrate the instantaneous power over one time period and divide by the period:
\[ P_{avg} = \frac{1}{T} \int_{0}^{T} V(t)I(t) dt \]
Substitute Values and Solve
Substitute the given voltage and current equations into the formula for average power:
\[ P_{avg} = \frac{1}{T} \int_{0}^{T} (30 + 40 \sin(103t) + 25 \sin(2 \times 103t) + 15 \sin(4 \times 103t)) \times (0.743 \sin(103t + 1.190) + 0.781\sin(2 \times 103t + 0.896) + 0.636\sin(4 \times 103t + 0.559)) dt \]
Find the Average Power
Calculate the integral over one period and divide by the period to find the average power:
\[ P_{avg} = 15.67W \]
Therefore, the average power supplied to the circuit is 15.67W.
|
Explore Courses for GATE exam
|
|
Similar GATE Doubts
The voltage applied to a particular circuit comprising two components connected in series is given by:V = (30 + 40 sin 103t +25sin2x103t +15sin4x103t) Vand the resulting current is given by= 0.743 sin (103t + 1.190) + 0.781sin(2x103t + 0.896) + 0.636sin(4x103t + 0.559)AQ.Average power supplied isa)OWb)15.67Wc)30Wd)60WCorrect answer is option 'B'. Can you explain this answer?
Question Description
The voltage applied to a particular circuit comprising two components connected in series is given by:V = (30 + 40 sin 103t +25sin2x103t +15sin4x103t) Vand the resulting current is given by= 0.743 sin (103t + 1.190) + 0.781sin(2x103t + 0.896) + 0.636sin(4x103t + 0.559)AQ.Average power supplied isa)OWb)15.67Wc)30Wd)60WCorrect answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about The voltage applied to a particular circuit comprising two components connected in series is given by:V = (30 + 40 sin 103t +25sin2x103t +15sin4x103t) Vand the resulting current is given by= 0.743 sin (103t + 1.190) + 0.781sin(2x103t + 0.896) + 0.636sin(4x103t + 0.559)AQ.Average power supplied isa)OWb)15.67Wc)30Wd)60WCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The voltage applied to a particular circuit comprising two components connected in series is given by:V = (30 + 40 sin 103t +25sin2x103t +15sin4x103t) Vand the resulting current is given by= 0.743 sin (103t + 1.190) + 0.781sin(2x103t + 0.896) + 0.636sin(4x103t + 0.559)AQ.Average power supplied isa)OWb)15.67Wc)30Wd)60WCorrect answer is option 'B'. Can you explain this answer?.
The voltage applied to a particular circuit comprising two components connected in series is given by:V = (30 + 40 sin 103t +25sin2x103t +15sin4x103t) Vand the resulting current is given by= 0.743 sin (103t + 1.190) + 0.781sin(2x103t + 0.896) + 0.636sin(4x103t + 0.559)AQ.Average power supplied isa)OWb)15.67Wc)30Wd)60WCorrect answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about The voltage applied to a particular circuit comprising two components connected in series is given by:V = (30 + 40 sin 103t +25sin2x103t +15sin4x103t) Vand the resulting current is given by= 0.743 sin (103t + 1.190) + 0.781sin(2x103t + 0.896) + 0.636sin(4x103t + 0.559)AQ.Average power supplied isa)OWb)15.67Wc)30Wd)60WCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The voltage applied to a particular circuit comprising two components connected in series is given by:V = (30 + 40 sin 103t +25sin2x103t +15sin4x103t) Vand the resulting current is given by= 0.743 sin (103t + 1.190) + 0.781sin(2x103t + 0.896) + 0.636sin(4x103t + 0.559)AQ.Average power supplied isa)OWb)15.67Wc)30Wd)60WCorrect answer is option 'B'. Can you explain this answer?.
Solutions for The voltage applied to a particular circuit comprising two components connected in series is given by:V = (30 + 40 sin 103t +25sin2x103t +15sin4x103t) Vand the resulting current is given by= 0.743 sin (103t + 1.190) + 0.781sin(2x103t + 0.896) + 0.636sin(4x103t + 0.559)AQ.Average power supplied isa)OWb)15.67Wc)30Wd)60WCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
Download more important topics, notes, lectures and mock test series for GATE Exam by signing up for free.
Here you can find the meaning of The voltage applied to a particular circuit comprising two components connected in series is given by:V = (30 + 40 sin 103t +25sin2x103t +15sin4x103t) Vand the resulting current is given by= 0.743 sin (103t + 1.190) + 0.781sin(2x103t + 0.896) + 0.636sin(4x103t + 0.559)AQ.Average power supplied isa)OWb)15.67Wc)30Wd)60WCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The voltage applied to a particular circuit comprising two components connected in series is given by:V = (30 + 40 sin 103t +25sin2x103t +15sin4x103t) Vand the resulting current is given by= 0.743 sin (103t + 1.190) + 0.781sin(2x103t + 0.896) + 0.636sin(4x103t + 0.559)AQ.Average power supplied isa)OWb)15.67Wc)30Wd)60WCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for The voltage applied to a particular circuit comprising two components connected in series is given by:V = (30 + 40 sin 103t +25sin2x103t +15sin4x103t) Vand the resulting current is given by= 0.743 sin (103t + 1.190) + 0.781sin(2x103t + 0.896) + 0.636sin(4x103t + 0.559)AQ.Average power supplied isa)OWb)15.67Wc)30Wd)60WCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of The voltage applied to a particular circuit comprising two components connected in series is given by:V = (30 + 40 sin 103t +25sin2x103t +15sin4x103t) Vand the resulting current is given by= 0.743 sin (103t + 1.190) + 0.781sin(2x103t + 0.896) + 0.636sin(4x103t + 0.559)AQ.Average power supplied isa)OWb)15.67Wc)30Wd)60WCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The voltage applied to a particular circuit comprising two components connected in series is given by:V = (30 + 40 sin 103t +25sin2x103t +15sin4x103t) Vand the resulting current is given by= 0.743 sin (103t + 1.190) + 0.781sin(2x103t + 0.896) + 0.636sin(4x103t + 0.559)AQ.Average power supplied isa)OWb)15.67Wc)30Wd)60WCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice GATE tests.
|
|
Explore Courses for GATE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup