Question

A circuit consisting of three resistance 12ohm, 18ohm and 36ohm respectively joined in parallel, and connected in series with the fourth resistance. The whole circuit is supplied at 60volts and it is found that the power dissipated in the 12 resistance is 36 Watts. Determine the value of the fourth resistor and the total power dissipated in the circuit

Answer 1

Since P = I^2 R, the current through the 12Ω resistor is √3 amps
Also, since P = E^2/R, the voltage drop across the 12Ω resistor is 12√3 volts
Since the resistors are in the ratio 2:3:6, the total current is
√3(1 + 2/3 + 1/2) = √3 * 13/6 amps
This same current must pass through the 4th resistor R.

Now, the total voltage drop is 60V, so the voltage across the unknown resistor R is (60-12√3). So the unknown R is (60-12√3)/(13/6 √3) ≈ 10.5Ω

Now you can find the circuit's power, since you know all the currents and resistances.

Answer 2

Given: R1 = 12 ohms,
R2 = 18 ohms,
R3 = 36 ohms.
R4 = ?

I1^2*R1 = 36,
I1^2 * 12 = 36,
I1 = 1.73A.

V1 = I1*R1 = 1.73 * 12 = 20.8 volts. = V2 = V3.

I2 = V2/R2 = 20.8/18 = 1.15A.
I3 = V3/R3 = 20.8/36 = 0.58A.

a. R4 = V4/I4 = (E-V1)/(I1+I2+I3) = (60-20.8)/3.46 = 11.3 ohms.

b. P = E * (I1+I2+I3) = 60 * 3.46 = ___Watts.

Answer 3

Yess

Answer 4

Yes