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  1. Asked: In:

    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?

    Baldev

    (a) When all the resistors are connected in parallel Then the 1/R=1/4Ω+1/8Ω+1/12Ω+1/24Ω=(6+3+2+1)/24Ω=12/24Ω=1/2Ω Thus, R = 2 Ω (b) When all the resistors are connected in series Thus, total resistance = 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω Thus, highest resistance = 48 Ω Lowest resistance = 2 Ω

    (a) When all the resistors are connected in parallel
    Then the 1/R=1/4Ω+1/8Ω+1/12Ω+1/24Ω=(6+3+2+1)/24Ω=12/24Ω=1/2Ω
    Thus, R = 2 Ω

    (b) When all the resistors are connected in series
    Thus, total resistance = 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω
    Thus, highest resistance = 48 Ω
    Lowest resistance = 2 Ω

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  2. Asked: In:

    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?

    Baldev

    (a) When resistors having resistance equal to 3 Ω and 6 Ω are connected in parallel and one having resistance equal to 2 Ω is connected in series Let total resistance due to resistors having resistance equal to 6 Ω and 3 Ω = R1 Therefore, 1/R1=1/6Ω+1/3Ω=1+2/6Ω=3/6Ω=1/2Ω Thus R1=2Ω Now, total effectiRead more

    (a) When resistors having resistance equal to 3 Ω and 6 Ω are connected in parallel and one having resistance equal to 2 Ω is connected in series

    Let total resistance due to resistors having resistance equal to 6 Ω and 3 Ω = R1
    Therefore, 1/R1=1/6Ω+1/3Ω=1+2/6Ω=3/6Ω=1/2Ω
    Thus R1=2Ω
    Now, total effective resistance in the circuit = R1 + 2 Ω = 2 Ω + 2 Ω = 4 Ω
    Hence, when resistors having resistance equal to 3 Ω and 6 Ω are connected in parallel and one having resistance equal to 2 Ω is connected in series, then the total effective resistance in the circuit = 4 Ω

    (b) When all the three resistance is connected in parallel then
    1/R=1/2Ω+1/3Ω+1/6Ω=3+2+1/6Ω=6/6Ω=1Ω
    Thus, R=1Ω
    When all the three resistance will be connected in parallel, then the total effective resistance in the circuit = 1 Ω

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  3. Asked: In:

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

    Baldev

    Advantages of connecting electrical appliances in parallel instead of connecting in series: (a) Voltage remains same in all the appliances. (b) Lower total effective resistance

    Advantages of connecting electrical appliances in parallel instead of connecting in series:
    (a) Voltage remains same in all the appliances.
    (b) Lower total effective resistance

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  4. Asked: In:

    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?

    Best Answer
    Baldev

    Given: R1 = 100Ω, R2 = 50 Ω, R3 = 500 Ω all are connected in parallel Potential difference = 220V Thus, 1/R=1/100Ω+1/50Ω+1/500Ω=(5+10+1)/500Ω=16/500Ω Therefore, R=500/16Ω=31.25Ω Electric current (I) through the circuit = V/R ⇒I=220V/31.25Ω=7.04A For electric iron Since it takes as well current as thRead more

    Given:
    R1 = 100Ω, R2 = 50 Ω, R3 = 500 Ω all are connected in parallel
    Potential difference = 220V
    Thus, 1/R=1/100Ω+1/50Ω+1/500Ω=(5+10+1)/500Ω=16/500Ω
    Therefore, R=500/16Ω=31.25Ω
    Electric current (I) through the circuit = V/R
    ⇒I=220V/31.25Ω=7.04A
    For electric iron
    Since it takes as well current as three appliances, thus electric current through it = 7.04A
    The electric current = 7.04 A and potential difference = 220 V
    Thus, Resistance of electric iron = Total resistance of three appliances = 31.25 Ω
    Thus, electric current through the electric iron = 7.04A
    Resistance of electric iron = 31.25 Ω

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  6. Asked: In:

    Judge the equivalent resistance when the following are connected in parallel

    Baldev

    Since 1/R=1/R1+1/R2+1/R3+..+1/Rn when resistors are connected in parallel (a) 1 Ω and 106 Ω Thus, 1/R=1/1Ω+1/106Ω=106+1/106Ω=107/106Ω Thus, R=106/107Ω=0.99Ω Thus, equivalent resistance of 1Ω and 106Ω are connected in parallel = 0.99Ω (b)  1 Ω and 103 Ω, and 106 Ω Thus, 1/R=1/1Ω+1/103Ω+1/106Ω=(10918+Read more

    Since 1/R=1/R1+1/R2+1/R3+..+1/Rn
    when resistors are connected in parallel
    (a) 1 Ω and 106 Ω
    Thus, 1/R=1/1Ω+1/106Ω=106+1/106Ω=107/106Ω
    Thus, R=106/107Ω=0.99Ω
    Thus, equivalent resistance of 1Ω and 106Ω are connected in parallel = 0.99Ω

    (b)  1 Ω and 103 Ω, and 106 Ω
    Thus, 1/R=1/1Ω+1/103Ω+1/106Ω=(10918+106+103)/103×106Ω=11127/10918Ω
    Thus, R=10918/11127Ω=1.02Ω
    Thus, equivalent resistance of  1 Ω, 103 Ω and 106 Ω are connected in parallel = 1.02Ω

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