📘 Class 10 Electricity NCERT Solutions

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)
Step 3: Find equivalent resistance
Step 4: Calculate ratio R/R'
Verification:
• Cutting wire increases resistance proportionally to length
• Parallel connection dramatically reduces equivalent resistance
• 25 times reduction is logical for 5 equal parallel resistors

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

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
Resistance remains constant (assuming temperature constant)

Step 2: Calculate power at 110 V
Using same resistance:
Alternative method (ratio method):
Since P ∝ V² (when R constant)
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

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
Step 3: Ratio of heat produced
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.

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:
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

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
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

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.

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
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)
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

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
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:
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

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

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

Answer:

Given: Three resistors, each R = 6 Ω

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

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

Answer:

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

Case 1: Coils used separately
Only one coil connected:
Case 2: Coils in series
Total resistance R_series = R + R = 24 + 24 = 48 Ω
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
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)

Answer:

Circuit (i): 6V battery with 1Ω and 2Ω in series
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
Circuit (ii): 4V battery with 12Ω and 2Ω in parallel
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
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.

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
Total current in parallel = I₁ + I₂
Method 2: Using resistances
Resistance of each lamp: R = V²/P
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

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

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
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

Answer:

(a) Why tungsten for lamp filaments?
(b) Why alloys for heating devices?
(c) Why not series for domestic circuits?
(d) Resistance vs cross-sectional area
(e) Why copper/aluminium for transmission?