Power MCQ Quiz - Objective Question with Answer for Power - Download Free PDF

Concept:

The power dissipated by the resistor is given by:

\(P=\frac{V^2}{R}=I^2R\)

V = Voltage across the resistor

R = Resistance value

Calculation:

The equivalent parallel combination of R₃ and R₂ will give:

\(R_{eq}=\frac{R_3R_2}{R_3+R_2}\)

With R₂ = 3R and R₃ = 2R, we get R_eq as:

\(R_{eq}=\frac{3R\times 2R}{3R+2R}=\frac{6R}{5}\)

The resulting circuit is redrawn as:

Now the voltage across R₁ using voltage division will be:

\(V_1=V\times \frac{R_1}{R_1+R_{eq}}\)

Putting on the respective values, we get:

\(V_1=V\times \frac{2R}{2R+6R/5}=\frac{10V}{16}\)

\(V_1=\frac{5V}{8}\)

Since R₂ and R₃ are parallel to each other, the voltage across R₃ will be the voltage across R_eq, i.e.

\(V_{R3}=V-\frac{5V}{8}=\frac{3V}{8}\)

The ratio of power between R₁ to R₃ is calculated as:

\(\frac{P_{R_1}}{P_{R_3}}=\frac{V_1^2/R_1}{V_{R3}^2/R_3}\)

R₁ = 2R and R₃ = 2R

\(\frac{P_{R_1}}{P_{R_3}}=\frac{V_1^2/R_1}{V_{R3}^2/R_3}=\frac{V_1^2R_3}{V_2^2R_1}\)

Putting the respective values, we get:

\(\frac{P_{R_1}}{P_{R_3}}=\frac{25\times2R\times 64}{64\times 2R\times 9}=\frac{25}{9}\)

Concept:

The power dissipated by the resistor is given by:

P = I²R

I = Current flowing across the resistor

R = Resistance of the resistor

The power dissipated by the resistor is directly proportional to the square of the current, i.e.

P ∝ I²

Analysis:

The net current flowing across the circuit is given by:

\(I=\frac{E}{R_{eq}}\)

R_eq = R₁ + R₂ + R₃ + R₄

\(I=\frac{E}{R_1+R_2+R_3+R_4}\)

The power dissipated in R₃ is:

P_R3 = I² × R₃

\(P_{R3}=(\frac{E}{R_1+R_2+R_3+R_4})^2 \times R_3\)

Observation:

• We observe that the power will decrease with an increase in R₁ or R₂ or R₄.
• With a decrease in E, the power dissipated will decrease.
• It is only when R₄ is short-circuited, i.e. when R₄ = 0, the power dissipated will increase.

Concept:

• A dBm is a unit of measurement used to indicate the ratio of power level with respect to a fixed reference level.
• With dBm, the reference level is 1 mW (milliwatt).

Power in dBm is expressed as:

\(P(dBm)=10log_{10}(\frac{P}{1m})\)

P = Power to be expressed in dBm

Application:

Given Output Power in dBm = 20

\(20=10log_{10}(\frac{P}{1m})\)

\(2=log_{10}(\frac{P}{1m})\)

\(10^2=\frac{P}{1m}\)

P = 0.1 W

The voltage across the resistor = 1 V_rms

Resistance Value = 100 Ω

The power dissipated by the resistor is given by:

\(P = \frac{V_{rms}^2}{R}\)

\(P = \frac{1^2}{100}=\frac{1}{100}=10^{-2}~W\)

We are asked to evaluate the power dissipated with respect to 1 mW reference power.

\(P=\frac{10^{-2}}{10^{-3}}=10\)

In dB, this is expressed as:

\(P_{dB}=10~log_{10}(P)=10~log_{10}\frac{10^{-2}}{10^{-3}}\)

P_dB = 10 log₁₀(10) = 10 dB

Concept:

Complex power of an AC circuit is given as:

\(S = V \times {I^*}\)

S= complex power

V = Voltage

I* = Conjugate of current

Calculation:

Given, V = (400 + j80) V

I = (60 – j20) = A

S = VI^* = (400 + j80) (60 + j20)

= 24000 + j8000 + j4800 - 1600

= 22400 + j12800

Active power, P = 22400 W

Q = 12800 VAR

Concept:

The power dissipated by the resistor is given by:

P = I²R

I = Current flowing across the resistor

R = Resistance of the resistor

The power dissipated by the resistor is directly proportional to the square of the current, i.e.

