Numerical Problems on Electric Power and Energy

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Home CBSE Class 10 Science Extra Qs Numerical Problems on Electric Power and Energy

Here we are providing numerical problems based on electrical power and energy. These problems are useful for students studying in class 10. Practice these questions to master these topics.

Class 10Numerical Problems
Useful forClass 10 Students
Topics CoveredElectric Power and Energy
No. of Questions70
TypeNumerical Problems

Numerical Problems on Electric Power and Energy

Numerical Problems on Electric Power and Energy

Class 10 Science · Electricity
SQ

Power, Voltage, and Current Practice Set 1

Attempt each question before revealing the answer
Ch 12 · Electricity
1
What will be the current drawn by an electric bulb of 40 W when it is connected to a source of 220V?
Answer 0.18 A 📝
Detailed Solution

Given Power $P = 40\text{ W}$ and Voltage $V = 220\text{ V}$. Using the formula $P = V \times I$:

$$I = \frac{P}{V} = \frac{40}{220}$$
$$I = \frac{2}{11} \approx 0.18\text{ A}$$
2
A bulb is rated as 250V; 0.4A. Find its power and resistance.
Answer $P = 100\text{ W}$, $R = 625\text{ }\Omega$ 📝
Detailed Solution

Given $V = 250\text{ V}$ and $I = 0.4\text{ A}$.

Power:

$$P = V \times I = 250 \times 0.4 = 100\text{ W}$$

Resistance: By Ohm’s Law ($V = IR$):

$$R = \frac{V}{I} = \frac{250}{0.4} = 625\text{ }\Omega$$
3
An electric bulb is connected to a 220V power supply line. If the bulb draw a current of 0.5A, calculate the power of the bulb.
Answer 110 W 📝
Detailed Solution

Given $V = 220\text{ V}$ and $I = 0.5\text{ A}$.

$$P = V \times I = 220 \times 0.5$$
$$P = 110\text{ W}$$
4
An electric bulb is connected to a 250 V generator. The current is 0.50 A. What is the power of the bulb?
Answer 125 W 📝
Detailed Solution

Given $V = 250\text{ V}$ and $I = 0.50\text{ A}$.

$$P = V \times I = 250 \times 0.50$$
$$P = 125\text{ W}$$
5
What current will be taken by a 920W appliance if the supply voltage is 230V?
Answer 4 A 📝
Detailed Solution

Given $P = 920\text{ W}$ and $V = 230\text{ V}$. Using $P = VI$:

$$I = \frac{P}{V} = \frac{920}{230}$$
$$I = 4\text{ A}$$
6
When an electric lamp is connected to 12V battery, it draws a current 0.5A. Find the power of the lamp.
Answer 6 W 📝
Detailed Solution

Given $V = 12\text{ V}$ and $I = 0.5\text{ A}$.

$$P = V \times I = 12 \times 0.5$$
$$P = 6\text{ W}$$
7
Calculate the power used in 2Ω resistor in each (i) a 6V battery in series with 1Ω and 2Ω resistor (ii) a 4V battery in parallel with 12Ω and 2Ω resistor.
Answer (i) 8 W (ii) 8 W 📝
Detailed Solution

(i) In Series: Total resistance $R_{eq} = 1\text{ }\Omega + 2\text{ }\Omega = 3\text{ }\Omega$. Current in the circuit $I = \frac{V}{R_{eq}} = \frac{6}{3} = 2\text{ A}$.
Since current remains the same in series, power in the $2\text{ }\Omega$ resistor is:

$$P = I^2 R = (2)^2 \times 2 = 4 \times 2 = 8\text{ W}$$

(ii) In Parallel: Voltage across parallel branches is the same as the battery voltage. So, the voltage across the $2\text{ }\Omega$ resistor is $4\text{V}$.

$$P = \frac{V^2}{R} = \frac{4^2}{2} = \frac{16}{2} = 8\text{ W}$$
8
A 100 W electric light bulb is connected to a 250 V supply. Determine (a) the current flowing in the bulb, and (b) the resistance of the bulb.
Answer (a) 0.4 A (b) $625\text{ }\Omega$ 📝
Detailed Solution

Given $P = 100\text{ W}$ and $V = 250\text{ V}$.

(a) Current:

$$I = \frac{P}{V} = \frac{100}{250} = 0.4\text{ A}$$

(b) Resistance:

$$R = \frac{V^2}{P} = \frac{250^2}{100} = \frac{62500}{100} = 625\text{ }\Omega$$
9
Calculate the power dissipated when a current of 4 mA flows through a resistance of 5 kΩ.
Answer 0.08 W (or 80 mW) 📝
Detailed Solution

Convert the units to SI units: $I = 4\text{ mA} = 0.004\text{ A}$ and $R = 5\text{ k}\Omega = 5000\text{ }\Omega$.

$$P = I^2 R$$
$$P = (0.004)^2 \times 5000 = 0.000016 \times 5000$$
$$P = 0.08\text{ W}$$
10
An electric kettle has a resistance of 30Ω. What current will flow when it is connected to a 240 V supply? Find also the power rating of the kettle.
Answer $I = 8\text{ A}$, $P = 1920\text{ W}$ (or $1.92\text{ kW}$) 📝
Detailed Solution

Given $R = 30\text{ }\Omega$ and $V = 240\text{ V}$.

