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Chapter 12: Electricity - NCERT Solutions

Get NCERT Solutions for Electricity Class 10. Step-by-step answers for Ohm's law and circuits from textbook. Meta tags: electricity ncert solutions, ohm's law solved, electrical circuits answers, resistance solutions, class 10 physics

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Each question block contains a "Show Answer" button. Click it to reveal the detailed answer. Only one answer is shown at a time — opening a new answer will hide the previous one. These solutions cover electric current, circuits, Ohm's law, resistance, series-parallel combinations, heating effect, and electric power.

Question 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

Answer: (d) 25

Explanation:
Step 1: Resistance of each part
Original resistance = R
Cut into 5 equal parts → Each part resistance = R/5

Step 2: Parallel combination
5 resistors each of R/5 connected in parallel
For parallel combination: 1/R' = 1/(R/5) + 1/(R/5) + 1/(R/5) + 1/(R/5) + 1/(R/5)

1/R' = 5/(R/5) = 5 × (5/R) = 25/R

Step 3: Find equivalent resistance

1/R' = 25/R
R' = R/25

Step 4: Calculate ratio R/R'

R/R' = R ÷ (R/25) = R × (25/R) = 25

Verification:
• Cutting wire increases resistance proportionally to length
• Parallel connection dramatically reduces equivalent resistance
• 25 times reduction is logical for 5 equal parallel resistors

Question 2

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

(a) I²R
(b) IR²
(c) VI
(d) V²/R

Answer: (b) IR²

Explanation:
Electrical power formulas:
1. P = VI (Basic definition: Power = Voltage × Current)
2. P = I²R (Using Ohm's law V = IR)
3. P = V²/R (Using Ohm's law I = V/R)

Why IR² is incorrect:
• According to Ohm's law: V = IR
• Power P = VI = (IR) × I = I²R
• IR² would be (I × R²) which has units of A·Ω², not watts
• Correct unit for power: watt (W) = J/s = V·A = A²·Ω

Derivation of correct formulas:

P = V × I (Basic definition) Using Ohm's law (V = IR): P = (IR) × I = I²R Using Ohm's law (I = V/R): P = V × (V/R) = V²/R

Quick check:
• I²R = (ampere)² × ohm = watt ✓
• IR² = ampere × (ohm)² ≠ watt ✗
• VI = volt × ampere = watt ✓
• V²/R = (volt)² / ohm = watt ✓

Question 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

Answer: (d) 25 W

Explanation:
Given:
• Rated voltage: V₁ = 220 V
• Rated power: P₁ = 100 W
• Operating voltage: V₂ = 110 V

Step 1: Find resistance of bulb
Using P = V²/R

R = V₁²/P₁ = (220)²/100 = 48400/100 = 484 Ω

Resistance remains constant (assuming temperature constant)

Step 2: Calculate power at 110 V
Using same resistance:

P₂ = V₂²/R = (110)²/484 = 12100/484 = 25 W

Alternative method (ratio method):
Since P ∝ V² (when R constant)

P₂/P₁ = (V₂/V₁)²
P₂/100 = (110/220)² = (1/2)² = 1/4
P₂ = 100 × (1/4) = 25 W

Important note:
• Bulb resistance assumed constant (actually changes with temperature)
• At lower voltage, bulb glows dimmer (less power = less heat = less light)
• Actual power might be slightly different due to temperature effect on resistance

Question 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

Answer: (c) 1:4

Explanation:
Given: Two identical wires:
• Same material → same resistivity ρ
• Same length → same length l
• Same diameter → same cross-sectional area A
∴ Both have same resistance R

Step 1: Equivalent resistances
• Series combination: R_series = R + R = 2R
• Parallel combination: 1/R_parallel = 1/R + 1/R = 2/R → R_parallel = R/2

Step 2: Heat produced (using H = V²t/R)
For same voltage V and time t, heat produced ∝ 1/R

