Electrical Energy and Power Practice Questions — JAMB UTME

Question 1Question →

An electric heater of resistance $20\,\Omega$ carries a current of $5\,\text{A}$ when operated. What is the electrical power dissipated by the heater?

500\,\text{W}

100\,\text{W}

4\,\text{W}

2000\,\text{W}

Show answer & explanation

Answer: $500\,\text{W}$

Answer

500\,\text{W}

According to Joule's law of electric heating, electrical power

P

dissipated in a resistor of resistance

R

carrying current

I

is given by

P = I^2 R

. Substituting the given values

I = 5\,\text{A}

and

R = 20\,\Omega

gives

P = (5\,\text{A})^2 \times 20\,\Omega = 25 \times 20 = 500\,\text{W}

.

Step-by-Step Solution

1

Identify the given quantities from the problem statement

Current

I = 5\,\text{A}

and resistance

R = 20\,\Omega

These parameters are required to calculate the power dissipated.

2

Select the appropriate formula for power in terms of current and resistance

P = I^2 R

Joule's heating law specifies that power dissipation is given by

P = I^2 R

.

3

Substitute the values and evaluate the numerical result

P = (5)^2 \times 20 = 25 \times 20 = 500\,\text{W}

Squaring

5

yields

25

, and multiplying by

20

gives the final power output of

500\,\text{W}

.

Key Concept

Electrical Power Dissipation in a Resistor

Estimated Time:45s

Question 2Question →

An electric current of $2.0\,\text{A}$ flows through a resistor of resistance $5.0\,\Omega$ for a duration of $10.0\,\text{seconds}$ . What is the total electrical energy, in Joules, dissipated by the resistor?

Show answer & explanation

Answer: 200

Answer

The total electrical energy dissipated by the resistor is

200\,\text{J}

.

According to Joule's law of heating, the electrical energy

E

converted into thermal energy when a current

I

flows through a resistor

R

for time

t

is given by

E = I^2 R t

. Substituting

I = 2.0\,\text{A}

,

R = 5.0\,\Omega

, and

t = 10.0\,\text{s}

gives

E = (2.0)^2 \times 5.0 \times 10.0 = 4.0 \times 5.0 \times 10.0 = 200\,\text{J}

.

Step-by-Step Solution

1

Identify the given physical values from the problem statement.

I = 2.0\,\text{A}

,

R = 5.0\,\Omega

,

t = 10.0\,\text{s}

.

These are the necessary parameters to compute electrical energy.

2

State the formula for electrical energy dissipated in a resistor (Joule's law).

E = I^2 R t

Electrical energy is equal to power multiplied by time, where power

P = I^2 R

.

3

Calculate the numerical value of the energy.

E = (2.0)^2 \times 5.0 \times 10.0 = 200\,\text{J}

Squaring the current gives

4.0\,\text{A}^2

, and multiplying by

5.0\,\Omega

and

10.0\,\text{s}

yields

200\,\text{J}

.

Key Concept

Joule's Law of Electrical Heating

Question 3Question →

Match each electrical quantity or operational scenario on the left with its corresponding mathematical expression on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Electrical power in terms of current and resistance corresponds to

I^2 R

. Electrical energy consumed across voltage

V

for time

t

corresponds to

\frac{V^2 t}{R}

. Rate of heat generation in terms of voltage and current corresponds to

V I

. Total electric charge transformed corresponds to

\frac{E}{V}

.

Each item is matched by applying the fundamental relationships of electrical energy (

E = V I t = I^2 R t = \frac{V^2 t}{R} = Q V

) and electrical power (

P = \frac{E}{t} = V I = I^2 R = \frac{V^2}{R}

).

Step-by-Step Solution

1

Analyze the first scenario (power with current and resistance)

Power formula derived from Ohm's law (

V = IR

) into

P = VI

gives

P = (IR)I = I^2 R

.

Relates current and resistance directly to power dissipation.

