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

Concept:

When voltage varies linearly with time, the average voltage is used to calculate total energy.

Energy delivered = Current × Average Voltage × Time

Given:

• Initial voltage V_start = 10 V
• Final voltage V_end = 9 V
• Current I = 1 A (constant)
• Time t = 1 hour = 3600 seconds

Step-by-step Calculation:

Average voltage = $\frac{V_{start}+V_{end}}{2}=\frac{10+9}{2}=9.5V$

Energy = I × V_avg × t = 1 × 9.5 × 3600 = 34200 J = 34.2 kJ

Concept: The energy delivered by a battery can be calculated using the formula:

Energy = ∫ P dt

Where:

• P is the instantaneous power delivered by the battery, given by P = V × I, where V is the terminal voltage and I is the current.
• dt represents the time interval over which the power is delivered.

Since the voltage drops linearly from 12V to 10V over the discharge period, we can use the average voltage to simplify the calculation of energy.

Calculation:

As the voltage drops linearly from 12V to 10V, the average voltage during this period is:

V_avg = (V_initial + V_final) / 2

Vavg  = (12 + 10) / 2 = 11V

Given that the discharge time is 20 minutes:

Time = 20 × 60 = 1200 seconds

Using the formula Energy = V_avg × I × Time, where:

Energy = 11 × 2 × 1200 = 26400 Joules = 26.4 kJ

Concept:

The power dissipated by the resistor is given by:

$P=\frac{V^2}{R}=I^{2}R$

V = Voltage across the resistor

R = Resistance value

Calculation:

The equivalent parallel combination of R₃ and R₂ will give:

$R_{eq}=\frac{R_3{R}_2}{R_3+R_2}$

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

$\frac{P_{R_1}}{P_{R_3}}=\frac{V_1^2/R_1}{V_{R3}^2/R_3}=\frac{V_1^2{R}_3}{V_2^2{R}_1}$

Putting the respective values, we get:

$\frac{P_{R_1}}{P_{R_3}}=\frac{25\times 2R\times 64}{64\times 2R\times 9}=\frac{25}{9}$

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:

$P(dBm)=10lo{g}_{10}(\frac{P}{1m})$

P = Power to be expressed in dBm

Application:

Given Output Power in dBm = 20

$20=10lo{g}_{10}(\frac{P}{1m})$

$2=lo{g}_{10}(\frac{P}{1m})$

${10}^2=\frac{P}{1m}$

P = 0.1 W

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:

$P_{dB}=10lo{g}_{10}(P)=10lo{g}_{10}\frac{{10}^{-2}}{{10}^{-3}}$

P_dB = 10 log₁₀(10) = 10 dB

Concept: The energy delivered by a battery can be calculated using the formula:

Energy = ∫ P dt

Where:

• P is the instantaneous power delivered by the battery, given by P = V × I, where V is the terminal voltage and I is the current.
• dt represents the time interval over which the power is delivered.

Since the voltage drops linearly from 12V to 10V over the discharge period, we can use the average voltage to simplify the calculation of energy.

Calculation:

As the voltage drops linearly from 12V to 10V, the average voltage during this period is:

V_avg = (V_initial + V_final) / 2

Vavg  = (12 + 10) / 2 = 11V

Given that the discharge time is 20 minutes:

Time = 20 × 60 = 1200 seconds

Using the formula Energy = V_avg × I × Time, where:

Energy = 11 × 2 × 1200 = 26400 Joules = 26.4 kJ

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:

$S=V\times I^{\ast }$

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:

$P_{dB}=10lo{g}_{10}(P)=10lo{g}_{10}\frac{{10}^{-2}}{{10}^{-3}}$

P_dB = 10 log₁₀(10) = 10 dB

Concept:

The power dissipated by the resistor is given by:

$P=\frac{V^2}{R}=I^{2}R$

V = Voltage across the resistor

R = Resistance value

Calculation:

The equivalent parallel combination of R₃ and R₂ will give:

$R_{eq}=\frac{R_3{R}_2}{R_3+R_2}$

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

$\frac{P_{R_1}}{P_{R_3}}=\frac{V_1^2/R_1}{V_{R3}^2/R_3}=\frac{V_1^2{R}_3}{V_2^2{R}_1}$

Putting the respective values, we get:

$\frac{P_{R_1}}{P_{R_3}}=\frac{25\times 2R\times 64}{64\times 2R\times 9}=\frac{25}{9}$

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:

$P(dBm)=10lo{g}_{10}(\frac{P}{1m})$

P = Power to be expressed in dBm

Application:

Given Output Power in dBm = 20

$20=10lo{g}_{10}(\frac{P}{1m})$

$2=lo{g}_{10}(\frac{P}{1m})$

${10}^2=\frac{P}{1m}$

P = 0.1 W

Concept:

The power dissipated by the resistor is given by:

$P=\frac{V^2}{R}=I^{2}R$

V = Voltage across the resistor

R = Resistance value

Calculation:

The equivalent parallel combination of R₃ and R₂ will give:

$R_{eq}=\frac{R_3{R}_2}{R_3+R_2}$

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

$\frac{P_{R_1}}{P_{R_3}}=\frac{V_1^2/R_1}{V_{R3}^2/R_3}=\frac{V_1^2{R}_3}{V_2^2{R}_1}$

Putting the respective values, we get:

$\frac{P_{R_1}}{P_{R_3}}=\frac{25\times 2R\times 64}{64\times 2R\times 9}=\frac{25}{9}$

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.

$P_L={\left(I_L^{\prime }\right)}^2{R}_L$

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

$I_L^{\prime }=0.5A$

$\therefore \frac{V_s}{I_L}=\frac{V_s^{\prime }}{I_L^{\prime }}=100$

$V_s{}^{\prime }=100I_L^{\prime }$

$V_s{}^{\prime }=100\times 0.5$

= 50 V