H_series = V²t/(2R) H_parallel = V²t/(R/2) = 2V²t/R

Step 3: Ratio of heat produced

H_series : H_parallel = [V²t/(2R)] : [2V²t/R] = (1/2) : 2 = 1 : 4

Alternative using current:
• Series: I_series = V/(2R)
• Parallel: I_parallel = V/(R/2) = 2V/R
Using H = I²Rt:
H_series : H_parallel = [V²/(4R²) × R × t] : [4V²/R² × (R/2) × t] = 1 : 4

Key insight: Parallel connection draws 4× more current than series, producing 4× more heat.

Question 5

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

Answer:

A voltmeter is always connected in parallel across the two points between which potential difference is to be measured.

Correct connection method:
1. Connect the positive terminal of voltmeter to the point at higher potential
2. Connect the negative terminal to the point at lower potential
3. The voltmeter must be connected across the component/circuit section

Why parallel connection?
• To measure potential difference between two points
• Parallel connection ensures voltmeter experiences same voltage as component
• Does not disturb the circuit current significantly (high resistance)

Circuit diagram representation:

┌───[Component]───┐
│ │
V │
│ │
└─────────────────┘
(Voltmeter in parallel)

Important characteristics of voltmeter:
1. High resistance (typically several kΩ or MΩ)
2. Draws negligible current from circuit
3. Connected externally across points of measurement

Contrast with ammeter:

Feature	Voltmeter	Ammeter
Connection	Parallel	Series
Resistance	Very high	Very low
Measures	Potential difference	Current
Effect on circuit	Negligible	Small

Question 6

A copper wire has diameter 0.5 mm and resistivity of 1.6 × 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?

Answer:

Part 1: Length for 10 Ω resistance
Given:
• Diameter d = 0.5 mm = 0.5 × 10⁻³ m = 5 × 10⁻⁴ m
• Radius r = d/2 = 2.5 × 10⁻⁴ m
• Resistivity ρ = 1.6 × 10⁻⁸ Ω m
• Required resistance R = 10 Ω

Formula: R = ρl/A, where A = πr²

l = (R × A)/ρ A = πr² = π × (2.5 × 10⁻⁴)² = π × 6.25 × 10⁻⁸ m² l = (10 × π × 6.25 × 10⁻⁸)/(1.6 × 10⁻⁸) l = (10 × π × 6.25)/(1.6) l = (62.5π)/(1.6) = 39.0625π l ≈ 39.0625 × 3.1416 ≈ 122.7 m

Part 2: Effect of doubling diameter
When diameter doubles: d' = 2d
• New radius r' = 2r
• New area A' = π(2r)² = 4πr² = 4A
• Resistance R' = ρl/A' = ρl/(4A) = R/4

R' = 10/4 = 2.5 Ω Change in resistance = R' - R = 2.5 - 10 = -7.5 Ω Percentage change = (7.5/10) × 100 = 75% decrease

Key relationships:
1. R ∝ l (directly proportional to length)
2. R ∝ 1/A (inversely proportional to area)
3. R ∝ 1/d² (inversely proportional to square of diameter)

Quick check: Doubling diameter → Area becomes 4 times → Resistance becomes ¼ times.

Question 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:

I (amperes)	0.5	1.0	2.0	3.0	4.0
V (volts)	1.6	3.4	6.7	10.2	13.2

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

Answer:

Step 1: Plotting V-I graph
• Take V on y-axis (vertical), I on x-axis (horizontal)
• Plot points: (0.5, 1.6), (1.0, 3.4), (2.0, 6.7), (3.0, 10.2), (4.0, 13.2)
• Draw best-fit straight line through origin

Step 2: Calculate resistance from graph
According to Ohm's law: V = IR, so R = V/I
We can calculate R for each data point:

