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A current from an AC source divides into two branches, A and B, in parallel. Branch A is an inductor 30μ H inductance and 1Ω resistance. Branche B is another inductor with inductance L and 1.5Ω resistance. For the ratio of currents in the branches to be independent of supply frequency, the value of L should be-
- a)30.5μ H
- b)20μ H
- c)45μ H
- d)29.5μ H
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A current from an AC source divides into two branches, A and B, in pa...
For independent of supply frequency
ω L - ω45μ = 0
L = 45μ H
Most Upvoted Answer
A current from an AC source divides into two branches, A and B, in pa...
To find the value of L, we need to analyze the circuit and understand the behavior of the currents in branches A and B.
Circuit Analysis:
- The circuit consists of two branches, A and B, connected in parallel to an AC source.
- Branch A has an inductor with an inductance of 30μH and a resistance of 1Ω.
- Branch B has an inductor with an unknown inductance L and a resistance of 1.5Ω.
Analysis of Branch A:
- In an AC circuit, the impedance of an inductor is given by Z = jωL, where j is the imaginary unit, ω is the angular frequency, and L is the inductance.
- The impedance of the inductor in branch A can be written as Z_A = jω(30μH) = j0.03ω.
- The current in branch A can be calculated using Ohm's Law: I_A = V/Z_A, where V is the voltage across the branch.
- Since the voltage across the two branches is the same, the current in branch A is inversely proportional to its impedance: I_A ∝ 1/Z_A.
Analysis of Branch B:
- Similarly, the impedance of the inductor in branch B can be written as Z_B = jωL.
- The current in branch B can be calculated using Ohm's Law: I_B = V/Z_B.
- The current in branch B is inversely proportional to its impedance: I_B ∝ 1/Z_B.
Condition for Ratio of Currents to be Independent of Frequency:
- For the ratio of currents in branches A and B to be independent of the supply frequency, the impedances of both branches should have the same frequency dependence.
- In other words, the imaginary parts of the impedances should cancel out, resulting in a constant ratio of currents.
- This implies that the imaginary parts of Z_A and Z_B should be equal: Im(Z_A) = Im(Z_B).
Calculating the Imaginary Parts:
- The imaginary part of Z_A is Im(Z_A) = 0.03ω.
- The imaginary part of Z_B is Im(Z_B) = ωL.
Setting up the Equation:
- Equating the imaginary parts of Z_A and Z_B, we get: 0.03ω = ωL.
- Simplifying the equation, we find: L = 0.03.
Therefore, the value of L should be 45μH (option C) for the ratio of currents in the branches to be independent of the supply frequency.
Circuit Analysis:
- The circuit consists of two branches, A and B, connected in parallel to an AC source.
- Branch A has an inductor with an inductance of 30μH and a resistance of 1Ω.
- Branch B has an inductor with an unknown inductance L and a resistance of 1.5Ω.
Analysis of Branch A:
- In an AC circuit, the impedance of an inductor is given by Z = jωL, where j is the imaginary unit, ω is the angular frequency, and L is the inductance.
- The impedance of the inductor in branch A can be written as Z_A = jω(30μH) = j0.03ω.
- The current in branch A can be calculated using Ohm's Law: I_A = V/Z_A, where V is the voltage across the branch.
- Since the voltage across the two branches is the same, the current in branch A is inversely proportional to its impedance: I_A ∝ 1/Z_A.
Analysis of Branch B:
- Similarly, the impedance of the inductor in branch B can be written as Z_B = jωL.
- The current in branch B can be calculated using Ohm's Law: I_B = V/Z_B.
- The current in branch B is inversely proportional to its impedance: I_B ∝ 1/Z_B.
Condition for Ratio of Currents to be Independent of Frequency:
- For the ratio of currents in branches A and B to be independent of the supply frequency, the impedances of both branches should have the same frequency dependence.
- In other words, the imaginary parts of the impedances should cancel out, resulting in a constant ratio of currents.
- This implies that the imaginary parts of Z_A and Z_B should be equal: Im(Z_A) = Im(Z_B).
Calculating the Imaginary Parts:
- The imaginary part of Z_A is Im(Z_A) = 0.03ω.
- The imaginary part of Z_B is Im(Z_B) = ωL.
Setting up the Equation:
- Equating the imaginary parts of Z_A and Z_B, we get: 0.03ω = ωL.
- Simplifying the equation, we find: L = 0.03.
Therefore, the value of L should be 45μH (option C) for the ratio of currents in the branches to be independent of the supply frequency.
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A current from an AC source divides into two branches, A and B, in parallel. Branch A is an inductor 30μ H inductance and 1Ω resistance. Branche B is another inductor with inductance L and 1.5Ω resistance. For the ratio of currents in the branches to be independent of supply frequency, the value of L should be-a)30.5μ Hb)20μ Hc)45μ Hd)29.5μ HCorrect answer is option 'C'. Can you explain this answer?
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A current from an AC source divides into two branches, A and B, in parallel. Branch A is an inductor 30μ H inductance and 1Ω resistance. Branche B is another inductor with inductance L and 1.5Ω resistance. For the ratio of currents in the branches to be independent of supply frequency, the value of L should be-a)30.5μ Hb)20μ Hc)45μ Hd)29.5μ HCorrect answer is option 'C'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about A current from an AC source divides into two branches, A and B, in parallel. Branch A is an inductor 30μ H inductance and 1Ω resistance. Branche B is another inductor with inductance L and 1.5Ω resistance. For the ratio of currents in the branches to be independent of supply frequency, the value of L should be-a)30.5μ Hb)20μ Hc)45μ Hd)29.5μ HCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A current from an AC source divides into two branches, A and B, in parallel. Branch A is an inductor 30μ H inductance and 1Ω resistance. Branche B is another inductor with inductance L and 1.5Ω resistance. For the ratio of currents in the branches to be independent of supply frequency, the value of L should be-a)30.5μ Hb)20μ Hc)45μ Hd)29.5μ HCorrect answer is option 'C'. Can you explain this answer?.
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