A short circuit occurs in the series R-L circuit with V = 20 kV, X = 8 ohm, and R = 0.8 ohm. The circuit breaker opens 3 cycles after fault inception. Determine (a)the rms ac fault current(???), (b)the rms “momentary” current(????), at ?=0.5cycle, which passes through the breaker before it opens, and (c)the rms asymmetrical fault current that the breaker interrupts.

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A short circuit occurs in the series R-L circuit with V = 20 kV, X = 8 ohm, and R = 0.8 ohm. The circuit breaker opens 3 cycles after fault inception. Determine (a)the rms ac fault current(???), (b)the rms “momentary” current(????), at ?=0.5cycle, which passes through the breaker before it opens, and (c)the rms asymmetrical fault current that the breaker interrupts. 

(Hint: \( t = \frac{\tau}{f} \) where \( \tau \) is time in cycles; \( T = \frac{L}{R} = \frac{X}{wR} = \frac{X}{2\pi f R} \))

In this hint, \( t \) is defined as the time in seconds, where \( \tau \) is the time in cycles and \( f \) is the frequency. It explains that the period \( T \) can be expressed in multiple ways: as the ratio of inductance \( L \) to resistance \( R \), as the ratio of reactance \( X \) to the product of angular frequency \( w \) and resistance \( R \), and as the ratio of reactance \( X \) to the product of \( 2\pi \), frequency \( f \), and resistance \( R \).
Transcribed Image Text:(Hint: \( t = \frac{\tau}{f} \) where \( \tau \) is time in cycles; \( T = \frac{L}{R} = \frac{X}{wR} = \frac{X}{2\pi f R} \)) In this hint, \( t \) is defined as the time in seconds, where \( \tau \) is the time in cycles and \( f \) is the frequency. It explains that the period \( T \) can be expressed in multiple ways: as the ratio of inductance \( L \) to resistance \( R \), as the ratio of reactance \( X \) to the product of angular frequency \( w \) and resistance \( R \), and as the ratio of reactance \( X \) to the product of \( 2\pi \), frequency \( f \), and resistance \( R \).
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