An inductor with L = 9.80 mH is connected across an ac source that has voltage amplitude 41.5 V . What value for the frequency o
1 answer:
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Answer:
a) <em>2.278 x 10^-5 volts</em>
b) <em>1.139 x 10^-6 Ampere</em>
c) <em>2.59 x 10^-11 W</em>
Explanation:
The radius of the wire r = 2 mm = 0.002 m
the number of turns N = 200 turns
direction of the magnetic field ∅ = 25°
magnetic field strength B = 0.02 T
varying time = 2 sec
The cross sectional area of the wire =
==> A = 3.142 x = 1.257 x 10^-5 m^2
Field flux Φ = BA cos ∅ = 0.02 x 1.257 x 10^-5 x cos 25°
==> Φ = 2.278 x 10^-7 Wb
The induced EMF is given as
E = NdΦ/dt
where dΦ/dt = (2.278 x 10^-7)/2 = 1.139 x 10^-7
E = 200 x 1.139 x 10^-7 = <em>2.278 x 10^-5 volts</em>
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b) If the two ends are connected to a resistor of 20 Ω, the current through the resistor is given as
= E/R
where R is the resistor
= (2.278 x 10^-5)/20 = <em>1.139 x 10^-6 Ampere</em>
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<em> </em>c) power delivered to the resistor is given as
P = E
P = (1.139 x 10^-6) x (2.278 x 10^-5) = <em>2.59 x 10^-11 W</em>