2

Analyze the second scenario (energy with voltage, resistance, and time)

Energy

E = P t

. Using

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

, energy becomes

E = \frac{V^2 t}{R}

.

Expresses energy consumption using voltage and resistance over a given time duration.

3

Analyze the third scenario (rate of heat generation with voltage and current)

Rate of heat generation is power

P = V I

.

Direct definition of electrical power as energy converted per unit time.

4

Analyze the fourth scenario (charge transferred from energy and voltage)

Since potential difference is energy per unit charge (

V = \frac{E}{Q}

), rearranging gives

Q = \frac{E}{V}

.

Relates fundamental definitions of potential difference, energy, and charge.

Key Concept

Formulas for Electrical Energy, Power, and Charge

Question 4Question →

An electric iron rated at $1200\,\text{W}$ is operated for $5\,\text{hours}$ each day for a period of $30\,\text{days}$ . If electrical energy costs $\text{₦}20.00$ per kilowatt-hour ( $\text{kWh}$ ), what is the total cost of electricity consumed by the iron over this period in Naira ( $\text{₦}$ )?

Show answer & explanation

Answer: 3600

Answer

The total cost of electricity consumed by the iron over the 30-day period is 3600 Naira.

To find the cost of electrical energy in commercial units, express power in kilowatts (

1.2\,\text{kW}

) and time in total hours (

150\,\text{h}

). The energy consumed is

1.2 \times 150 = 180\,\text{kWh}

. At a tariff of

\text{₦}20.00

per

\text{kWh}

, the total cost is

180 \times 20 = 3600\,\text{Naira}

.

Step-by-Step Solution

1

Convert the power rating of the appliance from watts to kilowatts

P = \frac{1200\,\text{W}}{1000} = 1.2\,\text{kW}

Commercial energy consumption is calculated in kilowatt-hours (kWh), requiring power in kilowatts.

2

Calculate the total operating time in hours

t = 5\,\text{hours/day} \times 30\,\text{days} = 150\,\text{hours}

The usage duration across the month must be expressed in total hours.

3

Determine electrical energy consumed in kWh

E = P \times t = 1.2\,\text{kW} \times 150\,\text{h} = 180\,\text{kWh}

Energy is the product of power in kilowatts and time in hours.

4

Calculate total cost of energy consumed

\text{Total Cost} = 180\,\text{kWh} \times \text{₦}20.00/\text{kWh} = \text{₦}3600

Total cost is found by multiplying energy in kWh by the unit tariff rate.

Key Concept

Commercial electrical energy unit (kWh) and billing calculation

Question 5Question →

Two electric lamps rated at $60\,\text{W}, 220\,\text{V}$ and $100\,\text{W}, 220\,\text{V}$ respectively are connected in series across a $220\,\text{V}$ supply line. What is the total electric power dissipated by the combination?

37.5\,\text{W}

80.0\,\text{W}

160.0\,\text{W}

40.0\,\text{W}

Show answer & explanation

Answer: $37.5\,\text{W}$

Answer

The total electric power dissipated by the two lamps in series is

37.5\,\text{W}

.

For two appliances designed for the same rated voltage

V

connected in series across that voltage

V

, the equivalent power is given by the formula

P_{\text{total}} = \frac{P_1 P_2}{P_1 + P_2}

. Substituting

P_1 = 60\,\text{W}

and

P_2 = 100\,\text{W}

gives

P_{\text{total}} = \frac{60 \times 100}{60 + 100} = 37.5\,\text{W}

.

Step-by-Step Solution

1

Calculate the resistance of each lamp from its rating.

R_1 = \frac{V^2}{P_1} = \frac{220^2}{60} = 806.67\,\Omega

and

R_2 = \frac{V^2}{P_2} = \frac{220^2}{100} = 484.00\,\Omega

Electrical devices are rated at a specific voltage, allowing their resistance to be determined via

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

.

2

Find the total resistance of the series circuit.

