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A series RLC circuit has Q of 100 and an impendence of (100 ± j0)Ω at its resonant angular frequency of 107 rad/s. The value of R and L respectively –
- a)100 Ω, 10-3H
- b)10 Ω, 102 H
- c)1000 Ω, 10 H
- d)100 Ω, 100 H
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
A series RLC circuit has Q of 100 and an impendence of (100 ± j...
ω = 107 rad/s
Q = 100
R = 100 Ω
for resonance condition,
⇒ LC = 10-14 ----(2)
By (1) and (2)
⇒ L = 10-3 H
Most Upvoted Answer
A series RLC circuit has Q of 100 and an impendence of (100 ± j...
+j200) ohms. Determine the values of R, L, and C.
We can start by using the formula for Q:
Q = R/(sqrt(L/C))
We know that Q = 100, so we can substitute that in:
100 = R/(sqrt(L/C))
We also know that the impedance is (100+j200) ohms. The impedance formula for a series RLC circuit is:
Z = R + j(wL - 1/(wC))
where w is the angular frequency. In this case, we can assume that the circuit is at resonance, so wL = 1/(wC). This simplifies the formula to:
Z = R
Since we know that the impedance is (100+j200) ohms, we can equate that to R:
R = 100
Now we can solve for L and C. From the impedance formula, we have:
100 + j(wL - 1/(wC)) = 100 + j200
This simplifies to:
wL - 1/(wC) = 200
We also know that wL = 1/(wC), so we can substitute that in:
1/(wC) - 1/(wC) = 200
0 = 200
This is obviously not true, so there must be an error in our assumptions. We assumed that the circuit is at resonance, but the given impedance does not match the impedance at resonance. We can try using the formula for impedance at resonance:
Z = R/(sqrt(1 - Q^2))
We know that Q = 100 and Z = (100+j200) ohms, so we can substitute that in:
(100+j200) = R/(sqrt(1 - 100^2))
Squaring both sides and simplifying, we get:
R = 10000 ohms
Now we can use the formula for impedance at resonance to solve for L and C:
Z = R = 10000 ohms
Z = sqrt(L/C)
L/C = (10000)^2
We need another equation to solve for L and C. We can use the formula for resonant frequency:
w = 1/sqrt(LC)
We know that the circuit is at resonance, so we can substitute in the given impedance:
w = sqrt(1/LC)
Now we have two equations with two unknowns:
L/C = (10000)^2
w = sqrt(1/LC)
We can solve for L and C by substituting w into the second equation:
w = sqrt(1/LC)
w^2 = 1/LC
LC = 1/w^2
Substituting L/C = (10000)^2, we get:
L = (10000)^2/w^2
C = 1/L(w^2)
Substituting w = sqrt(1/LC), we get:
L = (10000)^2/(1/LC)
C = L/(w^2)
Simplifying, we get:
L = 0.025 H
C = 10 nF
Therefore, the values of R, L, and C are:
R = 10000 ohms
L = 0.025 H
C = 10 nF
We can start by using the formula for Q:
Q = R/(sqrt(L/C))
We know that Q = 100, so we can substitute that in:
100 = R/(sqrt(L/C))
We also know that the impedance is (100+j200) ohms. The impedance formula for a series RLC circuit is:
Z = R + j(wL - 1/(wC))
where w is the angular frequency. In this case, we can assume that the circuit is at resonance, so wL = 1/(wC). This simplifies the formula to:
Z = R
Since we know that the impedance is (100+j200) ohms, we can equate that to R:
R = 100
Now we can solve for L and C. From the impedance formula, we have:
100 + j(wL - 1/(wC)) = 100 + j200
This simplifies to:
wL - 1/(wC) = 200
We also know that wL = 1/(wC), so we can substitute that in:
1/(wC) - 1/(wC) = 200
0 = 200
This is obviously not true, so there must be an error in our assumptions. We assumed that the circuit is at resonance, but the given impedance does not match the impedance at resonance. We can try using the formula for impedance at resonance:
Z = R/(sqrt(1 - Q^2))
We know that Q = 100 and Z = (100+j200) ohms, so we can substitute that in:
