Class 10 Electricity Important Questions : Solutions

Important Questions Notes

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 = /R
    ⇒ R = /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)
    = 4ρl/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 = /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
    R1V1²/P1 = 60²/120 = 300 Ω
    For B2,
    V2 = 100 V, P2 = 100 W
    R2V2²/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
    ⇒ Rp = 300/4 = 75 Ω
    So, Req 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. = IReq = 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
    ⇒ Rp = 10/3 Ω
  10. In reference to the given figure, find the current flowing through 20 Ω resistor.

    Sol:
    RAB = R1 + R2 = 5 + 20 = 25 Ω
    RCD = R1 + R2 = 10 + 40 = 50 Ω

    1/Req = 1/RAB + 1/RCD = 1/25 + 1/50 = 3/50
    ⇒ Req = 50/3 Ω

    I = 3 A
    So, V = IReq = 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.
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