Saturday, July 20, 2024


- Advertisement -spot_imgspot_img
- Advertisement -spot_img

This chapter explores the fundamental concepts of electricity, the invisible force that powers many aspects of our modern world. Here’s a breakdown of the key topics:

Electric Current:

  • Defined as the flow of electric charge (mainly electrons) through a conductor.
  • Measured in amperes (A).
  • Conventional current refers to the flow of positive charge, opposite to the actual electron flow.

Electric Circuits:

  • A closed path for electric current to flow.
  • Requires components like a source (battery), conducting wires, and sometimes a load (bulb, resistor).

Voltage (Electric Potential Difference):

  • The “pressure” that drives electric current flow in a circuit.
  • Measured in volts (V).
  • Represents the potential energy difference per unit charge between two points in a circuit.


  • Opposition to the flow of electric current in a conductor.
  • Measured in ohms (Ω).
  • Materials like metals have low resistance, while insulators have high resistance.

Ohm’s Law:

  • Relates voltage, current, and resistance in a circuit.
  • V = IR (Voltage = Current x Resistance)

Electric Power:

  • The rate at which electrical energy is transferred or consumed.
  • Measured in watts (W).
  • P = VI (Power = Voltage x Current)

Additional Topics:

  • Domestic Circuits: Understanding wiring systems in homes, with concepts like fuses and safety precautions.
  • Electromagnetism: The link between electricity and magnetism, leading to applications like electromagnets and motors.

Questions (Page 172)

1. What does an electric circuit mean ?

Ans : An electric circuit provides a pathway for electricity to travel, enabling various devices to function.

2. Define the unit of current.

Ans : The unit of electric current is the ampere (A). It represents the rate of flow of electric charge through a conductor.

3. Calculate the number of electrons constituting one coulomb of charge.

Ans : 

Number of electrons = 1 C / (1.602 x 10^-19 C/electron)

Number of electrons ≈ 6.24 x 10^18 electrons

Therefore, one coulomb of charge is equivalent to approximately 6.2 x 10^18 electrons.

Questions (Page 174)

1. Name a device that helps to maintain a potential difference across a conductor.

Ans : The device that helps to maintain a potential difference (voltage) across a conductor is a battery.

2. What is meant by saying that the potential difference between two points is IV?

Ans : The statement “the potential difference between two points is IV” refers to the relationship between voltage (V), current (I), and resistance (R) in an electrical circuit, specifically based on Ohm’s Law.

3. How much energy is given to each coulomb of charge passing through a 6 V battery ?

Ans : The energy given to each coulomb of charge passing through a 6 V battery is also 6 joules (J).

Questions (Page 181)

1. On what factors does the resistance of a conductor depend ?

Ans : The resistance of a wire depends on 4 things:

  1. Material: Metals conduct best, insulators worst.
  2. Length: Longer wires have more resistance.
  3. Thickness: Thicker wires have less resistance.
  4. Temperature: Hot wires generally have more resistance (except superconductors).

2. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source ? Why ?

Ans : Current will flow more easily through a thick wire of the same material compared to a thin wire when connected to the same source. This is because of the concept of resistance in conductors.

3. Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it ?

Ans : If the resistance of an electrical component remains constant and the potential difference (voltage) across it is halved, the current flowing through the component will also be halved.

This is based on Ohm’s Law, a fundamental principle in electricity that relates voltage (V), current (I), and resistance (R) in a circuit:

V = IR

4. Why are coils of electric toasters and electric irons are made of an-alloy rather than a pure metal ?

Ans : Coils in electric toasters and irons are made of alloys rather than pure metals for two main reasons:

  1. High Resistivity: Alloys generally have a higher resistivity compared to pure metals. Resistivity is a material’s opposition to the flow of electric current. When current passes through a coil, it generates heat due to resistance. A higher resistivity in the coil material leads to more efficient heating for these appliances.
  2. Melting Point: Pure metals often have lower melting points than alloys. In toasters and irons, the coils reach high temperatures during operation. Alloys are chosen because they have higher melting points, making them more resistant to melting and

5. Use the data in Table 12.2 (in NCERT Book on Page No. 207) to answer the following :

(i) Which among iron and mercury is a better conductor ?