R_{\text{total}} = R_1 + R_2 = 806.67 + 484.00 = 1290.67\,\Omega

Resistors in series add directly.

3

Calculate total power drawn from the

220\,\text{V}

supply.

P_{\text{total}} = \frac{V^2}{R_{\text{total}}} = \frac{220^2}{1290.67} = 37.5\,\text{W}

Total power is total voltage squared divided by equivalent resistance.

Key Concept

Power combination in series circuits

Estimated Time:1m 30s

Question 6Question →

Match each electrical operational scenario on the left with its corresponding effect on electrical energy or power on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Doubling current quadruples heat energy (

H \propto I^2

); halving voltage reduces power to one-quarter (

P \propto V^2

); connecting two identical resistors in parallel doubles total power (

P \propto 1/R_{\text{eq}}

); and operating a 500 W appliance for 2 hours consumes 1.0 kWh.

Each left-side scenario correctly aligns with fundamental laws of electrical energy and power: doubling current quadruples heat energy via

H \propto I^2

; halving voltage reduces power to one-quarter via

P \propto V^2

; parallel resistor connection halves total resistance and doubles power; and operating a 0.5 kW appliance for 2 hours consumes 1.0 kWh.

Step-by-Step Solution

1

Apply Joule's Law of Heating (

H = I^2 R t

) to determine the effect of changing current.

Doubling current from

I

to

2I

results in

H' = (2I)^2 R t = 4 I^2 R t = 4H

, so heat quadruples.

Heat generation depends on the square of current when resistance and time are constant.

2

Apply the voltage-power relationship (

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

) for changing voltage.

Halving potential difference to

V/2

yields

P' = \frac{(V/2)^2}{R} = \frac{V^2}{4R} = \frac{P}{4}

, so power drops to one-quarter.

Power is directly proportional to the square of potential difference across a constant resistance.

3

Calculate parallel resistance and total circuit power.

Equivalent resistance of two parallel resistors is

R/2

. Total power is

P_{\text{total}} = \frac{V^2}{R/2} = 2 \frac{V^2}{R}

, which is double the single resistor power.

Parallel combination reduces net resistance by half, thereby doubling total current and power drawn from a constant voltage source.

4

Calculate energy consumption in commercial units (kWh).

E = 0.5\,\text{kW} \times 2\,\text{h} = 1.0\,\text{kWh}

.

Commercial energy is computed by expressing power in kW and time in hours.

Key Concept

Mathematical relationships governing electrical power (

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

), Joule heating (

H = I^2 R t

), and commercial energy units (kWh).

Question 7Question →

An electric heating element rated at $1000\,\text{W}$ and $220\,\text{V}$ is connected to a $110\,\text{V}$ power line. Assuming the electrical resistance of the heating element remains constant, what is the power dissipated by the heater when operating at this reduced voltage?

250\,\text{W}

500\,\text{W}

1000\,\text{W}

2000\,\text{W}

Show answer & explanation

Answer: $250\,\text{W}$

Answer

250\,\text{W}

The electrical power rating of an appliance defines its fixed resistance via

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

. With

R = 48.4\,\Omega

, connecting the appliance to a

110\,\text{V}

supply yields

P = \frac{110^2}{48.4} = 250\,\text{W}

. Alternatively, because

P \propto V^2

for a constant resistance, halving the voltage reduces the power by a factor of

(1/2)^2 = 1/4

, giving

1000\,\text{W} \times \frac{1}{4} = 250\,\text{W}

.

Step-by-Step Solution

1

Calculate the resistance

R

of the heater from its rated values.

R = \frac{V_{\text{rated}}^2}{P_{\text{rated}}} = \frac{220^2}{1000} = \frac{48400}{1000} = 48.4\,\Omega

The resistance of a heating element is determined by its physical design and rated specifications.

2

Calculate the power

P_{\text{new}}

dissipated when connected to the

110\,\text{V}

supply.