P ∝ I²

Analysis:

The net current flowing across the circuit is given by:

\(I=\frac{E}{R_{eq}}\)

R_eq = R₁ + R₂ + R₃ + R₄

\(I=\frac{E}{R_1+R_2+R_3+R_4}\)

The power dissipated in R₃ is:

P_R3 = I² × R₃

\(P_{R3}=(\frac{E}{R_1+R_2+R_3+R_4})^2 \times R_3\)

Observation:

• We observe that the power will decrease with an increase in R₁ or R₂ or R₄.
• With a decrease in E, the power dissipated will decrease.
• It is only when R₄ is short-circuited, i.e. when R₄ = 0, the power dissipated will increase.

The voltage across the resistor = 1 V_rms

Resistance Value = 100 Ω

The power dissipated by the resistor is given by:

\(P = \frac{V_{rms}^2}{R}\)

\(P = \frac{1^2}{100}=\frac{1}{100}=10^{-2}~W\)

We are asked to evaluate the power dissipated with respect to 1 mW reference power.

\(P=\frac{10^{-2}}{10^{-3}}=10\)

In dB, this is expressed as:

\(P_{dB}=10~log_{10}(P)=10~log_{10}\frac{10^{-2}}{10^{-3}}\)

P_dB = 10 log₁₀(10) = 10 dB

Concept:

The power dissipated by the resistor is given by:

P = I²R

I = Current flowing across the resistor

R = Resistance of the resistor

The power dissipated by the resistor is directly proportional to the square of the current, i.e.

P ∝ I²

Analysis:

The net current flowing across the circuit is given by:

\(I=\frac{E}{R_{eq}}\)

R_eq = R₁ + R₂ + R₃ + R₄

\(I=\frac{E}{R_1+R_2+R_3+R_4}\)

The power dissipated in R₃ is:

P_R3 = I² × R₃

\(P_{R3}=(\frac{E}{R_1+R_2+R_3+R_4})^2 \times R_3\)

Observation:

• We observe that the power will decrease with an increase in R₁ or R₂ or R₄.
• With a decrease in E, the power dissipated will decrease.
• It is only when R₄ is short-circuited, i.e. when R₄ = 0, the power dissipated will increase.

Concept:

Complex power of an AC circuit is given as:

\(S = V \times {I^*}\)

S= complex power

V = Voltage

I* = Conjugate of current

Calculation:

Given, V = (400 + j80) V

I = (60 – j20) = A

S = VI^* = (400 + j80) (60 + j20)

= 24000 + j8000 + j4800 - 1600

= 22400 + j12800

Active power, P = 22400 W

Q = 12800 VAR

The voltage across the resistor = 1 V_rms

Resistance Value = 100 Ω

The power dissipated by the resistor is given by:

\(P = \frac{V_{rms}^2}{R}\)

\(P = \frac{1^2}{100}=\frac{1}{100}=10^{-2}~W\)

We are asked to evaluate the power dissipated with respect to 1 mW reference power.

\(P=\frac{10^{-2}}{10^{-3}}=10\)

In dB, this is expressed as:

\(P_{dB}=10~log_{10}(P)=10~log_{10}\frac{10^{-2}}{10^{-3}}\)

P_dB = 10 log₁₀(10) = 10 dB

Concept:

The power dissipated by the resistor is given by:

\(P=\frac{V^2}{R}=I^2R\)

V = Voltage across the resistor

R = Resistance value

Calculation:

The equivalent parallel combination of R₃ and R₂ will give:

\(R_{eq}=\frac{R_3R_2}{R_3+R_2}\)

With R₂ = 3R and R₃ = 2R, we get R_eq as:

\(R_{eq}=\frac{3R\times 2R}{3R+2R}=\frac{6R}{5}\)

The resulting circuit is redrawn as:

Now the voltage across R₁ using voltage division will be:

\(V_1=V\times \frac{R_1}{R_1+R_{eq}}\)

Putting on the respective values, we get:

\(V_1=V\times \frac{2R}{2R+6R/5}=\frac{10V}{16}\)

\(V_1=\frac{5V}{8}\)

Since R₂ and R₃ are parallel to each other, the voltage across R₃ will be the voltage across R_eq, i.e.

