Sinusoidal Steady State Analysis Questions and Answers

Q: In the given circuit, the average power dissipated in 6Ω resistor is :

The average power is 12 W.The source has a peak value of 12 V, and average power in a resistor depends on the rms value, \(P = V_{rms}^{2}/R\). For a pure resistor the frequency (377 rad/s, that is 60 Hz) does not matter.\[ \begin{aligned} V_{rms} &= \frac{12}{\sqrt{2}} \\ P &= \frac{V_{rms}^{2}}{R} = \frac{144/2}{6} = 12\ \text{W} \end{aligned} \]Using the peak value in place of the rms value gives 144/6 = 24 W, which is twice the true average. 4 W, 14 W and 16 W do not follow from the circuit.

Q: A power factor of 1 indicates a :

A power factor of 1 indicates a purely resistive circuit.Power factor is cos φ, where φ is the phase angle between voltage and current. It equals 1 when φ = 0, that is, when the voltage and current are in phase and the circuit as a whole behaves as a pure resistance. This happens in a circuit of resistors alone, and also in an RLC circuit at resonance, where the inductive and capacitive reactances cancel, which is why the answer speaks of the circuit rather than a single element.A purely reactive element or circuit has φ = 90° and a power factor of 0, so those options describe the opposite case.

Q: If the transfer function of the following network is\[ \frac{V_{o}(s)}{V_{i}(s)} = \frac{1}{2 + SCR}, \](Figure: a series resistor R from the input Vi to the output; across the output, a capacitor C and the load resistor RL are connected in parallel. Vo is taken across RL.)then the value the Load Resistor RL is :

The load resistor is R.The output is taken across C in parallel with \(R_{L}\), fed through the series resistor R, so the circuit is a voltage divider.\[ \begin{aligned} Z_{p} &= \frac{R_{L}}{1 + sCR_{L}} \\ \frac{V_{o}}{V_{i}} &= \frac{Z_{p}}{R + Z_{p}} \\ &= \frac{R_{L}}{R(1 + sCR_{L}) + R_{L}} \\ &= \frac{1}{1 + R/R_{L} + sCR} \end{aligned} \]For this to equal \(\frac{1}{2 + sCR}\), the constant terms must match: \(1 + R/R_{L} = 2\), so \(R_{L} = R\).With \(R_{L} = R/2\) the denominator would start with 3, with \(R_{L} = 2R\) it would start with 1.5, and with \(R/4\) it would start with 5.

Q: Consider the following network, impedance of this network as a function of the complex frequency 'S' consists of a certain number of zeros and poles. What is the location of poles ?

The poles are at s = -2 and s = ∞.Looking into the input, the 2 H inductor is in series with the 1 Ω shunt resistor in parallel with the branch of the 1 H inductor and the second 1 Ω resistor:\[ \begin{aligned} Z(s) &= 2s + \frac{1 \cdot (s + 1)}{1 + s + 1} \\ &= 2s + \frac{s + 1}{s + 2} \\ &= \frac{2s^{2} + 5s + 1}{s + 2} \end{aligned} \]The denominator gives a finite pole at s = -2. The numerator is of higher degree than the denominator, so Z(s) also grows without limit as s goes to infinity, which is a pole at infinity. Options without ∞ miss that pole, and +2 has the wrong sign: an RL network has its poles in the left half-plane.

Q: In an AC circuit, the power factor is the cosine of the :

The power factor is the cosine of the phase angle.In an AC circuit the current and voltage have the same frequency but may be shifted in time. The phase angle \(\phi\) is the angle between the voltage and current phasors, and the real power is\[ P = V_{rms} I_{rms} \cos\phi \]so \(\cos\phi\) is the fraction of the apparent power that does useful work. It equals 1 for a purely resistive load and falls towards zero as the load becomes purely inductive or capacitive.The voltage angle and current angle on their own depend on the arbitrary reference chosen; only their difference matters. 'Resistance angle' is not a defined quantity, although the impedance angle is equal to \(\phi\).

11.

In the given circuit, the average power dissipated in 6Ω resistor is :

A source V = 12 sin(377t) connected in a loop with current i and a 6 ohm resistor

A.4 Watt
B.12 Watt
C.14 Watt
D.16 Watt

Exam Year(s)

RSMSSB JE - 2025, RSMSSB - 2025,

Answer: B

The average power is 12 W.

