Class 10 Electricity Important Questions : Solutions

Important Questions Notes

Electricity Important Questions : Solutions

The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 120 V.
Sol:
Here, V = 60 V and I = 4 A
Using, R = V/I = 60/4 = 15 Ω
Now, V = 120 V
So, I = V/R = 120/15 = 8 A
The V-I graph of three resistors is shown below. Which resistor has maximum resistance?

Sol:
Resistor R₃ has maximum resistance.
An electric iron is marked 400 W and 220 V. What is the resistance when iron is hot? How long could it be used for ₹3 of electrical energy that costs 15 paise per unit.
Sol:
Given, P = 400 W and V = 220 V
Using, P = ^V²/_R
⇒ R = ^V²/_P = ^220²/₄₀₀ = 121 Ω
So, resistance when iron is hot = 121 Ω

Now, rate of electrical energy = 15 paise per unit
and total money = ₹3 = 300 paise
So, electrical energy consumed, E = ³⁰⁰/₁₅ = 20 unit = 20 kWh = 20000 Wh
Using, E = P × t
⇒ t = ^E/_P = ²⁰⁰⁰⁰/₄₀₀ = 50 h
So, It can be used for 50 hours.
A 20 cm long uniform wire of resistance 5 Ω is stretched to a uniform wire of 40 cm length. What will be the resistance of new wire?
Sol:
length = l = 20 cm
Area of cross-section = A
Resistance = R = ^ρl/_A = 5 Ω
New length = l’ = 40 cm = 2l
New area of cross-section = A’
When the wire is stretched, its volume remains the same.
So, A’ × l’ = A × l
⇒ A’ × 40 = A × 20
⇒ A’ = ^A/₂
New resistance
= R’
= ^ρl’/_A’
= ^ρ(2l)/_(^A/2)
= 4^ρl/_A
= 4R = 4 × 5 = 20 Ω
Therefore, resistance of the new wire is 20 Ω.
If R₁ and R₂ are the resistances of filaments of a 400 W and a 200 W lamp, designed to operate on the same voltage, then find the relationship between R₁ and R₂.
Sol:
For R₁, P₁ = 400 W
For R₂, P₂ = 200 W
Using, P = ^V²/_R
(V₁)² = P₁ x R₁ = 400R₁
and (V₂)² = P₂ x R₂ = 200R₂

Since, both filaments operate on the same voltage.
So, (V₁)² = (V₂)²
⇒ 400R₁ = 200R₂
⇒ 2R₁ = R₂
.
1. State Ohm’s law and write its expression.
  Sol:
  According to Ohm’s law, at constant temperature, the potential difference across the ends of a conductor is directly proportional to the amount of current flowing through it.
  Mathematically,
  V ∝ I
  V = IR where, R is constant of proportionality & is known as resistance.
2. Draw a circuit diagram to verify Ohm’s law.
  Sol: Circuit diagram for the verification of Ohm’s law:
The rating of bulb B₁ is 60 V, 12 W & bulb B₂ is 100 V, 100 W. Find the maximum emf of the battery so that all bulbs remain safe.

Sol:
For B₁,
V₁ = 60 V, P₁ = 120 W
R₁ = ^V₁²/_P₁ = ^60²/₁₂₀ = 300 Ω
For B₂,
V₂ = 100 V, P₂ = 100 W
R₂ = ^V₂²/_P₂ = ^100²/₁₀₀ = 100 Ω
Since, B₁ and B₂ are connected in parallel.
So, ¹/_{R_p} = ¹/_R₁ + ¹/_R₂ = ¹/₃₀₀ + ¹/₁₀₀ = ⁴/₃₀₀
⇒ R_p= ³⁰⁰/₄ = 75 Ω
So, R_eqacross circuit = 75 + 300 + 25 = 400 Ω

Current that can flow through B₁ = ^V₁/_R₁ = ⁶⁰/₃₀₀ = ¹/₅ A
Current that can flow through B₂ = ^V₂/_R₂ = ¹⁰⁰/₁₀₀ = 1 A
So, maximum current that can flow through the circuit so that all bulbs remain safe = ¹/₅ A

Therefore, required e.m.f. = IR_eq = ¹/₅ x 400 = 80 V
Find equivalent resistance between P and Q:

Sol:
Find equivalent resistance between P and Q:

Sol:
Here, all resistors are connected in parallel.
So, ¹/_{R_p} = ¹/_R₁ + ¹/_R₂ + ¹/_R₃ = ¹/₁₀ + ¹/₁₀ + ¹/₁₀ = ³/₁₀
⇒ R_p = ¹⁰/₃ Ω
In reference to the given figure, find the current flowing through 20 Ω resistor.

Sol:
R_AB = R₁ + R₂ = 5 + 20 = 25 Ω
R_CD = R₁ + R₂ = 10 + 40 = 50 Ω

¹/_{R_eq} = ¹/_{R_AB} + ¹/_{R_CD} = ¹/₂₅ + ¹/₅₀ = ³/₅₀
⇒ R_eq = ⁵⁰/₃ Ω

I = 3 A
So, V = IR_eq = 3 x ⁵⁰/₃ =50 V

In parallel, voltage remains same.
So, current flowing through AB, I = ^V/_{R_AB} = ⁵⁰/₂₅ = 2 A

In series, current flowing through each resistor remains same.
So, current flowing through 20 Ω resistor = 2 A.