I (A)	V (V)	R = V/I (Ω)
0.5	1.6	3.20
1.0	3.4	3.40
2.0	6.7	3.35
3.0	10.2	3.40
4.0	13.2	3.30

Step 3: Average resistance

R_avg = (3.20 + 3.40 + 3.35 + 3.40 + 3.30)/5 R_avg = 16.65/5 = 3.33 Ω

Step 4: Alternative method (graph slope)
Slope of V-I graph = ΔV/ΔI = Resistance
Using extreme points: (0.5, 1.6) and (4.0, 13.2)

Slope = (13.2 - 1.6)/(4.0 - 0.5) = 11.6/3.5 = 3.314 ≈ 3.31 Ω

Final answer: Resistance ≈ 3.33 Ω

Verification of Ohm's law:
• Graph is approximately straight line through origin
• Constant ratio V/I confirms ohmic conductor
• Small variations due to experimental errors

Question 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.

Answer:

Given:
• Voltage V = 12 V
• Current I = 2.5 mA = 2.5 × 10⁻³ A = 0.0025 A

Using Ohm's law: V = IR

R = V/I = 12/(2.5 × 10⁻³) = 12/0.0025 = 4800 Ω = 4.8 kΩ

Step-by-step calculation:
1. Convert mA to A: 2.5 mA = 2.5/1000 = 0.0025 A
2. Apply Ohm's law: R = V/I
3. R = 12 ÷ 0.0025 = 4800 Ω
4. Convert to kΩ: 4800/1000 = 4.8 kΩ

Alternative representation:

R = 12 V / 0.0025 A = 4800 V/A = 4800 Ω

Circuit interpretation:
• 4.8 kΩ is a relatively high resistance
• Explains why current is small (2.5 mA)
• Typical for electronic circuits (not power circuits)

Check: If resistance were lower, current would be higher. For example:
• If R = 1.2 kΩ = 1200 Ω, I = 12/1200 = 0.01 A = 10 mA

Question 9

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

Answer:

Given: Series circuit with:
• Battery voltage V = 9 V
• Resistors: R₁ = 0.2 Ω, R₂ = 0.3 Ω, R₃ = 0.4 Ω, R₄ = 0.5 Ω, R₅ = 12 Ω

Step 1: Total resistance in series

R_total = R₁ + R₂ + R₃ + R₄ + R₅ R_total = 0.2 + 0.3 + 0.4 + 0.5 + 12 R_total = 13.4 Ω

Step 2: Current in series circuit
Using Ohm's law: V = I × R_total

I = V/R_total = 9/13.4 ≈ 0.6716 A

Step 3: Current through 12 Ω resistor
In series circuit, same current flows through all components
∴ Current through 12 Ω resistor = Total current = 0.672 A (approx.)

Verification:
Voltage drops across each resistor:
• V₁ = I × R₁ = 0.672 × 0.2 = 0.1344 V
• V₂ = 0.672 × 0.3 = 0.2016 V
• V₃ = 0.672 × 0.4 = 0.2688 V
• V₄ = 0.672 × 0.5 = 0.3360 V
• V₅ = 0.672 × 12 = 8.064 V
Total = 0.1344 + 0.2016 + 0.2688 + 0.3360 + 8.064 = 9.0048 V ≈ 9 V ✓

Note: The small resistors (0.2-0.5 Ω) are negligible compared to 12 Ω. Approximate calculation: I ≈ 9/12.4 ≈ 0.726 A (close enough).

