Answer:
Explanation:
Initial separation of plate = d
final separation = 2d
The capacitance of the capacitor will reduce from C to C/2 because
capacitance = ε A / d
d is distance between plates.
As the batteries are disconnected , charge on the capacitor becomes fixed .
Initial charge on the capacitor
= Capacitance x potential difference
Q = C ΔV
Final charge will remain unchanged
Final charge = C ΔV
Final capacitance = C/2
Final potential difference = charge / capacitance
= C ΔV / C/2
= 2 ΔV
Potential difference is doubled after the pates are further separated.
Answer:
114.26
Explanation:
a)Formula for per unit impedance for change of base is
Zpu2= Zpu1×(kV1/kV2)²×(kVA2/kVA1)
Zpu2: New per unit impedance
Zpu1: given per unit impedance
kV1: give base voltage
kV2: New bas votlage
kVA1: given bas power
kVA2: new base power
In the question
Zpu2=??
Zpu1= 0.3
kV2=24kV
kV1= 13.8 kV
kVA2= 1MVA ×1000= 1000 kVA
kVA1=500kVA
Zpu2= 0.3(13.8/24)²×(1000/500)
Zpu2= 0.198
b) to find ohmic impedance we will first calculate base value of impedance(Zbase). So,
Zbase= kV²/MVA
Zbase= 13.8²/(500/1000)
Zbase=380.88
Now that we have base value of impedance, Zbase, we can calculate actual ohmic value of impedance(Zactual) by using the following formula:
Zpu=Zactual/Zbase
0.3= Zactual/380.88
Zactual= 114.26 ohms
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