(ii) Which material is the best conductor ?

Ans : 

(i) Resistivity of iron = 10.0 x 10-8 Ω m

Resistivity of mercury = 94.0 x 10-8 Ω m.

Thus iron is a better conductor because it has lower resistivity than mercury.

(ii) Because silver has the lowest resistivity (= 1.60 x 10-8 Ω m), therefore silver is the best conductor.

Questions (Page 185)

1. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.

Ans : 

2. Redraw the circuit of Questions 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter ?

Ans : 

Battery: Three cells connected in series, each with 2V (total voltage: 3 * 2V = 6V)


  • Resistor 1: 5 Ω
  • Resistor 2: 8 Ω
  • Resistor 3: 12 Ω

Questions (Page 188)

1. Judge the equivalent resistance when the following are connected in parallel :

(i) 1 Ω and 106 Ω,

(if) 1 Ω and 103 Ω and 106 Ω.

Ans : 

1. 1 Ω and 106 Ω connected in parallel:

For resistors in parallel, the reciprocals of their individual resistances add up to give the reciprocal of the equivalent resistance (1/Req).

Formula: 1/Req = 1/R1 + 1/R2

Given values:

  • R1 = 1 Ω
  • R2 = 106 Ω

Calculation: 1/Req = 1/1 Ω + 1/106 Ω ≈ 1 Ω (since 1/106 Ω is negligible)

Equivalent Resistance: Taking the reciprocal of both sides: Req ≈ 1 (1/Ω) = 1 Ω

Therefore, the equivalent resistance (Req) for 1 Ω and 106 Ω in parallel is approximately 1 Ω.

2. 1 Ω and 103 Ω and 106 Ω connected in parallel:

We can again use the same formula but add the reciprocals of all three resistances.

Formula: 1/Req = 1/R1 + 1/R2 + 1/R3

Given values:

  • R1 = 1 Ω
  • R2 = 103 Ω
  • R3 = 106 Ω

Calculation: 1/Req = 1/1 Ω + 1/103 Ω + 1/106 Ω ≈ 1.009 Ω (small approximation due to very low additional terms)

2. An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it ?

Ans : 


  • Lamp resistance (R1) = 100 Ω
  • Toaster resistance (R2) = 50 Ω
  • Water filter resistance (R3) = 500 Ω
  • Source voltage (V) = 220 V

Task 1: Total Current in the Circuit

Since the appliances are connected in parallel, the total current (I_total) is the sum of the individual currents flowing through each appliance.

1.1. Calculate Individual Currents:

We can use Ohm’s Law (V = IR) to find the current through each appliance:

  • Current through lamp (I1) = V / R1 = 220 V / 100 Ω = 2.2 A
  • Current through toaster (I2) = V / R2 = 220 V / 50 Ω = 4.4 A
  • Current through water filter (I3) = V / R3 = 220 V / 500 Ω = 0.44 A

1.2. Find Total Current:

I_total = I1 + I2 + I3 = 2.2 A + 4.4 A + 0.44 A = 7.04 A

Task 2: Iron Resistance for Same Current

We need to find the resistance (R_iron) of an iron that draws the same total current (7.04 A) from the same source (220 V).

2.1. Apply Ohm’s Law:

V = I_iron * R_iron (where I_iron is the current through the iron)

2.2. Solve for R_iron:

R_iron = V / I_iron = 220 V / 7.04 A ≈ 31.25 Ω

3. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series ?

