Test: Sinusoidal Steady-State Analysis - 1
10 Questions MCQ Test Topicwise Question Bank for Electrical Engineering | Test: Sinusoidal Steady-State Analysis - 1
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The phasor diagram shown in figure below is for a two-element series circuit having
It is clear that i leads v by 53.2° (< 90°).
Hence the two series elements must be R and C.
∴ tan θ = tan 53.3°
tan 45° < tan 53.3° < tan 60° or,
1 < tan 53.3° <1.732
Hence, tan 53.3° = 1.3367
For the circuit shown below, the voltage across the 3 Ω resistor V3 is
Using voltage division rule, we have
A part of a circuit shown below consists of a resistor, a capacitor and an inductor. At steady state, iR(t) = 10 sin t and vL{t) = 5 cost. The rms value of the current through the capacitor is
Using KCL at the given node,
In the circuit shown below, currents I and I1 are respectively
From given figure
Using phasor diagram shown below, the currents I and I1 are respectively
What should be the value of C for the circuit shown below such that the input power factor is unity for any frequency f of the source?
The input impedance Z of given circuit is
For p.f. to be unity,
In a series R-L circuit, the current and voltages are given by
i = cos (314t-20°) , v = 10 cos ( 314 t + 10°). The values of R and L are respectively
Here, i lags v by 30°
also
A series R-L circuit has resistance and reactance of 15 Ω and 10 Ω respectively. What should be the value of capacitor which when connected across the series combination in parallel, the system attains unity p.f.? (use f= 50 Hz)
The situation is shown in figure below with it's phasor diagram.
For unity p.f. operation,
The input p.f. of the circuit shown below is
or, Zin = 2.828∠-45°Ω
Hence, input p.f. = cos45° = 0.707 (lead)
The input admittance of the circuit shown below is
Hence
The driving point impedance
does not represents a passive one port network because
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