Current:

$$I = \frac{V}{R} = \frac{240}{30} = 8\text{ A}$$

Power rating:

$$P = V \times I = 240 \times 8 = 1920\text{ W}$$
SQ

Electrical Energy and Cost Practice Set 2

Apply $E = P \times t$ and commercial electricity formulas
Ch 12 · Electricity
11
A current of 5 A flows in the winding of an electric motor, the resistance of the winding being 100Ω. Determine (a) the p.d. across the winding, and (b) the power dissipated by the coil.
Answer (a) 500 V (b) 2500 W 📝
Detailed Solution

Given $I = 5\text{ A}$ and $R = 100\text{ }\Omega$.

(a) Potential Difference (p.d.):

$$V = IR = 5 \times 100 = 500\text{ V}$$

(b) Power dissipated:

$$P = I^2 R = (5)^2 \times 100 = 25 \times 100 = 2500\text{ W}$$
12
The current/voltage relationship for two resistors A and B is as shown in below Figure. Determine the value of the resistance of each resistor and also find the power dissipated through each resistor.
Answer Formula Based 📝
Detailed Solution

(Note: Since the figure is not provided, follow these standard steps to solve V-I graph problems).

1. Resistance: The resistance $R$ is calculated using Ohm’s law $R = V/I$. Find any distinct point $(I, V)$ on the line for resistor A and B. If it’s a V-I graph (Voltage on Y-axis), the slope is the resistance. If it’s an I-V graph, the inverse of the slope is the resistance.

2. Power: Once you determine the resistance $R$ from the graph and read the respective current $I$ or voltage $V$ from a coordinate point, use the power formulas:

$$P_A = V_A \times I_A \quad \text{and} \quad P_B = V_B \times I_B$$
13
The hot resistance of a 240 V filament lamp is 960Ω. Find the current taken by the lamp and its power rating.
Answer $I = 0.25\text{ A}$, $P = 60\text{ W}$ 📝
Detailed Solution

Given $V = 240\text{ V}$ and $R = 960\text{ }\Omega$.

Current:

$$I = \frac{V}{R} = \frac{240}{960} = \frac{1}{4} = 0.25\text{ A}$$

Power Rating:

$$P = V \times I = 240 \times 0.25 = 60\text{ W}$$
14
A 12 V battery is connected across a load having a resistance of 40Ω. Determine the current flowing in the load, the power consumed and the energy dissipated in 2 minutes.
Answer $I = 0.3\text{ A}$, $P = 3.6\text{ W}$, $E = 432\text{ J}$ 📝
Detailed Solution

Given $V = 12\text{ V}$, $R = 40\text{ }\Omega$, and time $t = 2\text{ minutes} = 120\text{ s}$.

$$I = \frac{V}{R} = \frac{12}{40} = 0.3\text{ A}$$
$$P = V \times I = 12 \times 0.3 = 3.6\text{ W}$$
$$Energy (E) = P \times t = 3.6\text{ W} \times 120\text{ s} = 432\text{ J}$$
15
A source of e.m.f. of 15 V supplies a current of 2 A for six minutes. How much energy is provided in this time?
Answer 10,800 J (or 10.8 kJ) 📝
Detailed Solution

Given $V = 15\text{ V}$, $I = 2\text{ A}$, and $t = 6\text{ mins} = 360\text{ seconds}$.

$$Energy (E) = V \times I \times t$$
$$E = 15 \times 2 \times 360 = 30 \times 360 = 10800\text{ J}$$
16
Electrical equipment in an office takes a current of 13 A from a 240 V supply. Estimate the cost per week of electricity if the equipment is used for 30 hours each week and 1 kWh of energy costs 7p.
Answer 655.2p (or £6.55) 📝
Detailed Solution

First, calculate the power in kilowatts (kW):

$$P = V \times I = 240 \times 13 = 3120\text{ W} = 3.12\text{ kW}$$

Energy used in one week ($E$) for $t = 30\text{ hours}$:

$$E = P \times t = 3.12\text{ kW} \times 30\text{ h} = 93.6\text{ kWh}$$

Total Cost:

$$\text{Cost} = 93.6\text{ kWh} \times 7\text{p/kWh} = 655.2\text{p}$$
17
An electric heater consumes 3.6 MJ when connected to a 250 V supply for 40 minutes. Find the power rating of the heater and the current taken from the supply.
Answer $P = 1500\text{ W}$, $I = 6\text{ A}$ 📝
Detailed Solution

Given $E = 3.6\text{ MJ} = 3,600,000\text{ J}$ and $t = 40\text{ mins} = 2400\text{ s}$.

