_____R____________
| | |
|+ | XC
E1 XL |
|- | |+
| | E2
___________________-
Z1 = R= 4 ohms <0 4+-j0
Z2 = XL= 6 ohms <90. 0+j6
Z3 = XC=8 ohms <-90. 0-j8
E1= 10 v < 0. 10 +j0
E2= 40 v < 60. 20+j34.64
Please help me find the current up through R using mesh analysis.
I am having trouble because of the voltage polar angles.
My derivative is
(Z1+Z2) -Z2
-Z2 (Z2+Z3)
= Z1Z2 + Z1Z3 + Z2Z3
I1= E1. -Z2
-E2. Z2+Z3
=Z2(E1-E2) +Z3E1/D
I2=Z1+Z2. E1
-Z2. -E2
=-Z2(e1+e2)-Z1E2/D
Once the numbers and j's and polar angles get thrown around I have trouble keeping track. I'm not even sure if my I1 or I2 numerator is distributed correctly.
Okay well the circuit picture didn't figure very well. R is in series with e1 and XL
XL is in series on the other side with XC and e2
I = VR/R = 10/4 = 2.5A. = Current through all components(R,L, and C).
VL = I*XL = 2.5 * 6 = j15.0 Volts.
Vc = I*Xc = 2.5 * 8 = -j20.0 Volts.
Z = R + j(XL-Xc)) = 4 + j(6-8) =
4 - j2 = 4.47 Ohms[-26.6o].
Es = I*Z = 2.5 * 4.47 = 11.2 Volts.
Pf = Cos(-26.6) = 0.894 = Power factor.
P = I^2*R = 2.5^2 * 4 = 25 Watts.
Note: If the voltage across the resistor
is incorrect, all voltages will be wrong.
To find the current flowing through R using mesh analysis, we need to apply Kirchhoff's voltage law (KVL) to the loops in the circuit.
First, let's label the current through R as I_R.
We will set up two mesh equations:
For mesh 1:
- Starting from the negative terminal of E1, we go through Z1, Z2, and then return to the positive terminal of E1.
- The voltage drops across Z1, Z2, and E1 must add up to zero.
- Using Ohm's law for impedance, we can write: -Z1 * I_R + (E1 - E2) - Z2 * I2 = 0 (equation 1)
For mesh 2:
- Starting from the negative terminal of E2, we go through Z2, Z3, and then return to the positive terminal of E2.
- The voltage drops across Z2, Z3, and E2 must add up to zero.
- Using Ohm's law for impedance, we can write: -(E1 - E2) - Z2 * I_R + Z3 * (I_R + I2) = 0 (equation 2)
Now, let's substitute the given values into the equations:
Z1 = 4 ohms < 0 degrees
Z2 = 6 ohms < 90 degrees
Z3 = 8 ohms < -90 degrees
E1 = 10 volts < 0 degrees
E2 = 40 volts < 60 degrees
Substituting these values into equations 1 and 2, we get:
-4 * I_R + (10 - 40) - 6 * I2 = 0
-(10 - 40) - 6 * I_R + 8 * (I_R + I2) = 0
Simplifying these equations gives:
-4 * I_R - 6 * I2 = 30 (equation 1')
-6 * I_R + 8 * I_R + 8 * I2 = -30 (equation 2')
Now we have a system of two equations with two unknowns (I_R and I2). We can solve this system using various methods such as substitution or elimination.
Using substitution method:
From equation 1', we can express I2 in terms of I_R:
I2 = (30 + 4 * I_R) / -6
Substituting this into equation 2', we get:
-6 * I_R + 8 * I_R + 8 * ((30 + 4 * I_R) / -6) = -30
Simplifying this equation, we can solve for I_R:
-6 * I_R + 8 * I_R - (8 * 30 + 32 * I_R) / 6 = -30
Multiply through by 6 to eliminate the fraction:
-36 * I_R + 48 * I_R - 8 * 30 - 32 * I_R = -180
(48 - 36 - 32) * I_R = -180 + 8 * 30
(I_R) * (-20) = -180 + 240
I_R = (240 - 180) / (-20)
I_R = 60 / (-20)
I_R = -3 Amps
So, the current flowing through R, I_R, is -3 Amps.