Question

An alternating voltage of 80 + j60 V is applied to a circuit and the current flowing is 4 – j2 A.
Find the (i) impedance, (ii) phase angle, (iii) power factor, and (iv) power consumed.

Answer 1

Answer:

V = (80 + j 60) = 100 ∠ 36.9°; I = − 4 + j 10 = 10.77 ∠ tan−1 (− 2.5) = 10.77 ∠ (180° − 68.2°) = 10.77 ∠ 111.8° (i) Z = V/I = 100 ∠36.9°/10.77∠ 111.8° = 9.28 ∠ − 74.9° = 9.28 (cos 74.9° − j sin 74.9°) = 2.42 − j 8.96 Ω Hence R = 2.42 Ω and XC = 8.96 Ω capacitive (ii) P = I2 R = 10.772 × 2.42 = 2.81 W (iii) Phase angle between voltage and current = 74.9° with current leading asRead more on Sarthaks.com - https://www.sarthaks.com/493166/an-a-c-voltage-80-j-60-volts-is-applied-to-a-circuit-and-the-current-flowing-is-4-j-10-amperes

Answer 2

Answer: Impedance is 10.77 when the alternating voltage of 80 + 60V is applied in the circuit.

Explanation: V = 80 + j 60 = 100

∠ 36.9°;

I = − 4 + j 10 = 10.77

∠ tan−1 − 2.5= 10.77 ∠ 180° − 68.2°= 10.77 ∠ 111.8°i

Z = V/I = 100 ∠36.9°/10.77∠ 111.8°= 9.28 ∠ − 74.9° = 9.28 cos 74.9° − j sin 74.9°= 2.42 − j 8.96 Ω

Hence R = 2.42 Ω and XC = 8.96 Ω.

Individual electronic parts, such as resistors, transistors, capacitors, inductors, and diodes, are connected by conductive wires or traces that allow an electric current to pass through them to form an electronic circuit. The difference in electric potential between two places is known as voltage, often referred to as electric pressure, electric tension, or potential difference. It translates into the amount of work required to transfer a test charge between two places in a static electric field.

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