Ans : Here are the main advantages of connecting electrical devices in parallel with a battery compared to connecting them in series:

1. Independent Operation:

  • In a parallel circuit, each device has its own path to the battery. This allows you to turn on or off individual devices without affecting the others. For example, you can turn off a light bulb without affecting the operation of a radio in a parallel circuit.
  • In a series circuit, all devices are connected in a single loop. If one device is turned off or malfunctions (burns out), it interrupts the entire circuit, and no devices will work.

2. Same Voltage Across Devices:

  • In a parallel circuit, the voltage across each device is equal to the voltage of the battery. This ensures that each device receives the same “electrical pressure” to function properly.
  • In a series circuit, the total voltage of the battery is divided among the connected devices. The voltage across each device depends on its individual resistance.

3. Individual Current Draw:

  • Each device in a parallel circuit draws current based on its own resistance. Devices with lower resistance will draw more current, while those with higher resistance will draw less.
  • In a series circuit, the same current flows through all connected devices. This can be problematic if a device with low resistance requires a high current, as it might overload other devices in the circuit with higher resistance.

4. Fault Tolerance (Improved Safety):

  • In a parallel circuit, a malfunctioning device (like a burned-out bulb) doesn’t affect the operation of other devices. The circuit remains functional as long as the battery and remaining devices are intact.
  • In a series circuit, a single device failure can disrupt the entire circuit, potentially leading to safety hazards like overheating or complete power loss.

4. How can three resistors of resistances 2Ω, 3 Ω, and 6Ω be connected to give a total resistance of (i) 4 Ω, (ii) 1 Ω ?

Ans : 

  • In parallel connection, the reciprocal of the total resistance is equal to the sum of the reciprocals of the individual resistances.
  • Mathematically: 1/R_total = 1/2 Ω + 1/3 Ω + 1/6 Ω
  • Solving for R_total: R_total = 1 / (1/2 + 1/3 + 1/6) Ω = 1 Ω

Therefore, for a total resistance of 1 Ω, connect the 2 Ω, 3 Ω, and 6 Ω resistors in parallel.

5. What is (i) the highest, (ii) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?

Ans : For the four coils with resistances 4 Ω, 8 Ω, 12 Ω, and 24 Ω:

(i) Highest Total Resistance:

This is achieved by connecting all the coils in series. In a series connection, the total resistance is the sum of the individual resistances.

Total Resistance (Highest) = R1 + R2 + R3 + R4 = 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω

(ii) Lowest Total Resistance:

This is achieved by connecting all the coils in parallel. In a parallel connection, the overall resistance is less than the smallest individual resistance.

Calculate the reciprocal of each resistance:

1/R1 = 1/4 Ω 1/R2 = 1/8 Ω 1/R3 = 1/12 Ω 1/R4 = 1/24 Ω

Add the reciprocals and take the reciprocal of the sum to find the total resistance:

Total Resistance (Lowest) = 1 / (1/4 Ω + 1/8 Ω + 1/12 Ω + 1/24 Ω) ≈ 1 / (11/24) Ω ≈ 2.18 Ω

Questions (Page 189)

1. Why does the cord of an electric heater not glow while the heating element does ?

Ans : The heater cord stays cool because it’s made of a low-resistance material (like copper) that lets current flow easily, generating minimal heat. The heating element, however, is made of a high-resistance material that creates friction for the current, converting that energy into heat, making it glow. Think of the cord as a wide river and the element as a narrow dam – the water (current) flows smoothly through the river (cord) but struggles at the dam (element), heating it up.

2. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

Ans : 

Heat (H) = Energy (W) = Charge (Q) x Voltage (V)

Here’s how to solve it:

  1. Charge (Q): 96000 coulombs (given)
  2. Voltage (V): 50 volts (given)
  3. Energy (W) or Heat (H): We need to find this.


H = Q x V = 96000 C * 50 V = 4,800,000 J (Joules)

Therefore, the heat generated is 4,800,000 joules.