Question 10

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

Answer:

Given:
• Voltage V = 220 V
• Required current I_total = 5 A
• Each resistor R = 176 Ω
• All resistors connected in parallel

Step 1: Required equivalent resistance
Using Ohm's law: V = I × R_eq

R_eq = V/I_total = 220/5 = 44 Ω

Step 2: Parallel combination formula
For n identical resistors R in parallel:
1/R_eq = n/R

R_eq = R/n n = R/R_eq = 176/44 = 4

Step 3: Verification
4 resistors of 176 Ω in parallel:
1/R_eq = 1/176 + 1/176 + 1/176 + 1/176 = 4/176
R_eq = 176/4 = 44 Ω ✓

Current through each resistor:
I_each = V/R = 220/176 = 1.25 A
Total current = 4 × 1.25 = 5 A ✓

Alternative method:
Current through one resistor = 220/176 = 1.25 A
Number needed for 5 A = 5/1.25 = 4

Practical significance:
• Parallel connection divides current
• Each resistor carries safe current (1.25 A)
• Total power = V × I_total = 220 × 5 = 1100 W

Question 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) 4 Ω.

Answer:

Given: Three resistors, each R = 6 Ω

(i) To get 9 Ω:
Connection: Two in parallel, then in series with the third

R₁(6Ω) ── R₂(6Ω) in parallel → R_parallel = 3Ω
Then in series with R₃(6Ω) → R_total = 3Ω + 6Ω = 9Ω

Calculation:
Two 6 Ω in parallel: 1/R_p = 1/6 + 1/6 = 2/6 = 1/3 → R_p = 3 Ω
This combination in series with third 6 Ω: R_total = 3 + 6 = 9 Ω ✓

(ii) To get 4 Ω:
Connection: Two in series, then in parallel with the third

R₁(6Ω) ── R₂(6Ω) in series → R_series = 12Ω
Then in parallel with R₃(6Ω) → 1/R_total = 1/12 + 1/6 = 1/4 → R_total = 4Ω

Calculation:
Two 6 Ω in series: R_s = 6 + 6 = 12 Ω
This 12 Ω in parallel with 6 Ω:
1/R_total = 1/12 + 1/6 = 1/12 + 2/12 = 3/12 = 1/4
R_total = 4 Ω ✓

Other possible combinations:

Target R	Connection	Calculation
18 Ω	All series	6 + 6 + 6 = 18 Ω
2 Ω	All parallel	1/R = 1/6 + 1/6 + 1/6 = 1/2 → R = 2 Ω
9 Ω	Two parallel + one series	6∥6 = 3, 3 + 6 = 9 Ω
4 Ω	Two series + one parallel	6+6=12, 12∥6 = 4 Ω

Question 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?

Answer:

Given:
• Supply voltage V = 220 V
• Each bulb rating: P = 10 W at 220 V
• Maximum allowable current I_max = 5 A
• All bulbs connected in parallel

Step 1: Current through one bulb
Using P = VI

I₁ = P/V = 10/220 = 1/22 ≈ 0.04545 A

Step 2: Maximum number of bulbs
In parallel, total current = sum of individual currents

n × I₁ ≤ I_max n ≤ I_max/I₁ = 5 ÷ (1/22) = 5 × 22 = 110

Step 3: Alternative method using resistance
Resistance of one bulb: R = V²/P = (220)²/10 = 48400/10 = 4840 Ω
Equivalent resistance for n parallel bulbs: R_eq = R/n = 4840/n Ω
Maximum current: I_max = V/R_eq = 220/(4840/n) = 220n/4840 = n/22
n/22 ≤ 5 → n ≤ 110

Final answer: Maximum 110 bulbs can be connected

Verification:
• 110 bulbs × 0.04545 A each = 4.9995 A ≈ 5 A ✓
• Total power = 110 × 10 W = 1100 W
• Using P = VI = 220 × 5 = 1100 W ✓

Safety consideration:
• Circuit breaker/fuse should be rated for 5 A
• Wiring must handle 5 A current
• In practice, leave some margin (use fewer than 110 bulbs)

Question 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?

