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

The correct answer is option 1) i.e. Bulb A

CONCEPT:​

• Power: The rate of work done by an electric current is called power. It is denoted by P. The SI unit of power is the watt (W).

Power dissipation is given by:

\(⇒ P = VI = \frac{V^2}{R}=I^2R\)

Where V is the potential difference across resistance, I is current flowing and R is resistance.

EXPLANATION:

Given that:

The power rating of bulb A, P_A = 50 W

The power rating of bulb B, PB = 100 W

We know, \(P =\frac{V^2}{R} ⇒ R =\frac{V^2}{P}\) 

• The resistance of bulb A, 

\(\Rightarrow R_A =\frac{V^2}{P_A} = \frac{200^2}{50} =800\: Ω\)

• The resistance of bulb B, 

\(⇒ R_B =\frac{V^2}{P_B} = \frac{200^2}{100} =400\: Ω\)

• The equivalent resistance provided by the bulbs connected in series,

⇒ R = 800 + 400 = 1200 Ω

• The current in the circuit, 

\(⇒ I =\frac{V}{R} = \frac{200}{1200} = 0.16\:A\)

The bulb which dissipates more power will glow brighter.

Power dissipated by the bulbs:

⇒ P_A = VI = I²R_A = 0.16² × 800 = 20.48 W

⇒ PB = VI = I2RB = 0.162 × 400 = 10.24 W

⇒ PA > PB

• Hence, bulb A will glow brighter.

Important Points  

• In a series connection, both bulbs will have the same current flowing through them.
• Therefore, the bulb with greater resistance will have a greater voltage drop across it and will have a higher power dissipation and brightness.

Key Points

• W stands for Watt, which is a unit of power.
• Power is defined as the rate at which energy is consumed or produced.
• A 11 W bulb consumes 11 joules of energy per second.
• The rating indicates the power consumption of the bulb when it is operated at the rated voltage.

Additional Information

• Watt (W):
  • It is the SI unit of power, equivalent to one joule per second.
  • Named after James Watt, the Scottish engineer.
• Power:
  • It is the rate at which energy is transferred or converted.
  • Mathematically, Power (P) = Energy (E) / Time (t).
• Energy:
  • It is the capacity to do work.
  • Measured in joules (J) in the International System of Units (SI).
• Electric Bulb:
  • An electric device that produces light when an electric current passes through it.
  • Rated by the power it consumes, typically mentioned in watts (W).

CONCEPT:

Electrical Power and Series vs. Parallel Circuits

• The power consumed by a resistor (like a bulb) is given by the formula:

  P = V² / R

• In a series circuit, the voltage is divided among the components, whereas in a parallel circuit, each component gets the full voltage of the battery.

EXPLANATION:

• To make the bulb glow the most, it must receive the maximum voltage possible, as power is directly proportional to the square of the voltage (P ∝ V²).
• In a parallel circuit, each branch gets the full voltage of the battery. This means the bulb will receive the full voltage and thus, will glow the most.
• If the bulb is in series with other components, the voltage is divided among them, reducing the voltage across the bulb.

Therefore, in a parallel circuit configuration, the bulb would glow the most.

The correct answer is: Option 1) PB = 124.8 W, PH = 33.25 W

Concepts:

The power consumed by a resistor is given by the formula: P = V² / R, where V is the voltage across the resistor, and R is its resistance.

For a room heater, we can calculate its resistance Rₕ using its rated power Pₕ and rated voltage Vₕ.

The power consumed by the bulb P_B is calculated in a similar way, using the voltage across the bulb.

The two devices (bulb and heater) are connected in series, so the current through both devices is the same.

Explanation:

For the heater, given:

Pₕ = 750 W, Vₕ = 220 V

Resistance of the heater Rₕ = Vₕ² / Pₕ = (220)² / 750 = 48400 / 750 = 64.53 Ω

For the bulb, given:

Pₗ = 200 W, Vₗ = 220 V

Resistance of the bulb Rₗ = Vₗ² / Pₗ = (220)² / 200 = 48400 / 200 = 242 Ω

Since both devices are in series, the total resistance in the circuit is:

Rₜ = Rₕ + Rₗ = 64.53 + 242 = 306.53 Ω

Now, the current in the circuit is:

I = V / Rₜ = 220 / 306.53 = 0.717 A

Power consumed by the heater (Pₕ) is:

Pₕ = I² × Rₕ = (0.717)² × 64.53 = 0.514 × 64.53 = 33.25 W

Power consumed by the bulb (Pₗ) is:

