<span>a.current varies throughout a parallel circuit.
Hope this helps!</span>
Answer:
A) If you want to achieve the SMALLEST possible resistance, you should attach the leads to the opposite faces that measure b) 5 cm by 8 cm.
B) If you want to achieve the LARGEST possible resistance, you should attach the leads to the opposite faces that measure a) 3 cm × 5 cm
Explanation:
Resistivity is directly proportional to lenght and inversely properly to cross sectional area.
For the first case, 5 cm by 8 cm gives the largest area and leave 3 cm as the lenght. The resistivity of the metal will be smallest in these dimensions.
For the second case, 3 cm by 5 cm gives the smallest area, leaving 8 cm as the lenght. This is the maximum arrangement that can give the largest resistance possible.
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