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
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.
Yess
Yes
To solve this problem, we need to break it down into steps:
Step 1: Find the equivalent resistance of the three resistors in parallel.
For resistors in parallel, we use the formula:
1/Req = 1/R1 + 1/R2 + 1/R3
where Req is the equivalent resistance, R1, R2, R3 are the resistances.
Using the given values, we can find the equivalent resistance (Req) of the three resistors in parallel:
1/Req = 1/12 + 1/18 + 1/36
1/Req = (3 + 2 + 1)/36
1/Req = 6/36
1/Req = 1/6
Therefore, Req = 6 ohms.
Step 2: Determine the total circuit resistance.
Since the fourth resistor is connected in series with the three resistors, the total circuit resistance (Rtotal) is the sum of the equivalent resistance (Req) and the fourth resistance (R4). Let's assume R4 as x ohms.
Rtotal = Req + R4
Rtotal = 6 + x
Step 3: Calculate the value of the fourth resistor (R4).
We know that the power dissipated in the 12-ohm resistor is 36 Watts. The power dissipated in a resistor can be calculated using the formula:
P = V^2 / R
where P is the power, V is the voltage, and R is the resistance.
Given V = 60 V and R = 12 ohms, we can calculate the power dissipated (P) using the given values:
36 = 60^2 / 12
36 = 3600 / 12
36 = 300
Since the power dissipated in the 12-ohm resistor is 36 Watts, we can conclude that R = 12 ohms.
Step 4: Calculate the total power dissipated in the circuit.
Now that we have the value of R4, we can find the total power dissipated in the circuit (Ptotal). Using the formula mentioned earlier:
Ptotal = V^2 / Rtotal
Ptotal = 60^2 / (6 + 12)
Ptotal = 3600 / 18
Ptotal = 200 Watts
Therefore, the value of the fourth resistor (R4) is 12 ohms, and the total power dissipated in the circuit is 200 Watts.
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.