Question
Question: A \(60\,Volt - 10\,Watt\) bulb is operated at \(100\,Volt - 60\,Hz\) A.C. The inductance required is...
A 60Volt−10Watt bulb is operated at 100Volt−60Hz A.C. The inductance required is
A. 2.56Henry
B. 0.32Henry
C. 0.64Henry
D. 1.28Henry
Solution
In order to solve this question we need to understand faraday’s law of electromagnetic induction which states that if flux through any loop is changed then it induces emf in that loop and current flows in loop in direction such that it opposes the cause of its generation. When a current is flowing in loop and suppose it starts decreasing then back emf induced in loop so as to oppose this change, this phenomena is known as inductance and device that exhibit this behavior is known as inductor. The back emf induced is directly proportional to rate of decrease of current and proportionality constant is shown by L and its SI unit is Henry.
Complete step by step answer:
Given, Power of bulb, P=10Watt.
And voltage across the bulb is, V=60Volt.
Let the current of the bulb be I.
So from the definition of power we get, P=VI.
I=VP
Putting values we get,
I=6010A
⇒I=61A
So the resistance develop across bulb is, R=IV
Putting values we get,
R=6160Ω
⇒R=360Ω
Ac voltage is given, V1=100Volt
So impedance required in circuit is,
Z=IV1
Putting values we get,
Z=61100
⇒Z=600Ω
Let inductor resistance be, XL.
So using formula of impedance,
Z=R2+XL2
Squaring on both sides, Z2=R2+XL2
XL2=Z2−R2
⇒XL=Z2−R2
Putting values we get, XL=(600)2−(360)2
XL=480Ω
Frequency of inductor is given as, f=250Hz
So, XL=2πfL
L=2πfXL
Putting values we get, L=(2×π×60)480Ω
∴L=1.2732Henry
So the correct option is D.
Note: It should be remembered that flux across loops can be defined as a scalar product of magnetic field vector and area vector. Flux across the loop is changed by changing magnetic field, or by changing area under loop and by changing orientation of loop in magnetic field. Also impedance in circuit is a more general form of resistance as it may be complex.