Answer:

Given:
• Voltage V = 220 V
• Each coil resistance R = 24 Ω

Case 1: Coils used separately
Only one coil connected:

I = V/R = 220/24 = 55/6 ≈ 9.17 A

Case 2: Coils in series
Total resistance R_series = R + R = 24 + 24 = 48 Ω

I = V/R_series = 220/48 = 55/12 ≈ 4.58 A

Case 3: Coils in parallel
Total resistance: 1/R_parallel = 1/R + 1/R = 1/24 + 1/24 = 2/24 = 1/12
R_parallel = 12 Ω

I = V/R_parallel = 220/12 = 55/3 ≈ 18.33 A

Summary table:

Connection	Equivalent Resistance	Current (A)	Power (W)
Single coil	24 Ω	9.17	2016.7
Series	48 Ω	4.58	1008.3
Parallel	12 Ω	18.33	4033.3

Observations:
1. Parallel connection draws maximum current (4× series current)
2. Series connection draws minimum current
3. Power ∝ 1/R for constant voltage
4. Parallel gives maximum heating (fast cooking)
5. Series gives gentle heating (slow cooking/warming)

Question 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.

Answer:

Circuit (i): 6V battery with 1Ω and 2Ω in series

6V — [1Ω] — [2Ω] — (complete circuit)

Step 1: Total resistance
R_total = 1 + 2 = 3 Ω

Step 2: Current in circuit
I = V/R_total = 6/3 = 2 A
(Same current through both resistors in series)

Step 3: Power in 2 Ω resistor
Using P = I²R

P₁ = I² × R₂Ω = (2)² × 2 = 4 × 2 = 8 W

Circuit (ii): 4V battery with 12Ω and 2Ω in parallel

4V —┬— [12Ω] —┐
└— [2Ω] —─┘

Step 1: Voltage across 2 Ω resistor
In parallel, voltage across each branch = battery voltage = 4 V

Step 2: Power in 2 Ω resistor
Using P = V²/R

P₂ = V²/R₂Ω = (4)²/2 = 16/2 = 8 W

Comparison:
• Circuit (i): Power in 2Ω = 8 W
• Circuit (ii): Power in 2Ω = 8 W
• Both circuits use same power (8 W) in the 2 Ω resistor

Verification for circuit (ii):
Current through 2Ω resistor: I₂Ω = V/R = 4/2 = 2 A
Power = I²R = (2)² × 2 = 8 W ✓

Interesting observation:
Different circuits, different configurations, but same power dissipation in the 2Ω resistor due to specific voltage/current values.

Question 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?

Answer:

Given:
• Lamp 1: P₁ = 100 W at V = 220 V
• Lamp 2: P₂ = 60 W at V = 220 V
• Connected in parallel to 220 V supply

Method 1: Using individual currents
For each lamp: I = P/V

I₁ = P₁/V = 100/220 = 5/11 ≈ 0.4545 A I₂ = P₂/V = 60/220 = 3/11 ≈ 0.2727 A

Total current in parallel = I₁ + I₂

I_total = 5/11 + 3/11 = 8/11 ≈ 0.7273 A

Method 2: Using resistances
Resistance of each lamp: R = V²/P

R₁ = (220)²/100 = 48400/100 = 484 Ω R₂ = (220)²/60 = 48400/60 = 806.67 Ω

Equivalent parallel resistance:
1/R_eq = 1/R₁ + 1/R₂ = 1/484 + 1/806.67
R_eq ≈ 302.5 Ω
Total current: I = V/R_eq = 220/302.5 ≈ 0.7273 A

Final answer: Total current drawn = 8/11 A ≈ 0.73 A

Power calculation:
Total power = P₁ + P₂ = 100 + 60 = 160 W
Verification: P = VI = 220 × (8/11) = 160 W ✓

Important note:
In parallel connection:
• Each lamp gets full 220 V (operates at rated power)
• Currents add up
• Powers add up
• This is how household appliances are connected

Question 16

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

Answer:

Energy formula: Energy (E) = Power (P) × Time (t)
Important: Time must be in consistent units (preferably hours)

