Question

ogf induced in a coil is 0.72 V when the current in the coil changes to 1.6 A in 0.25 s. What is the self-inductance L of the coil? An alternating voltage with a peak voltage of 12 V and a frequency 60 Hz is connected in series to a resistor with resistance R : 10 ?, a capacitor with capacitance C-1000 ?F, and an inductor with inductance L 500 mH. Determine the root-mean-square current in the circuit. (a)
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Answer #1

(B) e = L dI/dt

0.72 = L (4.6 - 1.6) / (0.25)

L = 0.06 H ......Ans

(a) R = 10 ohm

XL = 2 pi f L = 2 x pi x 60 x 0.500

XL = 188.5 Ohm

Xc = 1/ (2 pi f C ) = 2.65 ohm

Z = sqrt[ R^2 + (XL - XC)^2 ] = 186.1 ohm

I_peak = V / Z = 12 / 186.1 = 0.0645 A


I_rms = I_peak / sqrt(2)

= 0.0456 A

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