3. An electric iron of resistance 20Ω takes a current of 5 A. Calculate the heat developed in 30 s.

Ans : 

Heat (H) = I^2 * R * t


  • H is the heat generated (in Joules)
  • I is the current (in amperes)
  • R is the resistance (in ohms)
  • t is the time (in seconds)

Given values:

  • I = 5 A (current)
  • R = 20 Ω (resistance)
  • t = 30 s (time)


H = (5 A)^2 * 20 Ω * 30 s = 25 * 20 * 30 J = 15000 J

Therefore, the heat developed in the electric iron in 30 seconds is 15000 Joules.

Questions (192)

1. What determines the rate at which energy is delivered by a current ?

Ans : The rate at which energy is delivered by a current is determined by electric power.

2. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

Ans : 

P = V × I


  • P is the power (in watts)
  • V is the voltage (in volts)
  • I is the current (in amperes)

Given values:

  • V = 220 V
  • I = 5 A


P = 220 V × 5 A = 1100 W

Therefore, the electric motor consumes a power of 1100 watts.

2. Calculate Energy Consumed:

Energy consumption (E) is the product of power (P) and time (t). We need to convert the time from hours to seconds for consistency with the power unit (watts).

E = P × t

Time Conversion:

2 hours = 2 h * 60 minutes/hour * 60 seconds/minute = 7200 seconds


E = 1100 W × 7200 s = 7,920,000 J (Joules)

Alternatively, convert watts to joules per second (watts per second is also equal to joules):

1 watt = 1 J/s

Therefore, the energy consumed in 2 hours is:

E = P × t = 1100 W × 7200 s = 7,920,000 J

1. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio R/R’ is :

(a) 1/25

(b) 1/5

(c) 5

(d) 25

Ans :(d) 25

2. Which of the following terms does not represent electrical power in a circuit?

(a) I2R

(b) IR2

(c) VI

(d) V^2/2

Ans : (d) V^2/2

3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be :

(a) 100 W

(b) 75 W

(c) 50 W

(d) 25 W

Ans : (d) 25 W

4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be :

(a) 1 : 2

(b) 2 : 1

(c) 1 : 4

(d) 4 : 1

Ans : (c) 1 : 4.

5. How is a voltmeter connected in the circuit to measure the potential difference between two points ?

Ans : A voltmeter is a measuring instrument used to determine the potential difference (voltage) between two points in a circuit.

6. A copper wire has diameter 0.5 mm and resistivity of 1.6 x 10-8 Ω m. What will be the length of this wire to make its resistance 10 Ω ? How much does the resistance change if the diameter is doubled ?

Ans : 

If a wire of diameter doubled to it is taken, then area of cross-section becomes four times.
New resistance = 10/2 = 2.5 Ω, Thus the new resistance will be  ¼ times.
Decrease in resistance = (10 – 2.5) Ω = 7.5 Ω

7. The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below :

NCERT Solutions for Class 10 Science Chapter 12 Electricity Chapter End Questions Q7

Plot a graph between V and I and calculate the resistance of the resistor.

Ans :  

The graph between V and I for the above data is given below.
The slope of the graph will give the value of resistance.
Let us consider two points P and Q on the graph.
and from P along Y-axis, which meet at point R.
Now, QR = 10.2V – 34V = 6.8V
And PR = 3 – 1 = 2 ampere
NCERT Solutions for Class 10 Science Chapter 12 Electricity Chapter End Questions Q7.1
NCERT Solutions for Class 10 Science Chapter 12 Electricity Chapter End Questions Q7.2
Thus, resistance, R = 3.4 Ω

8. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.

