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 \( R \).

Given V- 20 kV R= 0.8 ohm X= 8 ohm For RL circuit, short circuit = bolted fault Instantanius quantitiesiac(t)=2VZsin (wt+α-θ)idc(t)=-2VZsin (α-θ)e(-tT) and rms quantitesIac=VZ ............................(1) Irms(t)=Iac2+Idc(t)2 =Iac2+2Iace(-tT) =Iac1+2e(-tT) Amp Irms(t) =k(T) Iac ................................(2)k(T) =1+2e(-tT) T=LR=XwR=X2πfRt=Γ/fwhere Γ =cylcesk(Γ) =1+2e-4πΓ(X/R) ......................(3) 1.rms ac fault current from equation 1Iac=VZ =VR2+X2 =20×10382+0.82Iac =2.487 A2. rms “momentary” current (Γ=0.5) From equation 3 XR=80.8=10k(Γ) =1+2e-4πΓ(X/R) =1+2e-4π(0.5)10 =1.438 from equation (2)Imomentary =k(Γ)Iac =1.438×2.487 Imomentary =3.5755 kA ..........at(0.5 cycles) 3.rms asymmetrical fault current that the breaker interrupts. as given breaker opens after 3 cycles i.e Γ=3 k(Γ) =1+2e-4πΓ(X/R) =1+2e-4π(3)10 =1.0227 from equation (2)Imomentary =k(Γ)Iac =1.0227×2.487 Imomentary =2.543 kA ..........at(3 cycles) = rms symmetrical fault current when breaker interrupts