Pₗ = I² × Rₗ = (0.717)² × 242 = 0.514 × 242 = 124.8 W

The correct answer is - 64 W

Key Points

• Power consumed by the bulb
  • The electric bulb is rated at 200 V and 100 W, which means it consumes 100 W of power when connected to a 200 V supply.
  • The power consumed by the bulb when connected to a different voltage can be calculated using the formula:
    P = V² / R
  • First, calculate the resistance (R) of the bulb when it operates at its rated values:
    R = V² / P
    R = (200)² / 100 = 400 Ω
  • Now, use the resistance to calculate the power consumed at 160 V:
    P = V² / R
    P = (160)² / 400 = 64 W
  • Therefore, the power consumed by the bulb when connected to a 160 V power supply is 64 W.

Additional Information

• Understanding Electric Power
  • Electric power is the rate at which electrical energy is transferred by an electric circuit. The SI unit of power is the watt (W).
  • Power (P) in a resistive load (such as a bulb) can be calculated using:
    P = V × I, where V is the voltage across the load and I is the current through the load.
  • For resistive loads, Ohm’s Law (V = I × R) can be used to relate voltage, current, and resistance.
  • Other relevant formulas include:
    P = I² × R and P = V² / R
• Voltage and Resistance in Bulbs
  • The rated voltage of a bulb indicates the voltage at which the bulb is designed to operate optimally.
  • Operating a bulb at a voltage lower or higher than its rated voltage can affect its power consumption and lifespan.
  • Resistance in a bulb usually remains constant, as it is determined by the material and geometry of the filament.

CONCEPT:

• Electric Power: The rate at which electrical energy is dissipated into other forms of energy is called electrical power i.e.,

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

Where V = Potential difference, R = Resistance and I = current.

CALCULATION:

Given - Potential difference (V) = 220 V, power of the bulb (P) = 100 W and actual voltage (V') = 110 V

• The resistance of the bulb can be calculated as,

​\(\Rightarrow R=\frac{V^2}{P}=\frac{(220)^2}{100}=484 \,\Omega\)

• The power consumed by the bulb.

\(\Rightarrow P=\frac{V^2}{R}=\frac{(110)^2}{484}=25 \,W\)

CONCEPT:

Electric power:

• The rate at which the electric energy is dissipated or consumed is termed as electric power.
• The electric power is given as,

\(⇒ P=IV=I^{2}R=\frac{V^{2}}{R}\)

Where P = electric power, V = voltage, I = current and R = resistance

CALCULATION:

Given P₁ = 200 W, P₂ = 100 W and V = 100 V,

We know that for an electric bulb the resistance is given as,

\(⇒ R=\frac{V^{2}}{P}\)

\(⇒ R\propto\frac{1}{P}\)     -----(1)

• For an electric bulb, the resistance of the bulb is inversely proportional to the power of the bulb.
• So the bulb which has more power will have low resistance, therefore the 100 W bulb will have more resistance compared to the 200 W bulb.
• The heat dissipated by the bulb is given as,

⇒ H = I²R     -----(2)

• Both the bulbs are connected in series so the current in both the bulbs will be equal. So the heat dissipated will be more in the bulb which has more resistance.
• Since the resistance of the 100 W bulb is more, so the heat dissipation of the 100 W bulb will be more.
• The brightness of the bulb depends on the heat dissipation by the bulb, so the 100 W bulb will have more brightness.

CONCEPT:

• Power: The rate of work done by an electric current is called power. It is denoted by P. The SI unit of power is the watt (W).

Power dissipation is given by:

Power (P) = V I = V2/R = I2 R

Where V is the potential difference across resistance, I is current flowing and R is resistance.

According to Ohm’s law:

V = I × R.

Where V = Voltage, I = current flowing through the circuit, R = Resistance

EXPLANATION:

Power (P) = V2/R

• From the above formula, keeping the potential same the power is inversely proportional to the resistance.
• The bulb which draws maximum power will glow more as compared to other and current is inversely proportional to resistance.
• Hence, the dim bulb will have higher resistance, and one which glows brighter than the other will have less resistance. So option 4 is correct.

Option 2 is correct, i.e. 5 ​kW h.

CONCEPT:

• Power (P): The rate of work done is called power.
  • The SI unit of power is the watt (W).

Power (P) = work done (W) / time taken (t)   (P = W / t = J / s)

• Energy: An object having the capability to do work is said to possess energy. The object which does the work loses energy and the object on which the work is done gains energy.
  • Its SI unit is Joule.

