Electronics - Parallel Circuits - Discussion

Yes, I want to know same thing. When P=VI and v=ir then howvit possible p=v/r.

Anyone explain the correct answer.

Total power reading of the wattmeter is 9.408.

If you solve for the individual powers in the branches:

P = V^2/R.
P1 = 2.304W , P2 = 1.152W , P3 = 0.768W &, P4 = 0.576W.

If you get the total power:
P1+P2+P3+P4 = 4.8 W which is smaller than the reading in the ammeter which definitely tells us that something is wrong with the circuit.

Now since there are ammeters connected in between the branches we can check if the individual values of branch currents match the readings.

First in R2, I2 = 48/2000 = 24mA.
In R3, I3 = 48/3000 = 16mA.
In R4, I4 = 48/4000 = 12mA.

From the Ammeter Connected in-branch R4 we can confirm that it is correct since it has the same reading of 0.012 or 12mA.

From the Ammeter Connection in-branch R2 the reading is 0.052 A or 52mA since we know that I2 = 24mA, via Kirchoff's Current Law 52mA=24mA+ I , I is the current that would go to the branch R3. I = 28mA.

Since we know that I3 = 16mA, Via Kirchoffs Current Law we can check that, I = I3 + I4; 28mA =16mA+12mA.

Since we confirmed that I3 and I4 is correct it also implies that the current in I2 is correct therefore all of the power ratings of P2, P3, and P4 is correct thus indicating that R1 is of the wrong value.