12. -10.49 points Tipler6 29.P072. My Notes Ask Your Teacher In a driven series

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12. -10.49 points Tipler6 29.P072. My Notes Ask Your Teacher In a driven series RLC circuit, the ideal generator has a peak emf equal to 197 V, the resistance is 60.1 , and the capacitance is 8.0 F. The inductance can be varied from 7.97 mH to 40.3 mH by the insertion of an iron core in the solenoid. The angular frequency of the generator is 2495 rad/s. If the capacitor voltage is not to exceed 150 V, find the following (a) the peak current (b) the ranges of inductance that are safe to use mH (lower) mH (higher) eBook Submit Answer Save Progress Practice Another Version

Explanation / Answer

impedance, z = sqrt(R2 + (XL - XC)2)

now for resonance, L = 1/w2C = 1/(2495 x 2495 x 8.01 x 10-6) = 20.1 mH which is in given range.

so to find peakk current we will use resonance condition of XL = XC i.e z = R

=> peak current = 197/60.1 = 3.28 A

b)

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