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Three resistors of 3Ω, 1.5Ω and 1Ω are connected in parallel. calculate the current in each of the resistors. the voltage of the battery they are connected

As we know that V=IR Potential across parallel remain same means potential drop across on 3 resistance will be same Current in 3 ohm is given by =V/R=4.5/3=1.5 A Current in 1.5 ohm is given by =4.5/1.5=3 A Current in 1 ohm resistance is given by=4.5 A

Then effective resistance of the 3 parallel resistors is = 1 / (1+1/3+2/3) = 1/2 ohm. Total effective resistance in the circuit is 8.5 + 0.5 = 9 ohms. Current in the 8.5 ohms = 4.5/9 = 0.5 amp. Voltage across 3 parallel resistors is = 0.5*0.5 = 0.25 Volts. Now current in 1 ohm resistance is = 0.25 amp. Current in 3 ohms resistance is = 0.25V/3 = 1/12 amp. The current in 1.5 ohms resistor is 0.25/1.5 = 1/6 amp.

ok nitya

Now,Our effective resistance of three parallel resistor become a single resistance and this resistance is connected to 8.5 ohm with series, then effective resistance will be change