Download 1.A 4.0 mH coil carries a current of 5.0 A. Find the energy stored in

1.A 4.0 mH coil carries a current of 5.0 A. Find the energy stored in the coil’s magnetic
field
AnswerGiven,
Current (I) = 5 Amp
Inductance (L) = 4.0 mH = 4 * 10-3 Henry
Now energy stored (E) = ½ L I2
=>E = 0.5 * 4 * 10-3 * 5 * 5 = 50 * 10-3 Joules = 50 m Joules
So energy stored in coils magnetic field is 50 milli Joules
2. A 15 turns square coil of area of 0.40 m2 is placed parallel to a magnetic field of 0.75 T.
The coil is flipped so its plane is perpendicular to the magnetic field in 0.050 s. What is the
magnitude of the average induced emf?
Answer:Given,
Magnetic field(B)= 0.75 T
Number of turns(N) = 15
Ф is th angle between the square coil and the magnetic filed
Ф1(Flux initial) = BA cos 0 = BA = 0.75 * .4 = 0.3
Ф2(Flux final) = BA cos 90 = 0
ΔФ = Ф2 - Ф1 = 0- 0.3 =-0.3
Emf (E) = - N ΔФ / t
=>E = (-15 * 0.75 * -0.3)/0.5 = 6.75 V
SO magnitude of average induced emf is 6.75 V
3. A series RLC circuit consist of a 100 ? resistor, a 10.0 µF capacitor, and a 0.350 H
inductor. The circuit is connected to a 120 V, 60 Hz supply. What is the rms current in the
circuit?
Answer:Given f = 60 hz
Voltage(V) =120 V
Resistance (R) = 100 Ω
Capacitor(C ) = 10 µF
Inductance(L) = 0.350 H
Angular frequency (w) = 2 f = 2 * 3.14 * 60 = 376.8 rad.
Impedance due to inductor(XL) = 2 fL = 2 * 3.14 * 60 * .35 = 131.88 Ω
Impedance due to Capacitor(XC) = 1/2 fC = 1/(2 * 3.14 * 60 * 10 * 10 -6 )= 265.4 Ω
Net impedance due to the combination of R,L and C = Z=
=>Z = 166.85 Ω
So current(I) =V / Z = 120/166.85 =0.72 A
So the RMS current flowing through the circuit is 0.72 Amp
4. The secondary coil of a neon sign transformer provides 7500 V at 10.0 mA. The primary
coil operates on 120V. What does the primary draw?
Answer:Given,
Voltage at primary(Vp) = 1200 V
Current at primary(Ip) = ?
Voltage at secondary(Vs) = 7500 V
Current at secondary(Is) = 10 mA =10-2 A
According to Transformer equation
Vs/Vp = Ip /Is
=>Ip = (Vs * Is) / Vp
=>Ip = (7500 *10-2 ) / 1200
=>Ip = 6.25 * 10-2 A = 62.5 mA
So the current in primary is 62.5 milliamperes.
5. What inductance is needed in series with a 4.7 µF capacitor for a resonant frequency of
10 kHz?
Answer:Given,
Inductance (L) = ?
Capacitance(C) = 4.7 µF = 4.7 * 10-6 farad
Resonant frequency (f) = 10 Khz = 104 Hz
Now
f = 1⁄2π
= 1⁄(2* 3.14
)
=>104 * 6.28 = 1⁄
=>
=>
1.59 * 10-5
1.59 * 10-5
=> L = 5.39 * 10-5 Henry
So inductance required is 5.39 * 10-5 H