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The resistance of each of the three wires joined as shown in figure is `4 Omega` and each one can have a maximum power of 20 watt (otherwise it will melt). What maximum power will the whole circuit dissipate ? |
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Answer» Let `I` be the current through a resistance wire of maximum power 20 watt. `:. I^(2)R = 20 or I^(2) xx 4 = 20 or I^(2) = 5` Effective resistance `R_(p)` between A and C will be `R_(p) = ( 4 xx 4)/(4 +4 ) = 2 Omega` Electric power dissipated as heat by the resistance between A and C `= I^(2)R_(p) = 5 xx 2 = 10` watt. Electric power dissipated as heat by the resistance between C and B `= I^(2)R = 5 xx 4 = 20` watt. `:.` Total electric power dissipated as heat by the whole circuit `= 10 +20 = 30` watt. |
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