\(V_{R3}=V-\frac{5V}{8}=\frac{3V}{8}\)

The ratio of power between R₁ to R₃ is calculated as:

\(\frac{P_{R_1}}{P_{R_3}}=\frac{V_1^2/R_1}{V_{R3}^2/R_3}\)

R₁ = 2R and R₃ = 2R

\(\frac{P_{R_1}}{P_{R_3}}=\frac{V_1^2/R_1}{V_{R3}^2/R_3}=\frac{V_1^2R_3}{V_2^2R_1}\)

Putting the respective values, we get:

\(\frac{P_{R_1}}{P_{R_3}}=\frac{25\times2R\times 64}{64\times 2R\times 9}=\frac{25}{9}\)

Concept:

• A dBm is a unit of measurement used to indicate the ratio of power level with respect to a fixed reference level.
• With dBm, the reference level is 1 mW (milliwatt).

Power in dBm is expressed as:

\(P(dBm)=10log_{10}(\frac{P}{1m})\)

P = Power to be expressed in dBm

Application:

Given Output Power in dBm = 20

\(20=10log_{10}(\frac{P}{1m})\)

\(2=log_{10}(\frac{P}{1m})\)

\(10^2=\frac{P}{1m}\)

P = 0.1 W

Concept:

The power dissipated by the resistor is given by:

\(P=\frac{V^2}{R}=I^2R\)

V = Voltage across the resistor

R = Resistance value

Calculation:

The equivalent parallel combination of R₃ and R₂ will give:

\(R_{eq}=\frac{R_3R_2}{R_3+R_2}\)

With R₂ = 3R and R₃ = 2R, we get R_eq as:

\(R_{eq}=\frac{3R\times 2R}{3R+2R}=\frac{6R}{5}\)

The resulting circuit is redrawn as:

Now the voltage across R₁ using voltage division will be:

\(V_1=V\times \frac{R_1}{R_1+R_{eq}}\)

Putting on the respective values, we get:

\(V_1=V\times \frac{2R}{2R+6R/5}=\frac{10V}{16}\)

\(V_1=\frac{5V}{8}\)

Since R₂ and R₃ are parallel to each other, the voltage across R₃ will be the voltage across R_eq, i.e.

\(V_{R3}=V-\frac{5V}{8}=\frac{3V}{8}\)

The ratio of power between R₁ to R₃ is calculated as:

\(\frac{P_{R_1}}{P_{R_3}}=\frac{V_1^2/R_1}{V_{R3}^2/R_3}\)

R₁ = 2R and R₃ = 2R

\(\frac{P_{R_1}}{P_{R_3}}=\frac{V_1^2/R_1}{V_{R3}^2/R_3}=\frac{V_1^2R_3}{V_2^2R_1}\)

Putting the respective values, we get:

\(\frac{P_{R_1}}{P_{R_3}}=\frac{25\times2R\times 64}{64\times 2R\times 9}=\frac{25}{9}\)

From the linearity property, we know that the ratio of current to voltage will remain constant, i.e.

\(\frac{{{V_s}}}{{{I_L}}}\) = Constant

Since the given circuit in the box in a linear resistive network, the above mentioned property will hold true.

Analysis:

For V_s = 20 V, I_L = 200 mA

\(\therefore \frac{{{V_s}}}{{{I_L}}} = \frac{{20}}{{200 \times {{10}^{ - 3}}}} = 100\)

Let the current through load is I’_L when the power absorbed is 2.5 W, i.e.

\({P_L} = {\left( {I_L'} \right)^2}{R_L}\)

\(2.5 = {\left( {I_L'} \right)^2} \times 10\)

\(I_L' = 0.5\;A\)

\(\therefore \frac{{{V_s}}}{{{I_L}}} = \frac{{V_s'}}{{I_L'}} = 100\)

\(V_s{'\;} = 100\;I_L'\)

\(V_s{'\;} = 100 \times 0.5\)

= 50 V

Applying KCL at V₀, we get:

\(\frac{{{V_0}}}{{50}} + \frac{{{V_0} - 90}}{{4 + 1}} - 4 = 0\) 

V₀ + 10 V₀ – 900 = 4 × 50

11 V₀ = 1100

V₀ = 100 V

Since V₀ in positive, the current source will be supplying power of value:

P_sup = V₀ × 4

P_sup = 4 × 100

P_sup = 400 W