Power:

$$P = \frac{E}{t} = \frac{3600000}{2400} = 1500\text{ W} \text{ (or } 1.5\text{ kW)}$$

Current: Given $V = 250\text{ V}$.

$$I = \frac{P}{V} = \frac{1500}{250} = 6\text{ A}$$
18
Determine the power dissipated by the element of an electric fire of resistance 20Ω when a current of 10 A flows through it. If the fire is on for 6 hours determine the energy used and the cost if 1 unit of electricity costs 7p.
Answer $P = 2000\text{ W}$, $E = 12\text{ kWh}$, Cost = 84p 📝
Detailed Solution

Power:

$$P = I^2 R = (10)^2 \times 20 = 100 \times 20 = 2000\text{ W} = 2\text{ kW}$$

Energy (in kWh or units):

$$E = P(\text{in kW}) \times t(\text{in hours}) = 2\text{ kW} \times 6\text{ h} = 12\text{ kWh}$$

Cost:

$$\text{Cost} = 12\text{ units} \times 7\text{p/unit} = 84\text{p}$$
19
A business uses two 3 kW fires for an average of 20 hours each per week, and six 150 W lights for 30 hours each per week. If the cost of electricity is 7p per unit, determine the weekly cost of electricity to the business.
Answer 1029p (or £10.29) 📝
Detailed Solution

Energy for the fires:

$$E_{fires} = 2 \text{ units} \times 3\text{ kW} \times 20\text{ h} = 120\text{ kWh}$$

Energy for the lights: Convert $150\text{ W}$ to $0.15\text{ kW}$.

$$E_{lights} = 6 \text{ units} \times 0.15\text{ kW} \times 30\text{ h} = 27\text{ kWh}$$

Total Energy and Cost:

$$E_{total} = 120 + 27 = 147\text{ kWh}$$
$$\text{Cost} = 147 \times 7\text{p} = 1029\text{p}$$
20
If 5 A, 10 A and 13 A fuses are available, state which is most appropriate for the following appliances which are both connected to a 240 V supply (a) Electric toaster having a power rating of 1 kW (b) Electric fire having a power rating of 3 kW.
Answer (a) 5 A fuse (b) 13 A fuse 📝
Detailed Solution

A fuse is selected so that its rating is slightly higher than the normal operating current of the appliance.

(a) Toaster ($1\text{ kW} = 1000\text{ W}$):

$$I = \frac{P}{V} = \frac{1000}{240} \approx 4.17\text{ A}$$

The most appropriate fuse is the 5 A fuse.

(b) Electric Fire ($3\text{ kW} = 3000\text{ W}$):

$$I = \frac{P}{V} = \frac{3000}{240} = 12.5\text{ A}$$

The most appropriate fuse is the 13 A fuse.

Numerical Problems on Electric Power and Energy (Continued)

Class 10 Science · Electricity
SQ

Power, Voltage, and Current Practice Set 3

Attempt each question before revealing the answer
Ch 12 · Electricity
21
The hot resistance of a 250 V filament lamp is $625\text{ }\Omega$. Determine the current taken by the lamp and its power rating.
Answer $I = 0.4\text{ A}$, $P = 100\text{ W}$ 📝
Detailed Solution

Given $V = 250\text{ V}$ and $R = 625\text{ }\Omega$.

Current:

$$I = \frac{V}{R} = \frac{250}{625} = 0.4\text{ A}$$

Power Rating:

$$P = V \times I = 250 \times 0.4 = 100\text{ W}$$
22
Determine the resistance of a coil connected to a 150 V supply when a current of (a) 75 mA (b) 300 mA flows through it. Determine the power dissipated through it.
Answer (a) $2000\text{ }\Omega, 11.25\text{ W}$ (b) $500\text{ }\Omega, 45\text{ W}$ 📝
Detailed Solution

Given $V = 150\text{ V}$.

(a) Current $I = 75\text{ mA} = 0.075\text{ A}$:

$$R = \frac{V}{I} = \frac{150}{0.075} = 2000\text{ }\Omega \text{ (or } 2\text{ k}\Omega\text{)}$$
$$P = V \times I = 150 \times 0.075 = 11.25\text{ W}$$

(b) Current $I = 300\text{ mA} = 0.3\text{ A}$:

$$R = \frac{V}{I} = \frac{150}{0.3} = 500\text{ }\Omega$$
$$P = V \times I = 150 \times 0.3 = 45\text{ W}$$
23
Determine the resistance of an electric fire which takes a current of 12A from a 240 V supply. Find also the power rating of the fire and the energy used in 20 h.
Answer $R = 20\text{ }\Omega$, $P = 2880\text{ W}$, $E = 57.6\text{ kWh}$ 📝
Detailed Solution

Given $I = 12\text{ A}$ and $V = 240\text{ V}$.