Ans : 

  • Voltage (V) = 12 V (battery voltage)
  • Current (I) = 2.5 mA = 2.5 x 10^-3 A (milliamps need to be converted to amperes)

Ohm’s Law:

The resistance (R) of a conductor is related to the voltage (V) across it and the current (I) flowing through it by the following formula:

R = V / I

Finding the Resistance:

Plug the known values into the formula:

R = 12 V / (2.5 x 10^-3 A)

Simplifying the calculation:

  • To avoid confusion with decimals, it’s often convenient to work with milliamps (mA) and convert the answer to kilohms (kΩ) at the end.

R = 12 V / 2.5 mA

R = (12 V) / (2.5 x 10^-3 A)  (convert mA to A)

R ≈ 4800 Ω

Converting to kilohms (kΩ):

R ≈ 4800 Ω * (1 kΩ / 1000 Ω) = 4.8 kΩ

Therefore, the resistance of the resistor is approximately 4.8 kilohms (kΩ).

9. A battery of 9V is connected in series with resistors of 0.2 O, 0.3 O, 0.4 Q, 0.5 Q and 12 £1, respectively. How much current would flow through the 12 Q resistor?

Ans : Total resistance, R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω – 13.4 Ω

Potential difference, V = 9 V

Current through the series circuit, I = V/R=12V/13.4 Ω 

 = 0.67 A

∵ There is no division of current in series. Therefore current through 12 Ω resistor = 0.67 A.

10. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line? 

Ans : 

Therefore, you will need four (4) 176 Ω resistors connected in parallel to carry 5 A on a 220 V line.

11. Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4Ω

Ans : Here, R1 = R2 = R3 = 6 Ω.

(i) Here, R1 = R2 = 6 Ω.

1/R_parallel = 1/6 Ω + 1/6 Ω = 1/3 Ω

Take the reciprocal of both sides to find R_parallel:

R_parallel = 3 Ω.

Total Resistance (R_total) = 6 Ω (single resistor) + 3 Ω (parallel combination)

R_total = 9 Ω.

(ii) Here, R1 = R2 = R3 = 6 Ω.

1/R_parallel = 1/6 Ω + 1/6 Ω + 1/6 Ω = 1/2 Ω

Take the reciprocal of both sides to find R_parallel:

R_parallel = 2 Ω.

12. Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A ?

Ans : 

  • We know the power rating of each bulb (P) is 10 W.
  • We know the voltage of the supply line (V) is 220 V.
  • We can use the formula P = VI to find the current (I) per bulb.

I = P / V = 10 W / 220 V = 0.045 A

n = Total allowable current / Current per bulb

n = 5 A / 0.045 A ≈ 111.11

13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases ?

Ans : 

(i) We can use Ohm’s Law (I = V / R) to calculate the current (I) for each coil:

  • Voltage (V) = 220 V (supply voltage)
  • Resistance (R) = 24 Ω (resistance of each coil)

Current through each coil (I_separate) = V / R = 220 V / 24 Ω ≈ 9.17 A

(ii) R_series = R_A + R_B = 24 Ω + 24 Ω = 48 Ω

Now, use Ohm’s Law again to find the total current (I_series):

  • Voltage (V) = 220 V
  • Total resistance (R) = R_series = 48 Ω

(iii) 1/R_parallel = 1/R_A + 1/R_B

Since both coils have the same resistance:

1/R_parallel = 1/24 Ω + 1/24 Ω = 1/12 Ω

Take the reciprocal of both sides to find R_parallel:

R_parallel = 12 Ω

  • Voltage (V) = 220 V
  • Total resistance (R) = R_parallel = 12 Ω

Current in parallel connection (I_parallel) = V / R = 220 V / 12 Ω ≈ 18.33 A

14. Compare the power used in the 2 Ω resistor in each of the following circuits

(i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and

(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

Ans : 

(i)  R_series = R1 + R2 R_series = 1 Ω + 2 Ω = 3 Ω

  • We can use the formula for power (P): P = V^2 / R
  • Apply the values: P_2Ω = (6 V)^2 / (2 Ω) = 18 W

(ii) Ohm’s Law (I = V / R) can be used to find the current (I_2Ω) flowing through the 2 Ω resistor: I_2Ω = V_battery / R_2Ω = 4 V / 2 Ω = 2 A

(P = V^2 / R): P_2Ω = (4 V)^2 / (2 Ω) = 8 W

Comparison of Power:

  • Power in 2 Ω resistor (circuit i): P_2Ω (series) = 18 W
  • Power in 2 Ω resistor (circuit ii): P_2Ω (parallel) = 8 W

15. 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 ?

