Activity 2: According to Ohm's low, the current (1) through an impedance (Z) is related to the voltage (V) across it as: V =/ x2Z. A voltage equal to 210e/0.2m Volts, is applied across a load of impedance (60 + jX) Ohms. The magnitude (i.e., modulus) of the current through it is measured as 3 Amperes. (2-a) Calculate all possible values of X. (2-b) Calculate the consumed reactive power (Q) using the formula: Q =; Imag {V x I'}. Assume phase (i.e., argument) of the current equal to 67°.
Activity 2: According to Ohm's low, the current (1) through an impedance (Z) is related to the voltage (V) across it as: V =/ x2Z. A voltage equal to 210e/0.2m Volts, is applied across a load of impedance (60 + jX) Ohms. The magnitude (i.e., modulus) of the current through it is measured as 3 Amperes. (2-a) Calculate all possible values of X. (2-b) Calculate the consumed reactive power (Q) using the formula: Q =; Imag {V x I'}. Assume phase (i.e., argument) of the current equal to 67°.
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Transcribed Image Text:Activity 2:
According to Ohm's low, the current (1) through an impedance (Z) is related to the voltage (V)
across it as: V =1 ×Z.
A voltage equal to 210e/0.2 Volts, is applied across a load of impedance (60 + jX) Ohms. The
magnitude (i.e., modulus) of the current through it is measured as 3 Amperes.
(2-a) Calculate all possible values of X.
(2-b) Calculate the consumed reactive power (Q) using the formula: Q = Imag {V x I'}.
Assume phase (i.e., argument) of the current equal to 67°.
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