P_{\text{new}} = \frac{V_{\text{new}}^2}{R} = \frac{110^2}{48.4} = \frac{12100}{48.4} = 250\,\text{W}

Electric power dissipated across a constant resistance varies with the square of the applied voltage.

Key Concept

Relationship between voltage, resistance, and electrical power dissipation

Estimated Time:1m 30s

Question 8Question →

An electric lamp rated at $60\,\text{W}$ is kept switched on for $50\,\text{seconds}$ . What is the total electrical energy consumed by the lamp during this time?

3000\,\text{J}

300\,\text{J}

1.2\,\text{J}

110\,\text{J}

Show answer & explanation

Answer: $3000\,\text{J}$

Answer

3000\,\text{J}

Electrical energy is calculated using the formula

E = P \times t

. Substituting the given values

P = 60\,\text{W}

and

t = 50\,\text{s}

gives

E = 60 \times 50 = 3000\,\text{J}

.

Step-by-Step Solution

1

Identify the given physical quantities.

Power

P = 60\,\text{W}

and time

t = 50\,\text{s}

.

Power is given in watts (

\text{J/s}

) and time is given in seconds.

2

Apply the electrical energy formula

E = P \times t

.

E = 60\,\text{W} \times 50\,\text{s} = 3000\,\text{J}

.

Total electrical energy is the product of power dissipation and time duration.

Key Concept

Electrical Energy and Power

Estimated Time:45s

Question 9Question →

An electric immersion heater with a resistance of $20\,\Omega$ carries a steady current of $3\,\text{A}$ . How much electrical energy is dissipated as heat by the heater in $10\,\text{seconds}$ ?

600\,\text{J}

1200\,\text{J}

1800\,\text{J}

180\,\text{J}

Show answer & explanation

Answer: $1800\,\text{J}$

Answer

The total electrical energy dissipated as heat by the heater is

1800\,\text{J}

.

According to Joule's law of heating, the electrical energy

E

converted into heat energy in a resistor is given by

E = I^2 R t

. Substituting

I = 3\,\text{A}

,

R = 20\,\Omega

, and

t = 10\,\text{s}

yields

E = (3)^2 \times 20 \times 10 = 9 \times 200 = 1800\,\text{J}

.

Step-by-Step Solution

1

Identify the given physical quantities

Resistance

R = 20\,\Omega

, Current

I = 3\,\text{A}

, Time

t = 10\,\text{s}

Recognize the required values needed to compute heat energy dissipated in a resistor.

2

Select the appropriate formula for heat energy

E = I^2 R t

Joule's Law of Heating relates electrical energy, current, resistance, and time.

3

Substitute the known values into the equation and calculate

E = (3)^2 \times 20 \times 10 = 9 \times 20 \times 10 = 1800\,\text{J}

Perform arithmetic multiplication to find the final heat energy in Joules.

Key Concept

Joule's Law of Heating

Question 10Question →

An electric ceiling fan operates on a $220\,\text{V}$ mains supply and draws a steady current of $0.5\,\text{A}$ . What is the electrical power rating of the fan?

110\,\text{W}

440\,\text{W}

220.5\,\text{W}

110\,\text{J}

Show answer & explanation

Answer: $110\,\text{W}$

Answer

The electrical power rating of the fan is

110\,\text{W}

.

Electrical power

P

delivered to a circuit element is given by

P = VI

, where

V

is potential difference and

I

is current. Substituting

220\,\text{V}

and

0.5\,\text{A}

gives

110\,\text{W}

.

Step-by-Step Solution

1

Identify the given physical quantities

Voltage

V = 220\,\text{V}

and current

I = 0.5\,\text{A}

.

These are the basic electrical parameters required to calculate power.

2

Apply the electrical power formula

P = V \times I

Electrical power is defined as the product of potential difference across a device and current flowing through it.

3

Substitute values and compute the power

P = 220\,\text{V} \times 0.5\,\text{A} = 110\,\text{W}

Multiplying potential difference by current yields the rate of energy conversion in Watts.

