Animated Solution for Physics - Electromagnetic Induction: An inductance coil has a reactance of 100Ω. When an AC signal of frequency 1000Hz is applied to the coil, the applied voltage leads the current by 45∘. The self-inductance of the coil is
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Visualized Solution
RealInductanceCoil
Z=100Ω
f=1000Hz
PhaseAngleRelation
ϕ=45∘
tanϕ=VRVL=RXL
EquatingReactanceandResistance
tan45∘=RXL
1=RXL⟹XL=R
ImpedanceFormula
Z=R2+XL2
SubstitutingValues
100=XL2+XL2
100=2XL
SolvingforXL
XL=2100
XL=502Ω
InductiveReactanceFormula
XL=2πfL
SubstitutingFrequency
502=2π(1000)L
SolvingforL
L=2000π502
L=40π2
FinalCalculation
L≈40×3.141.414
L≈1.125×10−2H
Conclusion
Option\ (a)\ is\ correct.
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The Sigma Insight: Alternating Current (AC) and Voltage
Solution Diagram
The problem of the "real" inductance coil is a classic trap in competitive physics. When a problem states that an "inductance coil has a reactance of 100Ω", it is tempting to immediately write XL=100Ω. However, a physical coil is made of wire, and wire has resistance. Therefore, a real coil is modeled as an ideal inductor L in series with a resistor R.
The "reactance" mentioned in the problem statement is actually referring to the total opposition to the current, which is the impedanceZ of the coil. Thus, we must start with the premise that Z=100Ω.
Decoding the Phase Angle
The problem gives us a crucial piece of information: the applied voltage leads the current by 45∘. In an AC circuit containing both resistance and inductance, the phase angle ϕ between the voltage and the current is given by the tangent relation:
tanϕ=RXL
Substituting ϕ=45∘, we get:
tan45∘=RXL
Since tan45∘=1, this beautifully simplifies our circuit model. It tells us that the inductive reactance is exactly equal to the resistance:
XL=R
Unlocking the Impedance
Now that we know XL=R, we can return to our initial realization about the impedance. The impedance Z of a series RL circuit is given by the Pythagorean sum of the resistance and the reactance:
Z=R2+XL2
We know Z=100Ω. Substituting R=XL into the equation, we get:
100=XL2+XL2
100=2XL
Solving for XL, we find the true inductive reactance of the coil:
XL=2100=502Ω
The Final Inductance Calculation
With the true inductive reactance in hand, finding the self-inductance L is a straightforward application of the reactance formula. We know that:
XL=2πfL
The frequency f is given as 1000Hz. Substituting our known values:
502=2π(1000)L
Now, we carefully isolate L:
L=2000π502=40π2
Plugging in the standard approximations 2≈1.414 and π≈3.14:
L≈40×3.141.414≈1.125×10−2H
This matches perfectly with option (a). The key takeaway here is to always read "coil" as a combination of L and R unless it is explicitly stated to be an "ideal" inductor!