For TV set:
P₁ = 250 W, t₁ = 1 hour

E₁ = P₁ × t₁ = 250 W × 1 h = 250 Wh

For toaster:
P₂ = 1200 W, t₂ = 10 minutes = 10/60 = 1/6 hour

E₂ = P₂ × t₂ = 1200 W × (1/6) h = 200 Wh

Comparison:
E₁ = 250 Wh, E₂ = 200 Wh
TV set uses more energy (250 Wh > 200 Wh)

In joules (SI units):
1 Wh = 3600 J
• E₁ = 250 × 3600 = 900,000 J = 900 kJ
• E₂ = 200 × 3600 = 720,000 J = 720 kJ
TV uses 900 kJ vs toaster 720 kJ

Key insight:
Although toaster has higher power (1200 W > 250 W), it operates for much shorter time (10 min vs 60 min).
Energy depends on both power AND time.

Practical analogy:
• TV: Like a slow, steady stream of water for 1 hour
• Toaster: Like a powerful gush of water for 10 minutes
Total volume (energy) depends on both flow rate (power) and time

Question 17

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

Answer:

Given:
• Resistance R = 8 Ω
• Current I = 15 A
• Time t = 2 hours (not needed for rate calculation)

Rate of heat development = Power
Using Joule's law of heating: P = I²R

P = (15)² × 8 = 225 × 8 = 1800 W

Alternative calculations:
1. Using P = VI and V = IR:
V = I × R = 15 × 8 = 120 V
P = V × I = 120 × 15 = 1800 W ✓

2. Using P = V²/R:
V = 120 V (from above)
P = (120)²/8 = 14400/8 = 1800 W ✓

Final answer: Rate of heat development = 1800 W
This means 1800 joules of heat energy produced every second.

Total heat energy in 2 hours:
Energy = Power × Time = 1800 W × 2 h = 3600 Wh = 3.6 kWh
In joules: 1800 J/s × 7200 s = 12,960,000 J = 12.96 MJ

Practical significance:
• 1800 W is a typical heater power
• Would consume 3.6 units of electricity in 2 hours
• Heating cost = 3.6 × electricity rate

Question 18

Explain the following:
(a) Why is the tungsten used almost exclusively for filament of electric lamps?
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why are copper and aluminium wires usually employed for electricity transmission?

Answer:

(a) Why tungsten for lamp filaments?

1. High melting point (3380°C) – can withstand high temperatures without melting
2. High resistivity – produces sufficient heat at normal voltages
3. Ductility – can be drawn into thin wires
4. Does not oxidize easily – lasts longer
5. High luminous efficiency – glows brightly when heated

(b) Why alloys for heating devices?

1. High resistivity – produces more heat for same current
2. High melting point – doesn't melt at operating temperatures
3. Does not oxidize readily – longer life at high temperatures
4. Examples: Nichrome (Ni+Cr+Fe), Constantan (Cu+Ni)
5. Comparison: Pure metals have lower resistivity, oxidize faster

(c) Why not series for domestic circuits?

1. Same current flows through all appliances – impractical
2. Different appliances need different currents
3. If one appliance fails, entire circuit breaks
4. Cannot operate independently – all must be on/off together
5. Voltage divides – appliances don't get rated voltage
6. Parallel advantages: Independent operation, same voltage, different currents

(d) Resistance vs cross-sectional area

R ∝ 1/A (inversely proportional) R = ρl/A

1. Thicker wire (larger A) → lower resistance
2. Thinner wire (smaller A) → higher resistance
3. Doubling area → halves resistance
4. Resistance ∝ 1/d² (d = diameter)
5. Practical: Power lines thick, heater coils thin

(e) Why copper/aluminium for transmission?

1. Low resistivity – minimal energy loss as heat
2. Good conductivity – efficient power transmission
3. Ductility – can be drawn into wires
4. Cost-effective – cheaper than silver (best conductor)
5. Abundant availability
6. Comparison: Copper better but heavier; Aluminium lighter but slightly higher resistance