Key Concept

Electrical Power (

P = VI

)

Estimated Time:45s

Question 11Question →

An electric water heater operating at a voltage of $240\,\text{V}$ has a heating element of resistance $48\,\Omega$ . It is used to heat $1.5\,\text{kg}$ of water from $20^\circ\text{C}$ to $100^\circ\text{C}$ . If the thermal efficiency of the heating process is $80\%$ , calculate the total electrical energy consumed by the heater in kilojoules ( $\text{kJ}$ ). [Take specific heat capacity of water = $4200\,\text{J}\cdot\text{kg}^{-1}\cdot\text{K}^{-1}$ ]

Show answer & explanation

Answer: 630

Answer

The total electrical energy consumed by the heater is

630\,\text{kJ}

.

The thermal energy required to raise the temperature of

1.5\,\text{kg}

of water by

80^\circ\text{C}

is

Q = 1.5 \times 4200 \times 80 = 504,000\,\text{J} = 504\,\text{kJ}

. Taking into account the

80\%

thermal efficiency, the total electrical energy consumed is

E_{\text{elec}} = \frac{504\,\text{kJ}}{0.80} = 630\,\text{kJ}

.

Step-by-Step Solution

1

Calculate the useful heat energy needed to heat the water.

Q = m c (T_2 - T_1) = 1.5 \times 4200 \times (100 - 20) = 504,000\,\text{J} = 504\,\text{kJ}

.

The thermal energy transferred to the water depends on its mass, specific heat capacity, and temperature increase.

2

Account for the efficiency of the heating element to find total electrical energy input.

E_{\text{elec}} = \frac{Q}{\text{Efficiency}} = \frac{504\,\text{kJ}}{0.80} = 630\,\text{kJ}

.

Since only

80\%

of the electrical energy is converted into useful heat energy for the water, the input electrical energy must be greater than the output heat energy.

Key Concept

Conversion of electrical energy to thermal energy and application of thermal efficiency.

Question 12Question →

An electric generator supplies power to a workshop through transmission lines having a total resistance of $2\,\Omega$ . The workshop operates electrical equipment drawing a power of $4\,\text{kW}$ at a terminal voltage of $200\,\text{V}$ . What is the total power generated by the generator?

4.8\,\text{kW}

4.0\,\text{kW}

4.04\,\text{kW}

24.0\,\text{kW}

Show answer & explanation

Answer: $4.8\,\text{kW}$

Answer

The total power generated by the generator is

4.8\,\text{kW}

.

The electrical current flowing through the system is

20\,\text{A}

based on the load power of

4000\,\text{W}

at

200\,\text{V}

. The power lost in transmission lines is

P = I^2 R = 20^2 \times 2 = 800\,\text{W}

(

0.8\,\text{kW}

). Therefore, the total electrical power generated by the source must be the sum of the load power and line losses, giving

4.0\,\text{kW} + 0.8\,\text{kW} = 4.8\,\text{kW}

.

Step-by-Step Solution

1

Calculate the current flowing through the circuit using the power and voltage at the workshop.

I = \frac{P_{\text{workshop}}}{V} = \frac{4000\,\text{W}}{200\,\text{V}} = 20\,\text{A}

Current is uniform across the series transmission line.

2

Calculate the power loss dissipated as heat in the transmission lines.

P_{\text{loss}} = I^2 R = (20\,\text{A})^2 \times 2\,\Omega = 400 \times 2 = 800\,\text{W} = 0.8\,\text{kW}

By Joule's law of heating, power dissipated in a resistor carrying current

I

is

I^2 R

.

3

Sum the useful power delivered to the workshop and the transmission power loss to obtain total generated power.

P_{\text{total}} = P_{\text{workshop}} + P_{\text{loss}} = 4.0\,\text{kW} + 0.8\,\text{kW} = 4.8\,\text{kW}

Total energy generated per unit time equals energy consumed by load plus energy lost.

