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QUESTION: 1

The phasor diagram shown in figure below is for a two-element series circuit having

Solution:

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

QUESTION: 2

For the circuit shown below, the voltage across the 3 Ω resistor V_{3} is

Solution:

Using voltage division rule, we have

QUESTION: 3

A part of a circuit shown below consists of a resistor, a capacitor and an inductor. At steady state, i_{R}(t) = 10 sin t and v_{L}{t) = 5 cost. The rms value of the current through the capacitor is

Solution:

Using KCL at the given node,

QUESTION: 4

In the circuit shown below, currents I and I_{1} are respectively

Solution:

From given figure

Using phasor diagram shown below, the currents I and I_{1} are respectively

QUESTION: 5

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?

Solution:

The input impedance Z of given circuit is

For p.f. to be unity,

QUESTION: 6

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

Solution:

Here, i lags v by 30°

also

QUESTION: 7

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)

Solution:

The situation is shown in figure below with it's phasor diagram.

For unity p.f. operation,

QUESTION: 8

The input p.f. of the circuit shown below is

Solution:

or, Z_{in} = 2.828∠-45°Ω

Hence, input p.f. = cos45° = 0.707 (lead)

QUESTION: 9

The input admittance of the circuit shown below is

Solution:

Hence

QUESTION: 10

The driving point impedance

does not represents a passive one port network because

Solution:

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