Resistance:

$$R = \frac{V}{I} = \frac{240}{12} = 20\text{ }\Omega$$

Power:

$$P = V \times I = 240 \times 12 = 2880\text{ W} = 2.88\text{ kW}$$

Energy (in 20 hours):

$$E = P(\text{in kW}) \times t(\text{in h}) = 2.88 \times 20 = 57.6\text{ kWh}$$
24
Determine the power dissipated when a current of 10 mA flows through an appliance having a resistance of $8\text{ k}\Omega$.
Answer 0.8 W 📝
Detailed Solution

Convert the units: $I = 10\text{ mA} = 0.01\text{ A}$ and $R = 8\text{ k}\Omega = 8000\text{ }\Omega$.

$$P = I^2 R = (0.01)^2 \times 8000$$
$$P = 0.0001 \times 8000 = 0.8\text{ W}$$
25
85.5 J of energy are converted into heat in nine seconds. What power is dissipated?
Answer 9.5 W 📝
Detailed Solution

Given Energy $E = 85.5\text{ J}$ and time $t = 9\text{ s}$. Power is the rate of energy transfer.

$$P = \frac{E}{t} = \frac{85.5}{9}$$
$$P = 9.5\text{ W}$$
26
A current of 4 A flows through a conductor and 10 W is dissipated. What p.d. exists across the ends of the conductor?
Answer 2.5 V 📝
Detailed Solution

Given $I = 4\text{ A}$ and $P = 10\text{ W}$. We know that $P = V \times I$.

$$V = \frac{P}{I} = \frac{10}{4}$$
$$V = 2.5\text{ V}$$
27
Find the power dissipated when:
(a) a current of 5 mA flows through a resistance of 20 kΩ
(b) a voltage of 400 V is applied across a 120 kΩ resistor
(c) a voltage applied to a resistor is 10 kV and the current flow is 4 mA.
Answer (a) 0.5 W (b) 1.33 W (c) 40 W 📝
Detailed Solution

(a) $I = 0.005\text{ A}$, $R = 20,000\text{ }\Omega$.

$$P = I^2R = (0.005)^2 \times 20000 = 0.000025 \times 20000 = 0.5\text{ W}$$

(b) $V = 400\text{ V}$, $R = 120,000\text{ }\Omega$.

$$P = \frac{V^2}{R} = \frac{400^2}{120000} = \frac{160000}{120000} \approx 1.33\text{ W}$$

(c) $V = 10,000\text{ V}$, $I = 0.004\text{ A}$.

$$P = VI = 10000 \times 0.004 = 40\text{ W}$$
28
A battery of e.m.f. 15 V supplies a current of 2 A for 5 min. How much energy is supplied in this time?
Answer 9000 J (or 9 kJ) 📝
Detailed Solution

Given $V = 15\text{ V}$, $I = 2\text{ A}$, and time $t = 5\text{ mins} = 300\text{ s}$.

$$E = V \times I \times t$$
$$E = 15 \times 2 \times 300 = 30 \times 300 = 9000\text{ J}$$
SQ

Electrical Energy and Cost Practice Set 4

Apply $E = P \times t$ and commercial electricity formulas
Ch 12 · Electricity
29
In a household during a particular week three 2 kW fires are used on average 25 h each and eight 100 W light bulbs are used on average 35 h each. Determine the cost of electricity for the week if 1 unit of electricity costs 7p.
Answer 1246p (or £12.46) 📝
Detailed Solution

Energy for fires:

$$E_{fires} = 3 \times 2\text{ kW} \times 25\text{ h} = 150\text{ kWh}$$

Energy for light bulbs ($100\text{ W} = 0.1\text{ kW}$):

$$E_{bulbs} = 8 \times 0.1\text{ kW} \times 35\text{ h} = 28\text{ kWh}$$

Total Energy and Cost:

$$E_{total} = 150 + 28 = 178\text{ kWh (units)}$$
$$\text{Cost} = 178 \times 7\text{p} = 1246\text{p}$$
30
Calculate the power dissipated by the element of an electric fire of resistance $30\text{ }\Omega$ when a current of 10 A flows in it. If the fire is on for 30 hours in a week determine the energy used. Determine also the weekly cost of energy if electricity costs 7.2p per unit.
Answer $P = 3000\text{ W}$, $E = 90\text{ kWh}$, Cost = 648p 📝
Detailed Solution

Power:

$$P = I^2R = (10)^2 \times 30 = 100 \times 30 = 3000\text{ W} = 3\text{ kW}$$

Energy:

$$E = P \times t = 3\text{ kW} \times 30\text{ h} = 90\text{ kWh}$$

Cost:

$$\text{Cost} = 90\text{ units} \times 7.2\text{p/unit} = 648\text{p}$$
31
A television set having a power rating of 120 W and electric lawnmower of power rating 1 kW are both connected to a 240 V supply. If 3 A, 5 A and 10 A fuses are available state which is the most appropriate for each appliance.
Answer TV: 3 A fuse. Lawnmower: 5 A fuse. 📝
Detailed Solution

The correct fuse should be rated slightly higher than the operating current.