Key Concept

Power Loss in Transmission Lines and Total Source Power

Question 13Question →

A battery of electromotive force (e.m.f.) $24\,\text{V}$ and internal resistance $2\,\Omega$ is connected across a parallel network consisting of two resistors of resistances $6\,\Omega$ and $12\,\Omega$ . What is the total electrical energy, in Joules, dissipated in the external circuit during an operating time of $5\,\text{minutes}$ ?

Show answer & explanation

Answer: 19200

Answer

The total electrical energy dissipated in the external circuit over 5 minutes is

19,200\,\text{J}

.

To compute the external energy dissipated, first combine the parallel resistors to get an equivalent external resistance of

4\,\Omega

. Adding the

2\,\Omega

internal resistance yields a total circuit resistance of

6\,\Omega

, which draws

4\,\text{A}

of current from the

24\,\text{V}

battery. The power delivered to the external load is

P_{ext} = I^2 R_p = 4^2 \times 4 = 64\,\text{W}

. Multiplying this power by the time duration in seconds (

5 \times 60 = 300\,\text{s}

) yields

19,200\,\text{J}

.

Step-by-Step Solution

1

Calculate the equivalent resistance of the external parallel circuit

R_p = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4\,\Omega

The two external resistors are connected in parallel.

2

Calculate the total current supplied by the battery

I = \frac{E}{R_p + r} = \frac{24}{4 + 2} = \frac{24}{6} = 4\,\text{A}

Ohm's law for a complete circuit accounts for both external resistance and internal resistance.

3

Determine the power dissipated exclusively in the external circuit

P_{ext} = I^2 R_p = (4)^2 \times 4 = 16 \times 4 = 64\,\text{W}

Electrical power dissipated across the external load depends on the total current squared times the equivalent external resistance.

4

Convert time from minutes to seconds and calculate total energy dissipated

t = 5 \times 60 = 300\,\text{s}

,

E = P_{ext} \times t = 64 \times 300 = 19,200\,\text{J}

Electrical energy is the product of power in Watts and time in seconds.

Key Concept

Electrical Energy and Power in Circuits with Internal Resistance

Question 14Question →

An electric lamp rated $60\,\text{W}$ operates normally when connected to a $240\,\text{V}$ mains supply. What is the electric current drawn by the lamp?

0.25\,\text{A}

4.0\,\text{A}

2.5\,\text{A}

14\,400\,\text{A}

Show answer & explanation

Answer: $0.25\,\text{A}$

Answer

The electric current drawn by the lamp is

0.25\,\text{A}

.

Electric power is defined by the formula

P = IV

, where

P

is power in watts,

I

is current in amperes, and

V

is potential difference in volts. Rearranging to solve for current yields

I = \frac{P}{V}

. Substituting

60\,\text{W}

for power and

240\,\text{V}

for voltage gives

I = \frac{60}{240} = 0.25\,\text{A}

.

Step-by-Step Solution

1

Identify given parameters and state the electric power formula.

Power

P = 60\,\text{W}

and voltage

V = 240\,\text{V}

. Power is related to current and voltage by

P = IV

.

Electric power dissipated by a component is the product of current and voltage across it.

2

Rearrange the formula to solve for current

I

.

I = \frac{P}{V}

Making current

I

the subject of the equation.

3

Substitute the values into the formula and solve.

I = \frac{60\,\text{W}}{240\,\text{V}} = 0.25\,\text{A}

Dividing

60

by

240

gives

0.25

amperes.

Key Concept

Relationship between Electrical Power, Voltage, and Current

Question 15Question →

An electric lamp of resistance $10\,\Omega$ is connected to a cell of electromotive force (e.m.f.) $12\,\text{V}$ and internal resistance $2\,\Omega$ . What is the electrical power dissipated as heat inside the cell?

2.0\,\text{W}

2.88\,\text{W}

10.0\,\text{W}

12.0\,\text{W}

Show answer & explanation

Answer: $2.0\,\text{W}$

Answer

The electrical power dissipated inside the cell is

2.0\,\text{W}

.

