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:
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Answer:
(a). Z = 54.54 ohm
(b). R = 36 ohm
(c). The circuit will be Capacitive.
Explanation:
Given data
I = 2.75 A
Voltage = 150 V
rad = 48.72°
(a). Impedance of the circuit is given by
Z = 54.54 ohm
(b). We know that resistance of the circuit is given by
Put the values of Z & in above formula we get
R = 36 ohm
(c). Since the phase angle is negative so the circuit will be Capacitive.