The source has a peak value of 12 V, and average power in a resistor depends on the rms value, \(P = V_{rms}^{2}/R\). For a pure resistor the frequency (377 rad/s, that is 60 Hz) does not matter.

\[ \begin{aligned} V_{rms} &= \frac{12}{\sqrt{2}} \\ P &= \frac{V_{rms}^{2}}{R} = \frac{144/2}{6} = 12\ \text{W} \end{aligned} \]

Using the peak value in place of the rms value gives 144/6 = 24 W, which is twice the true average. 4 W, 14 W and 16 W do not follow from the circuit.

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12.

A power factor of 1 indicates a :

A.Purely resistive element
B.Purely resistive circuit
C.Purely reactive element
D.Purely reactive circuit

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

A power factor of 1 indicates a purely resistive circuit.

Power factor is cos φ, where φ is the phase angle between voltage and current. It equals 1 when φ = 0, that is, when the voltage and current are in phase and the circuit as a whole behaves as a pure resistance. This happens in a circuit of resistors alone, and also in an RLC circuit at resonance, where the inductive and capacitive reactances cancel, which is why the answer speaks of the circuit rather than a single element.

A purely reactive element or circuit has φ = 90° and a power factor of 0, so those options describe the opposite case.

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13.

If the transfer function of the following network is

\[ \frac{V_{o}(s)}{V_{i}(s)} = \frac{1}{2 + SCR}, \]

(Figure: a series resistor R from the input V_i to the output; across the output, a capacitor C and the load resistor R_L are connected in parallel. V_o is taken across R_L.)

then the value the Load Resistor R_L is :

A.R/4
B.R/2
C.R
D.2R

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The load resistor is R.

The output is taken across C in parallel with \(R_{L}\), fed through the series resistor R, so the circuit is a voltage divider.

\[ \begin{aligned} Z_{p} &= \frac{R_{L}}{1 + sCR_{L}} \\ \frac{V_{o}}{V_{i}} &= \frac{Z_{p}}{R + Z_{p}} \\ &= \frac{R_{L}}{R(1 + sCR_{L}) + R_{L}} \\ &= \frac{1}{1 + R/R_{L} + sCR} \end{aligned} \]

For this to equal \(\frac{1}{2 + sCR}\), the constant terms must match: \(1 + R/R_{L} = 2\), so \(R_{L} = R\).

With \(R_{L} = R/2\) the denominator would start with 3, with \(R_{L} = 2R\) it would start with 1.5, and with \(R/4\) it would start with 5.

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14.

Consider the following network, impedance of this network as a function of the complex frequency 'S' consists of a certain number of zeros and poles. What is the location of poles ?

Ladder network: a 2 H series inductor, a 1 ohm shunt resistor, then a 1 H series inductor ending in a 1 ohm resistor; Z(S) is seen at the input

A.- 2
B.- 2, ∞
C.2
D.2, ∞

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The poles are at s = -2 and s = ∞.

Looking into the input, the 2 H inductor is in series with the 1 Ω shunt resistor in parallel with the branch of the 1 H inductor and the second 1 Ω resistor:

\[ \begin{aligned} Z(s) &= 2s + \frac{1 \cdot (s + 1)}{1 + s + 1} \\ &= 2s + \frac{s + 1}{s + 2} \\ &= \frac{2s^{2} + 5s + 1}{s + 2} \end{aligned} \]

The denominator gives a finite pole at s = -2. The numerator is of higher degree than the denominator, so Z(s) also grows without limit as s goes to infinity, which is a pole at infinity. Options without ∞ miss that pole, and +2 has the wrong sign: an RL network has its poles in the left half-plane.

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15.

In an AC circuit, the power factor is the cosine of the :

A.Voltage angle
B.Current angle
C.Phase angle
D.Resistance angle

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The power factor is the cosine of the phase angle.

In an AC circuit the current and voltage have the same frequency but may be shifted in time. The phase angle \(\phi\) is the angle between the voltage and current phasors, and the real power is

\[ P = V_{rms} I_{rms} \cos\phi \]

so \(\cos\phi\) is the fraction of the apparent power that does useful work. It equals 1 for a purely resistive load and falls towards zero as the load becomes purely inductive or capacitive.

The voltage angle and current angle on their own depend on the arbitrary reference chosen; only their difference matters. 'Resistance angle' is not a defined quantity, although the impedance angle is equal to \(\phi\).

Discuss in Forum

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