The total opposition to current in the circuit includes both the external lamp resistance and the internal resistance of the cell (

10\,\Omega + 2\,\Omega = 12\,\Omega

). This yields a circuit current of

1.0\,\text{A}

. Using Joule's law of heating (

P = I^2 r

), the power lost specifically inside the cell is

(1.0)^2 \times 2 = 2.0\,\text{W}

.

Step-by-Step Solution

1

Calculate the total resistance of the circuit.

R_{\text{total}} = R + r = 10\,\Omega + 2\,\Omega = 12\,\Omega

The internal resistance of the cell is in series with the external load resistance.

2

Determine the current flowing through the circuit.

I = \frac{E}{R_{\text{total}}} = \frac{12\,\text{V}}{12\,\Omega} = 1.0\,\text{A}

By Ohm's law applied to a complete circuit, current equals total e.m.f. divided by total resistance.

3

Calculate the power dissipated as heat in the internal resistance.

P_{\text{internal}} = I^2 r = (1.0\,\text{A})^2 \times 2\,\Omega = 2.0\,\text{W}

Joule heating power in a resistor is given by

P = I^2 r

.

Key Concept

Electrical Power Dissipation and Internal Resistance

Estimated Time:1m 30s

Question 16Question →

Match each electrical quantity or unit in List I with its corresponding formula or definition in List II.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Electrical Power (

P

) matches Product of potential difference and current (

V \times I

); Electrical Energy (

E

) matches Product of electrical power and time (

P \times t

); SI unit of Electric Power matches Watt (

\text{W}

); Commercial unit of Electrical Energy matches Kilowatt-hour (

\text{kWh}

).

Electrical power is defined by the product of potential difference and current (

P = VI

) and measured in Watts (

\text{W}

). Electrical energy is power multiplied by time (

E = Pt

) and commercially measured in Kilowatt-hours (

\text{kWh}

).

Step-by-Step Solution

1

Identify the formula for electrical power.

Electrical power

P = VI

.

Power represents the rate of electrical energy dissipation across a potential difference.

2

Identify the formula for electrical energy.

Electrical energy

E = P \times t

.

Energy is the work done over a period of time.

3

Determine the standard SI unit for power.

The SI unit of power is the Watt (

\text{W}

).

One Watt is defined as one Joule per second.

4

Determine the commercial unit for electrical energy.

The commercial unit of energy is the Kilowatt-hour (

\text{kWh}

).

1 kWh is equal to

3.6 \times 10^6\,\text{J}

, which is convenient for commercial electricity billing.

Key Concept

Electrical Energy and Power Definitions and Units

Question 17Question →

Match each electrical scenario or description in List I with its corresponding mathematical expression or unit in List II.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are: Energy dissipated in duration

t

matches

I^2 R t

; Power rating under voltage

V

matches

\frac{V^2}{R}

; Commercial unit equal to

3.6 \times 10^6\,\text{J}

matches 1 kWh; Rate of heat production matches

I V

.

Each description in List I corresponds directly to its standard physical formula or commercial unit in List II based on the definitions of electrical power (

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

) and electrical energy (

E = I^2 R t = P t

).

Step-by-Step Solution

1

Identify Joule's law of heating for energy dissipation over time.

Energy

E = I^2 R t

, matching the expression

I^2 R t

.

Joule's heating effect states that electrical energy transformed into heat energy is directly proportional to

I^2

,

R

, and

t

.

2

Relate power, voltage, and resistance.

Power

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

, matching the expression

\frac{V^2}{R}

.

Substituting Ohm's law

I = \frac{V}{R}

into power formula

P = I V

yields

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

.

3

Convert kilowatt-hours to joules for commercial unit matching.

1\,\text{kWh} = 1000\,\text{W} \times 3600\,\text{s} = 3.6 \times 10^6\,\text{J}

, matching 1 kWh.

Kilowatt-hour is the standard energy unit used by electric utility companies.

