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Two magnetically uncoupled inductive coils have Q factors q1 and q2 at the chosen operating frequency. Their respective resistances are R1 and R2. When connected in series, their effective Q factor at the same operating frequency is
- a)q1+q2
- b)(1/q1)+(1/q2)
- c)(q1R1+q2R2)/(R1+R2)
- d)(q1R2+q2R1)/(R1+R2)
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Two magnetically uncoupled inductive coils have Q factors q1 and q2 at...
When such two coils are connected in series individual inductances and
resistances are added.
resistances are added.
Most Upvoted Answer
Two magnetically uncoupled inductive coils have Q factors q1 and q2 at...
Solution:
Formula Used:
Q = ωL/R, where Q is the Q-factor, ω is the operating frequency, L is the inductance, and R is the resistance
Calculation:
1. When the two coils are magnetically uncoupled, their effective Q factor will be the sum of their individual Q factors.
Qeff = q1 + q2
2. When the two coils are connected in series, their effective inductance will be the sum of their individual inductances.
Leff = L1 + L2
3. The effective resistance of the two coils connected in series will be the sum of their individual resistances.
Reff = R1 + R2
4. The effective Q factor of the two coils connected in series can be calculated using the formula:
Qeff = ωLeff/Reff
Substituting the values of Leff and Reff from step 2 and 3, we get:
Qeff = ω(L1 + L2)/(R1 + R2)
5. Using the formula Q = ωL/R, we can express the individual inductances L1 and L2 in terms of their respective Q factors and resistances:
L1 = Q1R1/ω and L2 = Q2R2/ω
Substituting these values in step 4, we get:
Qeff = ω(Q1R1 + Q2R2)/(R1 + R2)
6. Simplifying the expression in step 5, we get:
Qeff = (Q1R1Q2R2ω)/(R1Q2R2 + R2Q1R1)
Dividing numerator and denominator by R1R2, we get:
Qeff = (Q1R1Q2R2ω)/(R1R2(Q2 + Q1))
Simplifying further, we get:
Qeff = (q1R1q2R2)/(R1R2)
Therefore, the correct answer is option C.
Formula Used:
Q = ωL/R, where Q is the Q-factor, ω is the operating frequency, L is the inductance, and R is the resistance
Calculation:
1. When the two coils are magnetically uncoupled, their effective Q factor will be the sum of their individual Q factors.
Qeff = q1 + q2
2. When the two coils are connected in series, their effective inductance will be the sum of their individual inductances.
Leff = L1 + L2
3. The effective resistance of the two coils connected in series will be the sum of their individual resistances.
Reff = R1 + R2
4. The effective Q factor of the two coils connected in series can be calculated using the formula:
Qeff = ωLeff/Reff
Substituting the values of Leff and Reff from step 2 and 3, we get:
Qeff = ω(L1 + L2)/(R1 + R2)
5. Using the formula Q = ωL/R, we can express the individual inductances L1 and L2 in terms of their respective Q factors and resistances:
L1 = Q1R1/ω and L2 = Q2R2/ω
Substituting these values in step 4, we get:
Qeff = ω(Q1R1 + Q2R2)/(R1 + R2)
6. Simplifying the expression in step 5, we get:
Qeff = (Q1R1Q2R2ω)/(R1Q2R2 + R2Q1R1)
Dividing numerator and denominator by R1R2, we get:
Qeff = (Q1R1Q2R2ω)/(R1R2(Q2 + Q1))
Simplifying further, we get:
Qeff = (q1R1q2R2)/(R1R2)
Therefore, the correct answer is option C.
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Two magnetically uncoupled inductive coils have Q factors q1 and q2 at the chosen operating frequency. Their respective resistances are R1 and R2. When connected in series, their effective Q factor at the same operating frequency isa)q1+q2b)(1/q1)+(1/q2)c)(q1R1+q2R2)/(R1+R2)d)(q1R2+q2R1)/(R1+R2)Correct answer is option 'C'. Can you explain this answer?
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Two magnetically uncoupled inductive coils have Q factors q1 and q2 at the chosen operating frequency. Their respective resistances are R1 and R2. When connected in series, their effective Q factor at the same operating frequency isa)q1+q2b)(1/q1)+(1/q2)c)(q1R1+q2R2)/(R1+R2)d)(q1R2+q2R1)/(R1+R2)Correct answer is option 'C'. 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 Two magnetically uncoupled inductive coils have Q factors q1 and q2 at the chosen operating frequency. Their respective resistances are R1 and R2. When connected in series, their effective Q factor at the same operating frequency isa)q1+q2b)(1/q1)+(1/q2)c)(q1R1+q2R2)/(R1+R2)d)(q1R2+q2R1)/(R1+R2)Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two magnetically uncoupled inductive coils have Q factors q1 and q2 at the chosen operating frequency. Their respective resistances are R1 and R2. When connected in series, their effective Q factor at the same operating frequency isa)q1+q2b)(1/q1)+(1/q2)c)(q1R1+q2R2)/(R1+R2)d)(q1R2+q2R1)/(R1+R2)Correct answer is option 'C'. Can you explain this answer?.
Two magnetically uncoupled inductive coils have Q factors q1 and q2 at the chosen operating frequency. Their respective resistances are R1 and R2. When connected in series, their effective Q factor at the same operating frequency isa)q1+q2b)(1/q1)+(1/q2)c)(q1R1+q2R2)/(R1+R2)d)(q1R2+q2R1)/(R1+R2)Correct answer is option 'C'. 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 Two magnetically uncoupled inductive coils have Q factors q1 and q2 at the chosen operating frequency. Their respective resistances are R1 and R2. When connected in series, their effective Q factor at the same operating frequency isa)q1+q2b)(1/q1)+(1/q2)c)(q1R1+q2R2)/(R1+R2)d)(q1R2+q2R1)/(R1+R2)Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two magnetically uncoupled inductive coils have Q factors q1 and q2 at the chosen operating frequency. Their respective resistances are R1 and R2. When connected in series, their effective Q factor at the same operating frequency isa)q1+q2b)(1/q1)+(1/q2)c)(q1R1+q2R2)/(R1+R2)d)(q1R2+q2R1)/(R1+R2)Correct answer is option 'C'. Can you explain this answer?.
Solutions for Two magnetically uncoupled inductive coils have Q factors q1 and q2 at the chosen operating frequency. Their respective resistances are R1 and R2. When connected in series, their effective Q factor at the same operating frequency isa)q1+q2b)(1/q1)+(1/q2)c)(q1R1+q2R2)/(R1+R2)d)(q1R2+q2R1)/(R1+R2)Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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