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In an LCR circuit, the capacitance is made one - fourth when in resonance, then how many times the inductance should be changed, so that the circuit remains in resonance?
Correct answer is '4'. Can you explain this answer?
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Verified Answer
In an LCR circuit, the capacitance is made one - fourth when in resona...
At resonance ; XL = Xc
Now
Thus change in inductance should be 4 times.
Now
Thus change in inductance should be 4 times.
Most Upvoted Answer
In an LCR circuit, the capacitance is made one - fourth when in resona...
Resonance in an LCR circuit occurs when the inductive reactance (XL) and the capacitive reactance (XC) cancel each other out, resulting in a purely resistive impedance. This happens when the angular frequency of the circuit (ω) is equal to the resonant frequency (ω0), which can be calculated using the formula:
ω0 = 1/√(LC)
Where L is the inductance and C is the capacitance of the circuit.
In this question, we are given that the capacitance is decreased to one-fourth of its original value. Let's call the original capacitance C1 and the new capacitance C2 (C2 = C1/4). We need to find out how many times the inductance should be changed to keep the circuit in resonance.
Let's assume the original inductance is L1 and the new inductance is L2. We can write the resonant frequency for both cases as:
ω01 = 1/√(L1C1) (original resonant frequency)
ω02 = 1/√(L2C2) (new resonant frequency)
According to the given condition, the circuit remains in resonance even after changing the capacitance. This means that the resonant frequency remains the same. So we can write:
ω01 = ω02
Substituting the values of ω0 and the given expressions for C2:
1/√(L1C1) = 1/√(L2(C1/4))
Simplifying the equation by squaring both sides and cross-multiplying:
L1C1 = L2(C1/4)
4L1C1 = L2C1
Dividing both sides by C1:
4L1 = L2
So, we can see that the inductance needs to be changed to four times its original value (L2 = 4L1) in order to maintain resonance in the circuit when the capacitance is reduced to one-fourth. Therefore, the correct answer is 4.
ω0 = 1/√(LC)
Where L is the inductance and C is the capacitance of the circuit.
In this question, we are given that the capacitance is decreased to one-fourth of its original value. Let's call the original capacitance C1 and the new capacitance C2 (C2 = C1/4). We need to find out how many times the inductance should be changed to keep the circuit in resonance.
Let's assume the original inductance is L1 and the new inductance is L2. We can write the resonant frequency for both cases as:
ω01 = 1/√(L1C1) (original resonant frequency)
ω02 = 1/√(L2C2) (new resonant frequency)
According to the given condition, the circuit remains in resonance even after changing the capacitance. This means that the resonant frequency remains the same. So we can write:
ω01 = ω02
Substituting the values of ω0 and the given expressions for C2:
1/√(L1C1) = 1/√(L2(C1/4))
Simplifying the equation by squaring both sides and cross-multiplying:
L1C1 = L2(C1/4)
4L1C1 = L2C1
Dividing both sides by C1:
4L1 = L2
So, we can see that the inductance needs to be changed to four times its original value (L2 = 4L1) in order to maintain resonance in the circuit when the capacitance is reduced to one-fourth. Therefore, the correct answer is 4.
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In an LCR circuit, the capacitance is made one - fourth when in resona...
At resonance ; XL = Xc
Now
Thus change in inductance should be 4 times.
Now
Thus change in inductance should be 4 times.
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In an LCR circuit, the capacitance is made one - fourth when in resonance, then how many times the inductance should be changed, so that the circuit remains in resonance?Correct answer is '4'. Can you explain this answer?
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