The energy transferred by the appliance using mains electricity is 17.3 KJ
<h3>Data obtained from the question </h3>
- Potential difference (V) = 230V
- Charge (Q) = 150 C
<h3>How to determine the energy transferred </h3>
The energy transferred can be obtained as follow:
E = ½QV
E = ½ × 150 × 230
E = 75 × 230
E = 17250 J
Divide by 1000 to express in kilojoules
E = 17250 / 1000
E = 17.3 KJ
Learn more about energy stored in a capacitor:
brainly.com/question/14739936
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
The answer to your question is 343 m/s