For the TV:

$$I = \frac{P}{V} = \frac{120}{240} = 0.5\text{ A} \implies \text{Use } 3\text{A fuse}$$

For the Lawnmower ($1\text{ kW} = 1000\text{ W}$):

$$I = \frac{P}{V} = \frac{1000}{240} \approx 4.17\text{ A} \implies \text{Use } 5\text{A fuse}$$
32
For a heater rated at 4kW and 220V, calculate: (a) the current (b) the resistance of the heater (c) the energy consumed in 2 hours and (d) the cost if 1kWh is priced at Rs. 4.60.
Answer (a) 18.18 A (b) $12.1\text{ }\Omega$ (c) 8 kWh (d) Rs. 36.80 📝
Detailed Solution

Given $P = 4000\text{ W}$ ($4\text{ kW}$) and $V = 220\text{ V}$.

(a) Current:

$$I = \frac{P}{V} = \frac{4000}{220} = \frac{400}{22} \approx 18.18\text{ A}$$

(b) Resistance:

$$R = \frac{V}{I} = \frac{220}{400/22} = \frac{4840}{400} = 12.1\text{ }\Omega$$

(c) Energy Consumed (in 2 hours):

$$E = P \times t = 4\text{ kW} \times 2\text{ h} = 8\text{ kWh}$$

(d) Cost:

$$\text{Cost} = 8 \times 4.60 = \text{Rs. } 36.80$$
33
A radio set of 60W runs for 50hrs. How much electrical energy consumed?
Answer 3 kWh (or 10.8 MJ) 📝
Detailed Solution

Given $P = 60\text{ W} = 0.06\text{ kW}$ and $t = 50\text{ h}$.

In commercial units:

$$E = P \times t = 0.06\text{ kW} \times 50\text{ h} = 3\text{ kWh}$$

In SI units (Joules), where $t = 50 \times 3600 = 180000\text{ s}$:

$$E = 60\text{ W} \times 180000\text{ s} = 10,800,000\text{ J} = 10.8\text{ MJ}$$
34
A current of 4A flows through a 12V car headlight bulb for 10min. How much energy transfer occurs during this time?
Answer 28,800 J (or 28.8 kJ) 📝
Detailed Solution

Given $I = 4\text{ A}$, $V = 12\text{ V}$, and $t = 10\text{ mins} = 600\text{ s}$.

$$E = V \times I \times t$$
$$E = 12 \times 4 \times 600 = 48 \times 600 = 28800\text{ J}$$
35
Calculate the energy transferred by a 5A current flowing through a resistor of $2\text{ }\Omega$ for 30min.
Answer 90,000 J (or 90 kJ) 📝
Detailed Solution

Given $I = 5\text{ A}$, $R = 2\text{ }\Omega$, and time $t = 30\text{ mins} = 1800\text{ s}$. Using the Joule’s heating formula:

$$E = I^2Rt$$
$$E = (5)^2 \times 2 \times 1800 = 25 \times 2 \times 1800$$
$$E = 50 \times 1800 = 90000\text{ J}$$

Numerical Problems on Electric Power and Energy (Continued)

Class 10 Science · Electricity
SQ

Electrical Energy and Cost Practice Set 5

Attempt each question before revealing the answer
Ch 12 · Electricity
36
A bulb is rated at 200V-100W. What is its resistance? 5 such bulbs burn for 4 hrs. What is the electrical energy consumed? Calculate the cost if the rate is Rs. 4.60 per unit.
Answer $R = 400\text{ }\Omega$, $E = 2\text{ kWh}$, Cost = Rs. 9.20 📝
Detailed Solution

Resistance: Using $P = \frac{V^2}{R}$

$$R = \frac{V^2}{P} = \frac{200^2}{100} = \frac{40000}{100} = 400\text{ }\Omega$$

Energy Consumed: Total power for 5 bulbs $= 5 \times 100\text{ W} = 500\text{ W} = 0.5\text{ kW}$. Time $= 4\text{ h}$.

$$E = P \times t = 0.5\text{ kW} \times 4\text{ h} = 2\text{ kWh (units)}$$

Cost:

$$\text{Cost} = 2\text{ units} \times 4.60 = \text{Rs. } 9.20$$
37
A refrigerator having a power rating of 350W operates for 10hours a day. Calculate the cost of electrical energy to operate it for a month of 30days. The rate of electrical energy is Rs. 3.40 per KWh.
Answer Rs. 357 📝
Detailed Solution

Power $P = 350\text{ W} = 0.35\text{ kW}$. Total time $t = 10\text{ hours/day} \times 30\text{ days} = 300\text{ hours}$.

