✅ Q1. A wire of resistance R cut into 5 equal parts ⇒ each part has resistance R/5.
When connected in parallel,
Equivalent resistance:
\frac{1}{R’} = \sum \frac{1}{R/5} = 5 × \frac{5}{R} = \frac{25}{R}
Thus, R’ = R/25 ⇒ R / R’ = 25
🎯 Answer: (d) 25
✅ Q2.
Power: P = I^2R = VI = V^2/R,
but IR^2 is dimensionally incorrect.
🎯 Answer: (b) IR²
✅ Q3. Bulb rated at 220V, 100W ⇒ resistance:
R = \frac{V^2}{P} = \frac{220^2}{100} = 484Ω
At 110V, power:
P’ = \frac{V^2}{R} = \frac{110^2}{484} ≈ 25W
🎯 Answer: (d) 25W
✅ Q4.
Heat ∝ I^2R. In series, total resistance = 2R, current = V/2R.
In parallel, total resistance = R/2, current = 2V/R.
Heat ratio: (V^2/2R) : (2V^2/R) = 1:4
🎯 Answer: (c) 1:4
✅ Q5.
🎯 A voltmeter is connected in parallel across the two points where potential difference is to be measured.
✅ Q6.
Given: diameter = 0.5 mm = 0.0005m, radius r = 0.00025m.
Resistivity ρ = 1.6×10^{-8}, R = 10Ω
Using R = ρ\frac{L}{A}, A = πr^2:
L = \frac{R×πr^2}{ρ} ≈ \frac{10×π×(0.00025)^2}{1.6×10^{-8}} ≈ 1.23 m
If diameter is doubled ⇒ radius becomes 2r, area becomes 4A, resistance becomes R/4 = 2.5Ω.
✅ Q7.
Given table:
|
I (A) |
0.5 |
1.0 |
2.0 |
3.0 |
4.0 |
|
V (V) |
1.6 |
3.4 |
6.7 |
10.2 |
13.2 |
Plot V vs I, slope of graph gives R.
From data:
R ≈ V/I ≈ 3.3Ω (approximate average slope).
✅ Q8.
V = 12V, I = 2.5mA = 0.0025A,
R = V/I = 12 / 0.0025 = 4800Ω
✅ Q9.
Total resistance in series:
R = 0.2+0.3+0.4+0.5+12 = 13.4Ω
Total current:
I = V/R = 9/13.4 ≈ 0.672A
✅ Q10.
We want I = 5A, V = 220V, each resistor R = 176Ω.
For n resistors in parallel:
\frac{220}{176/n} = 5 ⇒ n = \frac{220×5}{176} ≈ 6.25 ≈ 7
So at least 7 resistors are needed.
✅ Q11.
To get:
- 9Ω: two 6Ω in series + one 6Ω in parallel ⇒ (6+6)//6 = 12×6/(12+6) = 72/18 = 4Ω → actually wrong.
Actually:
- (i) 9Ω: two in parallel (3Ω) + one in series: 3+6=9Ω
- (ii) 4Ω: all three in parallel ⇒ R = 6/3 = 2Ω, no — better: two in series =12Ω, in parallel with third: R = (12×6)/(12+6)=4Ω
✅ Q12.
Each bulb: P = 10W, V = 220V, current per bulb:
I = P/V = 10/220 ≈ 0.045A
Max current = 5A ⇒
Number of bulbs = 5/0.045 ≈ 111
✅ Q13.
Coils: A & B, each 24Ω, 220V:
- A or B alone: I = V/R = 220/24 ≈ 9.17A
- in series: R = 48Ω, I = 220/48 ≈ 4.58A
- in parallel: R = 12Ω, I = 220/12 ≈ 18.33A
✅ Q14.
Case (i): series of 1Ω + 2Ω, total 3Ω, current: 6/3=2A
Power in 2Ω: P = I^2R = 2^2×2 =8W
Case (ii): 2Ω & 12Ω parallel on 4V:
Potential across both =4V.
Power in 2Ω: P = V^2/R =4^2/2=8W
So, same power (8W) in both cases.
Here are the answers to your last set of questions from Chapter: Electricity (Class 10, NCERT, CBSE):
✅ Q15.
Two lamps in parallel:
- Lamp 1: P_1 = 100 W, V = 220 V,
Current drawn:
I_1 = \frac{P_1}{V} = \frac{100}{220} ≈ 0.455 A - Lamp 2: P_2 = 60 W, V = 220 V,
Current drawn:
I_2 = \frac{P_2}{V} = \frac{60}{220} ≈ 0.273 A
Total current:
I = I_1 + I_2 ≈ 0.455 + 0.273 ≈ 0.728 A
🎯 Total current ≈ 0.73 A
✅ Q16.
Energy used E = P × t
- TV: P = 250 W, t = 1 hr = 3600 s,
E_1 = 250 × 3600 = 9 × 10^5 J - Toaster: P = 1200 W, t = 10 min = 600 s,
E_2 = 1200 × 600 = 7.2 × 10^5 J
🎯 TV uses more energy than toaster.
✅ Q17.
Rate of heat developed = power = I^2 R
Given I = 5 A, R = 44Ω:
P = I^2 R = 5^2 ×44 = 25 ×44 = 1100 W
🎯 Rate of heat developed = 1100 W
✅ Q18. Explain:
(a)
🎯 Tungsten is used for filaments because:
- It has a high melting point (~3380°C).
- It does not oxidize easily at high temperatures (when enclosed in an inert atmosphere).
- Can withstand high temperatures and emit bright light.
(b)
🎯 Conductors of heating devices are made of alloys, because:
- Alloys have higher resistivity than pure metals → produce more heat.
- They do not oxidize or melt at high temperatures easily.
(c)
🎯 Why series arrangement is not used in domestic circuits:
- In series, if one appliance fails, the entire circuit breaks.
- Each appliance gets less voltage → they may not function properly.
- Current through all appliances is same → no independent control.
(d)
🎯 How resistance varies with area of cross-section:
R ∝ \frac{1}{A}
Resistance decreases when cross-sectional area increases.
(e)
🎯 Why copper & aluminium are used for transmission:
- They are good conductors of electricity (low resistivity).
- They are ductile and can be drawn into wires.
- They are relatively cheap and widely available.
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