Class 10 Electricity Important Questions : Solutions

Important Questions Notes Previous Years Questions

Electricity Important Questions : Solutions

1. 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
   
2. The V-I graph of three resistors is shown below. Which resistor has maximum resistance?
   
   Sol:
   Resistor R₃ has maximum resistance.
   
3. 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.
   
4. 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.
   
5. 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
   
6. .
   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:
      
      
7. 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/_Rp = ¹/_R1 + ¹/_R2 = ¹/₃₀₀ + ¹/₁₀₀ = ⁴/₃₀₀
   ⇒ R_p = ³⁰⁰/₄ = 75 Ω
   So, R_eq across 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
   
8. Find equivalent resistance between P and Q:
   
   Sol:
   
   
9. Find equivalent resistance between P and Q:
   
   Sol:
   Here, all resistors are connected in parallel.
   So, ¹/_Rp = ¹/_R1 + ¹/_R2 + ¹/_R3 = ¹/₁₀ + ¹/₁₀ + ¹/₁₀ = ³/₁₀
   ⇒ R_p = ¹⁰/₃ Ω
   
10. 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/_Req = ¹/_RAB + ¹/_RCD = 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/_RAB = ⁵⁰/₂₅ = 2 A
    
    In series, current flowing through each resistor remains same.
    So, current flowing through 20 Ω resistor = 2 A.