Energy:

$$E = P \times t = 0.35\text{ kW} \times 300\text{ h} = 105\text{ kWh}$$

Cost:

$$\text{Cost} = 105\text{ units} \times 3.40 = \text{Rs. } 357$$
38
What will be the current drawn by an electric bulb of 40W when it is connected to a source of 220V?
Answer 0.18 A 📝
Detailed Solution

Given Power $P = 40\text{ W}$ and Voltage $V = 220\text{ V}$.

$$I = \frac{P}{V} = \frac{40}{220} = \frac{2}{11} \approx 0.1818\text{ A}$$
39
An electric bulb is rated 220V and 100W. When it is operated on 110V, find the power consumed.
Answer 25 W 📝
Detailed Solution

The resistance of the bulb remains constant. Let’s find $R$ at rated values ($V = 220\text{ V}, P = 100\text{ W}$):

$$R = \frac{V^2}{P} = \frac{220^2}{100} = \frac{48400}{100} = 484\text{ }\Omega$$

Now, calculate power consumed at new voltage $V’ = 110\text{ V}$:

$$P’ = \frac{V’^2}{R} = \frac{110^2}{484} = \frac{12100}{484} = 25\text{ W}$$

(Shortcut: If voltage is halved, the power becomes one-fourth: $100 / 4 = 25\text{ W}$).

40
An electric heater draws a current of 10A from a 220V supply. What is the cost of using the heater for 5 hrs everyday for 30days if the cost of 1 unit is Rs. 5.20?
Answer Rs. 1716 📝
Detailed Solution

Power:

$$P = V \times I = 220 \times 10 = 2200\text{ W} = 2.2\text{ kW}$$

Energy: Total time $t = 5 \times 30 = 150\text{ hours}$.

$$E = P \times t = 2.2\text{ kW} \times 150\text{ h} = 330\text{ kWh}$$

Cost:

$$\text{Cost} = 330\text{ units} \times 5.20 = \text{Rs. } 1716$$
41
In house two 60W electric bulbs are lighted for 4 hrs and three 100W bulbs for 5 hrs everyday. Calculate the electrical energy consumed in 30days.
Answer 59.4 kWh (units) 📝
Detailed Solution

Daily energy for 60W bulbs: $E_1 = 2 \times 60\text{ W} \times 4\text{ h} = 480\text{ Wh}$

Daily energy for 100W bulbs: $E_2 = 3 \times 100\text{ W} \times 5\text{ h} = 1500\text{ Wh}$

Total daily energy: $480 + 1500 = 1980\text{ Wh} = 1.98\text{ kWh}$

Energy for 30 days:

$$E = 1.98\text{ kWh/day} \times 30\text{ days} = 59.4\text{ kWh}$$
42
An electric motor takes 5A current from a 220V supply line. Calculate the power of the motor and electrical energy consumed by it in 2 hrs.
Answer $P = 1100\text{ W}$, $E = 2.2\text{ kWh}$ 📝
Detailed Solution

Power:

$$P = V \times I = 220 \times 5 = 1100\text{ W} = 1.1\text{ kW}$$

Energy:

$$E = P \times t = 1.1\text{ kW} \times 2\text{ h} = 2.2\text{ kWh}$$
43
An electric iron consumes energy at a rate of 840 W when heating is at the maximum rate and 360 W when the heating is at the minimum. The voltage is 220V. What are the current and the resistance in each case?
Answer Max: $3.82\text{ A}, 57.6\text{ }\Omega$ | Min: $1.64\text{ A}, 134.15\text{ }\Omega$ 📝
Detailed Solution

Maximum heating rate ($P = 840\text{ W}, V = 220\text{ V}$):

$$I = \frac{P}{V} = \frac{840}{220} \approx 3.82\text{ A}$$
$$R = \frac{V}{I} = \frac{220}{3.82} \approx 57.6\text{ }\Omega$$

Minimum heating rate ($P = 360\text{ W}, V = 220\text{ V}$):

$$I = \frac{P}{V} = \frac{360}{220} \approx 1.64\text{ A}$$
$$R = \frac{V}{I} = \frac{220}{1.64} \approx 134.15\text{ }\Omega$$
SQ

Electrical Energy and Cost Practice Set 6

Attempt each question before revealing the answer
Ch 12 · Electricity
44
An electric refrigerator rated 400 W operates 8 hour/day. What is the cost of the energy to operate it for 30 days at Rs 3.00 per kW h?
Answer Rs. 288 📝
Detailed Solution

Power $P = 400\text{ W} = 0.4\text{ kW}$. Total time $t = 8 \times 30 = 240\text{ hours}$.

