the 4 240Ω resistors have a combined resistance of 60Ω
Add that to the 40Ω in series, and the total is 100Ω
P = E^2/R, so the parallel group dissipates
960W
Now just figure the energy of the heater.
Four resistors , each of 240ohms are connected in parallel , and the group then joined in series with a heater of 40ohms resistance. Then complete circuit is connected to a 240volts supply . Calculate total resistance in the circuit , the power dissipated in the parallel group and the energy dissipated by the heater in joule per minute
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