4

Determine the general definition of electrical power as the rate of energy transfer.

Power

P = I V

, matching the expression

I V

.

Electric power is defined as the product of current

I

and potential difference

V

.

Key Concept

Formulas and units for electrical energy, electric power, Joule's law of heating, and kilowatt-hours.

Estimated Time:1m 30s

Question 18Question →

An electric immersion heater rated at $1.5\,\text{kW}$ is operated on a $240\,\text{V}$ mains supply for $14\,\text{minutes}$ . Calculate the total electrical energy consumed by the heater during this period, in megajoules ( $\text{MJ}$ ).

Show answer & explanation

Answer: 1.26

Answer

The total electrical energy consumed by the heater is

1.26\,\text{MJ}

.

Electrical energy consumed is obtained by multiplying electrical power by time (

E = P \times t

). Converting power to watts (

1.5\,\text{kW} = 1500\,\text{W}

) and time to seconds (

14\,\text{min} = 840\,\text{s}

) yields

E = 1500 \times 840 = 1,260,000\,\text{J} = 1.26\,\text{MJ}

.

Step-by-Step Solution

1

Convert power rating from kilowatts to watts

P = 1.5\,\text{kW} = 1500\,\text{W}

The standard SI unit of power for energy calculation in Joules is Watts.

2

Convert time duration from minutes to seconds

t = 14\,\text{minutes} \times 60\,\text{seconds/minute} = 840\,\text{seconds}

The standard SI unit of time in Joule calculations is seconds.

3

Calculate energy consumed in Joules using

E = P \times t

E = 1500\,\text{W} \times 840\,\text{s} = 1,260,000\,\text{J}

Electrical energy is the product of power in watts and time in seconds.

4

Convert energy from Joules to Megajoules

E = \frac{1,260,000}{10^6} = 1.26\,\text{MJ}

One Megajoule (

1\,\text{MJ}

) is equivalent to

10^6\,\text{Joules}

.

Key Concept

Electrical Energy Consumption

Question 19Question →

An electric water heater with an internal heating element of resistance $40\,\Omega$ is connected to a $200\,\text{V}$ mains power supply. If the heater is operated for $15\,\text{minutes}$ each day, what is the total electrical energy consumed by the heater over a period of $30\,\text{days}$ ?

7.5\,\text{kWh}

30.0\,\text{kWh}

450.0\,\text{kWh}

37.5\,\text{kWh}

Show answer & explanation

Answer: $7.5\,\text{kWh}$

Answer

The total electrical energy consumed over 30 days is

7.5\,\text{kWh}

.

The electrical power rating of the water heater is

P = \frac{V^2}{R} = \frac{200^2}{40} = 1000\,\text{W} = 1.0\,\text{kW}

. Operating for

15\,\text{minutes}

(

0.25\,\text{hours}

) daily for

30\,\text{days}

yields a total time of

7.5\,\text{hours}

. The total energy consumed is

1.0\,\text{kW} \times 7.5\,\text{h} = 7.5\,\text{kWh}

.

Step-by-Step Solution

1

Calculate the electric power rating of the heater

P = \frac{V^2}{R} = \frac{200^2}{40} = \frac{40000}{40} = 1000\,\text{W} = 1.0\,\text{kW}

Electric power dissipated in a resistance

R

connected across potential difference

V

is given by

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

.

2

Calculate the total operating time in hours

t = 30 \times \frac{15}{60} = 7.5\,\text{hours}

Commercial electrical energy is measured in kilowatt-hours (

\text{kWh}

), so time must be converted from minutes to hours.

3

Calculate total electrical energy consumed

E = P \times t = 1.0\,\text{kW} \times 7.5\,\text{h} = 7.5\,\text{kWh}

Electrical energy consumed is the product of power in kilowatts and total time in hours.

Key Concept

Calculation of commercial electrical energy consumption in kilowatt-hours using electric power and operating time.

Electrical Energy and Power

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