CALCULATION:
Given that:

time (t) = 5 hours

Power(P) = 200 W

Total number of devices = 5

Power (P) = work done (W)/time taken (t)

Energy = Work done (W) = Power (P) × time taken (t) = 200 × 5 × 5 

Energy = 1000 × 5 = 5000 Watt hour = 5 KWh

Additional Information

• Watt is a small unit that's why kilowatt-hour is used as the unit for electrical energy.
• 1 unit of electric energy: When one kilowatt load works for 1 hour then the energy consumed is called 1 unit of electricity.

1 Unit of electricity = 1 KWh = 1000 Watt-hour = 3.6 × 106 J

1 Kilo-watt = 1000 Watt

• 1 Watt: The energy consumption rate of 1 joule per second is called 1 watt.

CONCEPT:

• Resistance: The property of electrical materials and electrical instruments that opposes the flow of current through them is called resistance.
  • It is denoted by R. The SI unit of resistance is the ohm (Ω).
• Power: The rate of work done by an electric current is called power. It is denoted by P. The SI unit of power is the watt (W).

Power dissipation is given by:

Power (P) = V I = V2/R = I2 R

Where V is the potential difference across resistance, I is current flowing and R is resistance.

The heat dissipated in a circuit is given by:

Heat (H) = I2 R t

CALCULATION:

Power (P) = V I = V2/R = I2 R

So option 3 is correct.

CONCEPT:

• Power: The rate of work done by the electric energy is called power. It is denoted by P.
  • The SI unit of power is the watt (W).
  • Watt is a small unit that's why kilowatt-hour is used as the unit for electrical energy.
• 1 unit of electric energy: When one-kilowatt load works for 1 hour then the energy consumed is called 1 unit of electricity.

1 Unit of electricity = 1 KWh = 1000 Watt-hour = 3.6 × 106 J

1 Kilo-watt = 1000 Watt

• 1 Watt: The energy consumption rate of 1 joule per second is called 1 watt.

EXPLANATION:

• kWh is the unit of electrical energy which is equal to work done by electrical energy. So option 4 is correct.

CONCEPT:

• Resistance: The property of electrical materials and electrical instruments that opposes the flow of current through them is called resistance.
  • It is denoted by R. The SI unit of resistance is the ohm (Ω).
• Power: The rate of work done by an electric current is called power. It is denoted by P. The SI unit of power is the watt (W).

Power dissipation is given by:

Power (P) = V I = V2/R = I2 R

Where V is the potential difference across resistance, I is current flowing and R is resistance.

CALCULATION:

The electric power of an electric appliance is given by:

Power (P) = V I = I V

So option 2 is correct.

Concept:

• Ohm's law: It states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperature, remain constant.
• Mathematically, current-voltage can be written as V = IR, where V = Voltage, I = Current, R = Resistance
• For equivalent resistance in series combinations
  • Req = R1 + R2 + - -

• The heat dissipated by a resistor in a given time t is, H = I2 Rt, where I = current, R = Resistance, t = time

Calculation:

Given, Resistance, R₁ = 4 Ω , R₂ = 8 Ω,

Battery, V = 12 volt

The resistor is connected in series,

In series combination, R_eq = 4 + 8 = 12 Ω 

Current in the circuit, \(I=\frac{V}{R}=\frac{12}{12}=1~A\)

The heat dissipated by 8 Ω resistor in 10 s is,

H = I²Rt

H = (1)² × 8 × 10 = 80 J

Hence, the heat dissipated is 80 J.

The correct answer is option "4"

Concept:

• Electric Power: The rate at which electrical energy is dissipated into other forms of energy is called electrical power i.e.,

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

Where V = Potential difference, R = Resistance and I = current.

Calculation:

Given - Potential difference (V) = 220 V, power of the bulb (P) = 80 W and actual voltage (V') = 110 V

• The resistance of the bulb can be calculated as,

​\(\Rightarrow R=\frac{V^2}{P}=\frac{(220)^2}{80}=605 \,\Omega\)

• The power consumed by the bulb.

\(\Rightarrow P=\frac{V^2}{R}=\frac{(110)^2}{605}=20 \,W\)

CONCEPT:

• Electric Power: The rate at which electrical energy is dissipated into other forms of energy is called electrical power i.e.,

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

Where V = Potential difference, R = Resistance and I = current.

CALCULATION:

Given - Potential difference (V) = 200 V and current (I) = 0.5 A

• Electric power can be written as,

​⇒ P = VI

⇒ P = 200 × 0.5 = 100 W