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Taruna

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  1. Asked: October 20, 2020In: Class 10

    An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

    Taruna
    Added an answer on February 3, 2021 at 8:21 am

    Here, I = 5 A, V = 220 V, t = 2h = 7,200 s Power, P = V I = 220 x 5 = 1100 W Energy consumed = P x t = 1100 W x 7200 s = 7,920,000 J

    Here, I = 5 A, V = 220 V, t = 2h = 7,200 s
    Power, P = V I = 220 x 5 = 1100 W
    Energy consumed = P x t = 1100 W x 7200 s = 7,920,000 J

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  2. Asked: October 20, 2020In: Class 10

    What determines the rate at which energy is delivered by a current?

    Taruna
    Added an answer on February 3, 2021 at 8:17 am

    Resistance of the circuit determines the rate at which energy is delivered by a current..

    Resistance of the circuit determines the rate at which energy is delivered by a current..

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  3. Asked: October 20, 2020In: Class 10

    An electric iron of resistance 20 ohm takes a current of 5 A. Calculate the heat developed in 30 s.

    Taruna
    Added an answer on February 3, 2021 at 8:16 am

    Here, R = 20 Ω, i = 5 A, t = 3s Heat developed, H = I2 R t = 25 x 20 x 30 = 15,000 J = 1.5 x 104 J

    Here, R = 20 Ω, i = 5 A, t = 3s
    Heat developed, H = I2 R t = 25 x 20 x 30 = 15,000 J = 1.5 x 104 J

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  4. Asked: October 20, 2020In: Class 10

    Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

    Taruna
    Added an answer on February 3, 2021 at 8:15 am

    Here, Q = 96,000 C, t =1 hour = 1 x 60 x 60 sec = 3,600 s, V = 50 V Heat generated, H = VQ = 50Vx 96,000 C = 48,00,000 J = 4.8 x 106 J

    Here, Q = 96,000 C, t =1 hour = 1 x 60 x 60 sec = 3,600 s, V = 50 V
    Heat generated, H = VQ = 50Vx 96,000 C = 48,00,000 J = 4.8 x 106 J

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  5. Asked: October 20, 2020In: Class 10

    Why does the cord of an electric heater not glow while the heating element does?

    Taruna
    Added an answer on February 3, 2021 at 8:14 am

    Heat generated in a circuit is given by I2R t. The heating element of an electric heater made of nichrome glows because it becomes red-hot due to the large amount of heat produced on passing current because of its high resistance, but the cord of the electric heater made of copper does not glow becaRead more

    Heat generated in a circuit is given by I2R t. The heating element of an electric heater made of nichrome glows because it becomes red-hot due to the large amount of heat produced on passing current because of its high resistance, but the cord of the electric heater made of copper does not glow because negligible heat is produced in it by passing current because of its extremely low resistance.

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  6. Asked: October 20, 2020In: Class 10

    What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 ohm, 8 ohm, 12 ohm, 24 ohm?

    Taruna
    Added an answer on February 3, 2021 at 8:12 am

    (i) Highest resistance can be obtained by connecting the four coils in series. Then, R = 4Ω + 8Ω + 12Ω + 24Ω = 48Ω (ii) Lowest resistance can be obtained by connecting the four coils in parallel. 1/r = 1/4 +1/8+1/12+1/24=12/24=1/2 R= 2ohm.

    (i) Highest resistance can be obtained by connecting the four coils in series.
    Then, R = 4Ω + 8Ω + 12Ω + 24Ω = 48Ω
    (ii) Lowest resistance can be obtained by connecting the four coils in parallel.

    1/r = 1/4 +1/8+1/12+1/24=12/24=1/2

    R= 2ohm.

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  7. Asked: October 20, 2020In: Class 10

    How can three resistors of resistances 2 ohm, 3 ohm, and 6 ohm be connected to give a total resistance of (a) 4 ohm, (b) 1 ohm?

    Taruna
    Added an answer on February 3, 2021 at 8:10 am

    (i) We can get a total resistance of 4Ω by connecting the 2Ω resistance in series with the parallel combination of 3Ω and 6Ω. R= R1 + (R2 x R3)/ (R2 + R3) = 2 + (3 x 6)/ (3+6) = 4 ohm. (ii) We can obtain a total resistance of 1Ω by connecting resistors of 2 Ω, 3 Ω and 6 Ω in parallel. 1/R = 1/R1 + 1Read more

    (i) We can get a total resistance of 4Ω by connecting the 2Ω resistance in series with the parallel combination of 3Ω and 6Ω.

    R= R1 + (R2 x R3)/ (R2 + R3)

    = 2 + (3 x 6)/ (3+6) = 4 ohm.

    (ii) We can obtain a total resistance of 1Ω by connecting resistors of 2 Ω, 3 Ω and 6 Ω in parallel.

    1/R = 1/R1 + 1/R2 = 1/2 +1/3+ 1/6 =1 ohm.

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  8. Asked: October 20, 2020In: Class 10

    What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

    Taruna
    Added an answer on February 3, 2021 at 6:09 am

    Advantages of connecting electrical devices in parallel with the battery are : In parallel circuits, if an electrical appliance stops working due to some defect, then all other appliances keep working normally. In parallel circuits, each electrical appliance has its own switch due to which it can beRead more

    Advantages of connecting electrical devices in parallel with the battery are :

    1. In parallel circuits, if an electrical appliance stops working due to some defect, then all other appliances keep working normally.
    2. In parallel circuits, each electrical appliance has its own switch due to which it can be turned on turned off independently, without affecting other appliances.
    3. In parallel circuits, each electrical appliance gets the same voltage (220 V) as that of the power supply line.
    4. In the parallel connection of electrical appliances, the overall resistance of the household circuit is reduced due to which the current from the power supply is high.
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  9. Asked: October 20, 2020In: Class 10

    An electric lamp of 100 ohm, a toaster of resistance 50 ohm, and a water filter of resistance 500 ohm are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

    Taruna
    Added an answer on February 3, 2021 at 6:07 am

    Resistance of electric lamp, R1 = 100 Ω Resistance of toaster, R2 = 50 Ω Resistance of water filter, R3 = 500 Ω Equivalent resistance Rp of the three appliances connected in parallel, is 1/100+1/50+1/500 =5+10+1/500 = 16/500 ⟹ 𝑅 = 500/16 =31.25 Ω Resistance of electric iron = Equivalent resistance oRead more

    Resistance of electric lamp, R1 = 100 Ω
    Resistance of toaster, R2 = 50 Ω
    Resistance of water filter, R3 = 500 Ω
    Equivalent resistance Rp of the three appliances connected in parallel, is

    1/100+1/50+1/500 =5+10+1/500 = 16/500 ⟹ 𝑅 = 500/16 =31.25 Ω

    Resistance of electric iron = Equivalent resistance of the three appliances connected in parallel = 31.25 Ω
    Applied voltage, V = 220 V

    𝐼=𝑉/𝑅 = 220 𝑉/31.25 Ω = 7.04 𝐴

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  10. Asked: October 20, 2020In: Class 10

    Judge the equivalent resistance when the following are connected in parallel

    Taruna
    Added an answer on February 3, 2021 at 6:02 am

    When the resistances are connected in parallel, the equivalent resistance is smaller than the smallest individual resistance. (i) Equivalent resistance < 1 Ω. (ii) Equivalent resistance < 1 Ω.

    When the resistances are connected in parallel, the equivalent resistance is smaller than the smallest individual resistance.
    (i) Equivalent resistance < 1 Ω.
    (ii) Equivalent resistance < 1 Ω.

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