(100+j200) = R/(sqrt(1 - 100^2))
Squaring both sides and simplifying, we get:
R = 10000 ohms
Now we can use the formula for impedance at resonance to solve for L and C:
Z = R = 10000 ohms
Z = sqrt(L/C)
L/C = (10000)^2
We need another equation to solve for L and C. We can use the formula for resonant frequency:
w = 1/sqrt(LC)
We know that the circuit is at resonance, so we can substitute in the given impedance:
w = sqrt(1/LC)
Now we have two equations with two unknowns:
L/C = (10000)^2
w = sqrt(1/LC)
We can solve for L and C by substituting w into the second equation:
w = sqrt(1/LC)
w^2 = 1/LC
LC = 1/w^2
Substituting L/C = (10000)^2, we get:
L = (10000)^2/w^2
C = 1/L(w^2)
Substituting w = sqrt(1/LC), we get:
L = (10000)^2/(1/LC)
C = L/(w^2)
Simplifying, we get:
L = 0.025 H
C = 10 nF
Therefore, the values of R, L, and C are:
R = 10000 ohms
L = 0.025 H
C = 10 nF
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Community Answer
A series RLC circuit has Q of 100 and an impendence of (100 ± j...
At series resonance the value of impedance is R so the value of R is 100ohm .
now we know Q= resonance frequency* L/R
100 = 10^7*L/100
100*100/10^7=L
L=10^-3
So the value R and L is 100 ohms and 10^-3 Henry respectively.
now we know Q= resonance frequency* L/R
100 = 10^7*L/100
100*100/10^7=L
L=10^-3
So the value R and L is 100 ohms and 10^-3 Henry respectively.
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Question Description
A series RLC circuit has Q of 100 and an impendence of (100 ± j0)Ω at its resonant angular frequency of 107rad/s. The value of R and L respectively –a)100 Ω, 10-3Hb)10 Ω, 102Hc)1000 Ω, 10 Hd)100 Ω, 100 HCorrect answer is option 'A'. Can you explain this answer? for Electrical Engineering (EE) 2026 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A series RLC circuit has Q of 100 and an impendence of (100 ± j0)Ω at its resonant angular frequency of 107rad/s. The value of R and L respectively –a)100 Ω, 10-3Hb)10 Ω, 102Hc)1000 Ω, 10 Hd)100 Ω, 100 HCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A series RLC circuit has Q of 100 and an impendence of (100 ± j0)Ω at its resonant angular frequency of 107rad/s. The value of R and L respectively –a)100 Ω, 10-3Hb)10 Ω, 102Hc)1000 Ω, 10 Hd)100 Ω, 100 HCorrect answer is option 'A'. Can you explain this answer?.
A series RLC circuit has Q of 100 and an impendence of (100 ± j0)Ω at its resonant angular frequency of 107rad/s. The value of R and L respectively –a)100 Ω, 10-3Hb)10 Ω, 102Hc)1000 Ω, 10 Hd)100 Ω, 100 HCorrect answer is option 'A'. Can you explain this answer? for Electrical Engineering (EE) 2026 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A series RLC circuit has Q of 100 and an impendence of (100 ± j0)Ω at its resonant angular frequency of 107rad/s. The value of R and L respectively –a)100 Ω, 10-3Hb)10 Ω, 102Hc)1000 Ω, 10 Hd)100 Ω, 100 HCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A series RLC circuit has Q of 100 and an impendence of (100 ± j0)Ω at its resonant angular frequency of 107rad/s. The value of R and L respectively –a)100 Ω, 10-3Hb)10 Ω, 102Hc)1000 Ω, 10 Hd)100 Ω, 100 HCorrect answer is option 'A'. Can you explain this answer?.
Solutions for A series RLC circuit has Q of 100 and an impendence of (100 ± j0)Ω at its resonant angular frequency of 107rad/s. The value of R and L respectively –a)100 Ω, 10-3Hb)10 Ω, 102Hc)1000 Ω, 10 Hd)100 Ω, 100 HCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE). Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free.
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