Ans : 

We can use the formula for power (P) to find the current (I) for each lamp:

  • Power (P) for lamp 1: 100 W
  • Power (P) for lamp 2: 60 W
  • Voltage (V) for both lamps: 220 V (since they are in parallel)

Formula: P = V * I

Current for lamp 1 (I_1):

I_1 = P_1 / V = 100 W / 220 V ≈ 0.45 A (rounded to 2 decimal places)

Current for lamp 2 (I_2):

I_2 = P_2 / V = 60 W / 220 V ≈ 0.27 A (rounded to 2 decimal places)

2. Total Current in Parallel Connection:

In a parallel connection, the total current (I_total) is the sum of the currents flowing through each individual branch (lamp in this case):

I_total = I_1 + I_2

Total Current:

I_total = 0.45 A + 0.27 A ≈ 0.72 A (rounded to 2 decimal places)

Therefore, the total current drawn from the line is approximately 0.72 amperes (A).

16. Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes ?

Ans : 

Energy consumption (E) is calculated using the formula:

E = P * t


  • E is energy consumption (in watt-hours, Wh)
  • P is power rating (in watts, W)
  • t is time (in hours, h)
  1. Calculate Energy Consumption:

TV Set:

  • Power (P) = 250 W
  • Time (t) = 1 hour (1 h)

E_TV = 250 W * 1 h = 250 Wh


  • Power (P) = 1200 W
  • Time (t) = 10 minutes (convert to hours for consistent units)
    t_toaster = 10 minutes * (1 hour / 60 minutes) = 1/6 hour

E_toaster = 1200 W * (1/6 hour) = 200 Wh

17. An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.

Ans : 

Here, R = 8 Ω, 1 = 15 A, t = 2 h

The rate at which heat is developed in the heater is equal to the power.

Therefore, P = I2 R = (15)2 x 8 = 1800 Js-1

18. Explain the following:

(i) Why is tungsten used almost exclusively for filament of electric lamps ?

(ii) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal ?

(in) Why is the series arrangement not used for domestic circuits ?

(iv) How does the resistance of a wire vary with its area of cross-section ?

(v) Why are copper and aluminium wires usually employed for electricity transmission?

Ans : 

(i) Tungsten Filament:

  • High melting point withstands high operating temperatures without melting.
  • Low vapor pressure prevents filament from evaporating and blackening the bulb.
  • Relatively high resistivity promotes efficient conversion of electricity to heat for light emission.

(ii) Alloys in Heating Elements:

  • Pure metals may have excessively high melting points or resistivity for optimal heating.
  • Alloys offer a balance of properties like lower melting point, adjusted resistivity for efficient heat generation, and better mechanical strength.

(iii) Parallel for Domestic Circuits:

  • Series circuits: A single device failure disrupts the entire circuit. Voltage is divided, potentially causing issues for appliances with different voltage needs.
  • Parallel circuits: Independent operation of appliances, consistent voltage for all devices.

(iv) Resistance and Area:

Resistance is inversely proportional to the area of the wire’s cross-section. A thicker wire (larger area) has lower resistance.

(v) Copper & Aluminum for Transmission:

  • High conductivity minimizes energy loss during transmission.
  • Copper (better conductivity) or Aluminum (lighter and cheaper) depending on the application’s needs.
- Advertisement -spot_imgspot_img
Latest news
- Advertisement -spot_img
Related news
- Advertisement -spot_imgspot_img