Energy:

$$E = P \times t = 0.4\text{ kW} \times 240\text{ h} = 96\text{ kWh}$$

Cost:

$$\text{Cost} = 96 \times 3.00 = \text{Rs. } 288$$
45
An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Answer $P = 1100\text{ W}$, $E = 2.2\text{ kWh}$ 📝
Detailed Solution

(This is a duplicate of Question 42).

Power: $P = V \times I = 220 \times 5 = 1100\text{ W} = 1.1\text{ kW}$.

Energy: $E = P \times t = 1.1\text{ kW} \times 2\text{ h} = 2.2\text{ kWh}$.

46
Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Answer 0.73 A 📝
Detailed Solution

In a parallel circuit, the total power is the sum of individual powers.

$$P_{total} = P_1 + P_2 = 100\text{ W} + 60\text{ W} = 160\text{ W}$$

The total current drawn is:

$$I = \frac{P_{total}}{V} = \frac{160}{220} = \frac{16}{22} \approx 0.727\text{ A}$$
47
Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
Answer The 250 W TV set 📝
Detailed Solution

Energy used by TV:

$$E_{TV} = 250\text{ W} \times 1\text{ h} = 250\text{ Wh}$$

Energy used by Toaster: $10\text{ minutes} = \frac{10}{60}\text{ h} = \frac{1}{6}\text{ h}$.

$$E_{Toaster} = 1200\text{ W} \times \frac{1}{6}\text{ h} = 200\text{ Wh}$$

Since $250\text{ Wh} > 200\text{ Wh}$, the TV set uses more energy.

48
Two bulbs A and B are rated 100W – 120V and 10W – 120V respectively. They are connected across a 120V source in series. Which will consume more energy?
Answer Bulb B (10W bulb) 📝
Detailed Solution

The resistance of a bulb is given by $R = \frac{V^2}{P}$. Since voltage ratings are identical, resistance is inversely proportional to rated power. Therefore, Bulb B (10W) has a much higher resistance than Bulb A (100W).

When connected in series, the same current ($I$) flows through both bulbs. The power actually consumed in the circuit is $P_{consumed} = I^2 R$.

Since $I$ is constant for both and $R_B > R_A$, Bulb B will consume more power and thus more energy.

49
Two bulbs A and B are rated 100W – 120V and 10W – 120V respectively. They are connected across a 120V source in series. Find the current in each bulb. Which will consume more energy.
Answer $I = 0.0757\text{ A}$. Bulb B consumes more energy. 📝
Detailed Solution

First, find the resistance of each bulb:

$$R_A = \frac{120^2}{100} = \frac{14400}{100} = 144\text{ }\Omega$$
$$R_B = \frac{120^2}{10} = \frac{14400}{10} = 1440\text{ }\Omega$$

Since they are in series, equivalent resistance $R_{eq} = 144 + 1440 = 1584\text{ }\Omega$.

The current in the series circuit (same for both bulbs) is:

$$I = \frac{V}{R_{eq}} = \frac{120}{1584} \approx 0.0757\text{ A}$$

As established in the previous question, because $P = I^2 R$ and $R_B > R_A$, Bulb B (10W) consumes more energy.

50
An electric kettle is rated at 230V, 1000W. What is the resistance of its element? What maximum current can pass through its element?
Answer $R = 52.9\text{ }\Omega$, $I = 4.35\text{ A}$ 📝
Detailed Solution

Given $V = 230\text{ V}$ and $P = 1000\text{ W}$.

Maximum Current:

$$I = \frac{P}{V} = \frac{1000}{230} \approx 4.35\text{ A}$$

Resistance:

$$R = \frac{V^2}{P} = \frac{230^2}{1000} = \frac{52900}{1000} = 52.9\text{ }\Omega$$

Numerical Problems on Electric Heating (Joule’s Law)

Class 10 Science · Electricity
SQ

Joule’s Heating Practice Set 1

Apply $H = I^2Rt$, $H = VIt$, and $H = \frac{V^2}{R}t$
Ch 12 · Electricity
60
100 J of heat are produced each second in a $4\text{ }\Omega$ resistance. Find the potential difference across the resistor.
Answer 20 V 📝
Detailed Solution

Heat produced per second is the electrical power ($P$). Thus, $P = 100\text{ J/s} = 100\text{ W}$.
Given $R = 4\text{ }\Omega$. Using the formula $P = \frac{V^2}{R}$:

$$V^2 = P \times R = 100 \times 4 = 400$$
$$V = \sqrt{400} = 20\text{ V}$$
61
Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Answer $4.8 \times 10^6\text{ J}$ (or $4800\text{ kJ}$) 📝
Detailed Solution

Given Charge $Q = 96000\text{ C}$ and Voltage $V = 50\text{ V}$. Time is extraneous information here because work/heat relates directly to charge and potential difference.

$$H = W = V \times Q$$
$$H = 50 \times 96000 = 4,800,000\text{ J} = 4.8 \times 10^6\text{ J}$$
62
An electric iron of resistance $20\text{ }\Omega$ takes a current of 5 A. Calculate the heat developed in 30 s.
Answer $15,000\text{ J}$ (or $15\text{ kJ}$) 📝
Detailed Solution

