A circuit,consisting of 3 resistances 12,18 & 36 ohms respectively joined in parallel is connected in series with fourth resistance.the whole is supplied at 60 volt & it is found that the power dissipated in the 12Ohm resistance is 36 watt. Determine the value of the fourth resistance & total power dissipated in the group.
P1 = V^2 / R1,
P1 = V^2 / 12 = 36 W.
V^2 = 12*36 = 432,
V=20.8 Volts. = Voltage across R1,R2,R3
I1 = V1 / R1 = 20.8 / 12 = 1.73A.
I2 = 20.8 / 18 = 1.16A.
I3 = 20.8 / 36 = 0.578A.
I4=I1 + I2 + I3 = 1.73 + 1.16 + 0.578 =
3.47A.
R4=V4 / I4 = (60 - 20.8) / 3.47 = 11.3
Ohms.
Pt=E*It = E*I4 = 60 * 3.47 = 208 Watts.
Draw a circuit diagram showing correct connection of all resistors and supply voltage
Well, it seems like you have quite the electrifying problem there! Let me try to untangle it for you.
Let's start by calculating the equivalent resistance of the three resistors in parallel. To find the equivalent resistance of resistors in parallel, you use the formula:
1/Req = 1/R1 + 1/R2 + 1/R3
Using the values you provided, we can substitute the resistances into the equation:
1/Req = 1/12 + 1/18 + 1/36
Now, let me calculate it... *clownishly tinkering with a calculator*
After a quick calculation, I found that the equivalent resistance (Req) is 6 ohms. So, the resistance of the fourth resistor (let's call it R4) must be equal to 6 ohms. Easy peasy!
As for the total power dissipated in the group, we need to calculate the total current flowing through the circuit. Since the resistances are all connected in series, the total current remains the same throughout the circuit.
Using Ohm's Law (V = I * R), we can find the total current (I) by dividing the total voltage (60 volts) by the total resistance (Req + R4):
I = V / (Req + R4)
I = 60 / (6 + 6)
I = 60 / 12
I = 5 amperes
Now that we have the total current, we can find the total power (P) dissipated in the group using the formula P = I^2 * R. Since the power in the 12-ohm resistance is given as 36 watts, we can calculate the total power dissipated:
P = I^2 * R
P = (5)^2 * (12 + 6)
P = 25 * 18
P = 450 watts
So, the value of the fourth resistance is 6 ohms, and the total power dissipated in the circuit is 450 watts. I hope that electrifying explanation brightened your day!
To determine the value of the fourth resistance and the total power dissipated in the circuit, we can use the following steps:
Step 1: Calculate the equivalent resistance (Req) of the three resistances (12, 18, and 36 ohms) in parallel.
The formula for calculating the equivalent resistance of resistances in parallel is:
1/Req = 1/R1 + 1/R2 + 1/R3
Substituting the values into the equation:
1/Req = 1/12 + 1/18 + 1/36
Let's simplify the equation:
1/Req = (3 + 2 + 1)/36
1/Req = 6/36
1/Req = 1/6
To eliminate the reciprocal, we take the inverse of both sides of the equation:
Req = 6 ohms
Step 2: Calculate the current flowing through the resistor with resistance 12 ohms, using Ohm's Law (V = IR).
We are given that the power dissipated in the 12 ohm resistance is 36 watts.
Power (P) = I^2 * R
Substituting the given values:
36 = I^2 * 12
Solving for the current (I):
I^2 = 36 / 12
I^2 = 3
I = √3
I ≈ 1.732 A (rounded to three decimal places)
Step 3: Calculate the voltage drop across the resistor with resistance 12 ohms using Ohm's Law (V = IR).
Voltage (V) = I * R
Substituting the known values:
V = 1.732 * 12
V ≈ 20.784 V (rounded to three decimal places)
Step 4: Calculate the voltage drop across the fourth resistance.
To find the value of the fourth resistance, subtract the voltage drop across the resistor with resistance 12 ohms from the total supply voltage.
Voltage drop across the fourth resistance = Total supply voltage - Voltage drop across the 12 ohm resistance
We are given that the total supply voltage is 60 volts.
Voltage drop across the fourth resistance = 60 - 20.784
Voltage drop across the fourth resistance ≈ 39.216 V (rounded to three decimal places)
Step 5: Calculate the value of the fourth resistance using Ohm's Law (V = IR).
We know the voltage drop across the fourth resistance and the current flowing through it.
Voltage drop (V) = Current (I) * Resistance (R)
Substituting the known values:
39.216 = 1.732 * R
Solving for the fourth resistance (R):
R = 39.216 / 1.732
R ≈ 22.646 ohms (rounded to three decimal places)
So, the value of the fourth resistance is approximately 22.646 ohms.
Step 6: Calculate the total power dissipated in the group.
To find the total power dissipated in the circuit, we can sum up the power dissipated in each individual resistor using the formula P = I^2 * R.
Total Power (Ptotal) = Power in the 12 ohm resistance + Power in the 18 ohm resistance + Power in the 36 ohm resistance + Power in the fourth resistance
Given:
Power in the 12 ohm resistance = 36 watts
Current (I) = 1.732 A (rounded from previous calculations)
Power in the 12 ohm resistance = I^2 * 12
P12 = (1.732)^2 * 12
P12 ≈ 35.7944 watts (rounded to four decimal places)
Power in the 18 ohm resistance = (I^2 * R)
P18 = (1.732)^2 * 18
P18 ≈ 53.6916 watts (rounded to four decimal places)
Power in the 36 ohm resistance = (I^2 * R)
P36 = (1.732)^2 * 36
P36 ≈ 107.3832 watts (rounded to four decimal places)
Power in the fourth resistance = (I^2 * R)
P4 = (1.732)^2 * 22.646
P4 ≈ 67.1514 watts (rounded to four decimal places)
Total Power (Ptotal) = P12 + P18 + P36 + P4
Ptotal ≈ 35.7944 + 53.6916 + 107.3832 + 67.1514
Ptotal ≈ 263.0206 watts (rounded to four decimal places)
So, the total power dissipated in the group is approximately 263.0206 watts.