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13 November, 00:16

A 20V battery is connected to a complex circuit with three branches. The first branch has a 10Ω light bulb and a switch in series. The second branch has a 10Ω and 20Ω light bulb in series. The third branch has only the battery. Solve the circuit.

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  1. 13 November, 00:37
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    voltage of battery, V = 20 v

    first branch: a 10 ohm bulb and a switch

    second branch: a 10 ohm bulb in series with a 20 ohm bulb

    Total resistance of the second branch as they are connected in series

    Rs = R1 + R2

    Rs = 10 + 20 = 30 ohm

    Now they are connected in parallel, so the effective resistance of the circuit is R.

    R = 10 x 30 / (10 + 30) = 300 / 40 = 7.5 ohm

    total current in the circuit, i = V / R = 20 / 7.5 = 2.67 Amphere

    Voltage across first branch, V1 = V = 20 V

    Current in first branch, i1 = V / R' = 20 / 10 = 2 A

    Voltage across second branch, V2 = V = 20 V

    current in second branch, i2 = i - i1 = 2.667 - 2 = 0.667 A
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