Given $R = 20\text{ }\Omega$, $I = 5\text{ A}$, and $t = 30\text{ s}$. By Joule’s law of heating:

$$H = I^2 \times R \times t$$
$$H = (5)^2 \times 20 \times 30 = 25 \times 600$$
$$H = 15000\text{ J}$$
63
A p.d. of 250V is applied across a resistance of $500\text{ }\Omega$ in an electric iron. Calculate (i) current (ii) heat energy produced in joules in 10s.
Answer (i) 0.5 A (ii) 1250 J 📝
Detailed Solution

Given $V = 250\text{ V}$ and $R = 500\text{ }\Omega$.

(i) Current:

$$I = \frac{V}{R} = \frac{250}{500} = 0.5\text{ A}$$

(ii) Heat Energy (for $t = 10\text{ s}$):

$$H = V \times I \times t = 250 \times 0.5 \times 10 = 1250\text{ J}$$
64
Calculate the heat produced when 96000C of charge is transferred in 1 hour through a p.d. of 50V.
Answer $4.8 \times 10^6\text{ J}$ 📝
Detailed Solution

(Note: This is a duplicate of Question 61).

$$H = V \times Q = 50 \times 96000 = 4,800,000\text{ J} = 4.8 \times 10^6\text{ J}$$
65
A resistance of $40\text{ }\Omega$ and one of $60\text{ }\Omega$ are arranged in series across 220V supply. Find the heat in joules produced by this combination of resistances in half a minute.
Answer 14,520 J 📝
Detailed Solution

Total resistance in series $R_{eq} = 40 + 60 = 100\text{ }\Omega$.
Given $V = 220\text{ V}$ and time $t = 0.5\text{ min} = 30\text{ s}$.

$$H = \frac{V^2}{R_{eq}} \times t = \frac{(220)^2}{100} \times 30$$
$$H = \frac{48400}{100} \times 30 = 484 \times 30 = 14520\text{ J}$$
SQ

Joule’s Heating Practice Set 2

Attempt each question before revealing the answer
Ch 12 · Electricity
66
When a current of 4A passes through a certain resistor for 10min, $2.88 \times 10^4\text{ J}$ of heat are produced. Calculate (a) power of the resistor (b) the voltage across the resistor.
Answer (a) 48 W (b) 12 V 📝
Detailed Solution

Given $I = 4\text{ A}$, time $t = 10\text{ mins} = 600\text{ s}$, and heat $H = 2.88 \times 10^4\text{ J} = 28,800\text{ J}$.

(a) Power:

$$P = \frac{H}{t} = \frac{28800}{600} = 48\text{ W}$$

(b) Voltage: Using $P = V \times I$

$$V = \frac{P}{I} = \frac{48}{4} = 12\text{ V}$$
67
A heating coil has a resistance of $200\text{ }\Omega$. At what rate will heat be produced in it when a current of 2.5 A flows through it?
Answer 1250 J/s (or 1250 W) 📝
Detailed Solution

(Note: Corrected “2.5 $\Omega$” in prompt to 2.5 A for current)

The rate at which heat is produced is the Power ($P$). Given $R = 200\text{ }\Omega$ and $I = 2.5\text{ A}$.

$$P = I^2R = (2.5)^2 \times 200$$
$$P = 6.25 \times 200 = 1250\text{ W (or J/s)}$$
68
An electric heater of resistance $8\text{ }\Omega$ takes a current of 15A from the mains supply line. Calculate the rate at which heat is developed in the heater.
Answer 1800 J/s (or 1800 W) 📝
Detailed Solution

Rate of heat development is Power ($P$). Given $R = 8\text{ }\Omega$ and $I = 15\text{ A}$.

$$P = I^2R = (15)^2 \times 8$$
$$P = 225 \times 8 = 1800\text{ W}$$
69
A resistance of $25\text{ }\Omega$ is connected to a 12V battery. Calculate the heat energy in joule generated per minute.
Answer 345.6 J 📝
Detailed Solution

Given $R = 25\text{ }\Omega$, $V = 12\text{ V}$, and time $t = 1\text{ minute} = 60\text{ s}$.

$$H = \frac{V^2}{R} \times t = \frac{(12)^2}{25} \times 60$$
$$H = \frac{144}{25} \times 60 = 5.76 \times 60 = 345.6\text{ J}$$
70
How much heat will an instrument of 12W produce in one minute if it is connected to a battery of 12V?
Answer 720 J 📝
Detailed Solution

Given Power $P = 12\text{ W}$ and time $t = 1\text{ minute} = 60\text{ s}$. (The voltage is extra information).

$$H = P \times t = 12\text{ W} \times 60\text{ s} = 720\text{ J}$$

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