Answer :
To solve for the values of [tex]\( R \)[/tex] and [tex]\( X \)[/tex] in the given circuit, we need to use the information provided:
1. Impedance of the coil [tex]\( Z_{\text{coil}} = 8 + j6 \, \Omega \)[/tex]
2. Equivalent impedance [tex]\( Z_{\text{eq}} = 5 \Omega \)[/tex] (with 0 degrees phase angle, meaning purely real)
Let's denote the capacitive reactance by [tex]\( X \)[/tex]. The impedance of a capacitor is given by [tex]\( -jX \)[/tex]. The series combination of the coil and the capacitor will then have an impedance [tex]\( Z_{\text{series}} \)[/tex] defined as:
[tex]\[ Z_{\text{series}} = Z_{\text{coil}} + (-jX) \][/tex]
[tex]\[ Z_{\text{series}} = (8 + j6) + (-jX) \][/tex]
[tex]\[ Z_{\text{series}} = 8 + j(6 - X) \][/tex]
Next, we need to find the equivalent impedance of the parallel combination of [tex]\( Z_{\text{series}} \)[/tex] and the resistor [tex]\( R \)[/tex]. The impedance for two components in parallel follows the formula:
[tex]\[ \frac{1}{Z_{\text{eq}}} = \frac{1}{Z_{\text{series}}} + \frac{1}{R} \][/tex]
Given [tex]\( Z_{\text{eq}} = 5 \, \Omega \)[/tex], we can write:
[tex]\[ \frac{1}{5} = \frac{1}{8 + j(6 - X)} + \frac{1}{R} \][/tex]
This equation involves complex numbers, so we must separate it into real and imaginary parts. The first term on the right-hand side is the reciprocal of a complex number:
[tex]\[ \frac{1}{8 + j(6 - X)} = \frac{8 - j(6 - X)}{(8)^2 + (6 - X)^2} = \frac{8 - j(6 - X)}{64 + 36 - 12X + X^2} = \frac{8 - j(6 - X)}{100 - 12X + X^2} \][/tex]
Multiplying both numerator and denominator by the conjugate:
[tex]\[ \frac{1}{8 + j(6 - X)} = \frac{8 - j(6 - X)}{100 - 12X + X^2} \][/tex]
Now we separate into real and imaginary parts:
[tex]\[ \text{Real part: } \frac{8}{100 - 12X + X^2} \][/tex]
[tex]\[ \text{Imaginary part: } \frac{-(6 - X)}{100 - 12X + X^2} \][/tex]
We come back to our equation:
[tex]\[ \frac{1}{5} = \frac{8}{100 - 12X + X^2} + \frac{1}{R} \][/tex]
Because [tex]\( Z_{\text{eq}} \)[/tex] is purely real, the imaginary part must be zero:
[tex]\[ \frac{-(6 - X)}{100 - 12X + X^2} = 0 \][/tex]
[tex]\[ -(6 - X) = 0 \][/tex]
[tex]\[ X = 6 \][/tex]
Now substituting [tex]\( X = 6 \)[/tex] into the real part:
[tex]\[ \frac{1}{5} = \frac{8}{100 - 12(6) + 6^2} + \frac{1}{R} \][/tex]
[tex]\[ \frac{1}{5} = \frac{8}{100 - 72 + 36} + \frac{1}{R} \][/tex]
[tex]\[ \frac{1}{5} = \frac{8}{64} + \frac{1}{R} \][/tex]
[tex]\[ \frac{1}{5} = \frac{1}{8} + \frac{1}{R} \][/tex]
Solving for [tex]\( R \)[/tex]:
[tex]\[ \frac{1}{R} = \frac{1}{5} - \frac{1}{8} \][/tex]
[tex]\[ \frac{1}{R} = \frac{8 - 5}{40} \][/tex]
[tex]\[ \frac{1}{R} = \frac{3}{40} \][/tex]
[tex]\[ R = \frac{40}{3} = 13.33 \, \Omega \][/tex]
Thus, the values are:
- The resistor [tex]\( R = 13.33 \, \Omega \)[/tex]
- The capacitive reactance [tex]\( X = 6 \, \Omega \)[/tex]
1. Impedance of the coil [tex]\( Z_{\text{coil}} = 8 + j6 \, \Omega \)[/tex]
2. Equivalent impedance [tex]\( Z_{\text{eq}} = 5 \Omega \)[/tex] (with 0 degrees phase angle, meaning purely real)
Let's denote the capacitive reactance by [tex]\( X \)[/tex]. The impedance of a capacitor is given by [tex]\( -jX \)[/tex]. The series combination of the coil and the capacitor will then have an impedance [tex]\( Z_{\text{series}} \)[/tex] defined as:
[tex]\[ Z_{\text{series}} = Z_{\text{coil}} + (-jX) \][/tex]
[tex]\[ Z_{\text{series}} = (8 + j6) + (-jX) \][/tex]
[tex]\[ Z_{\text{series}} = 8 + j(6 - X) \][/tex]
Next, we need to find the equivalent impedance of the parallel combination of [tex]\( Z_{\text{series}} \)[/tex] and the resistor [tex]\( R \)[/tex]. The impedance for two components in parallel follows the formula:
[tex]\[ \frac{1}{Z_{\text{eq}}} = \frac{1}{Z_{\text{series}}} + \frac{1}{R} \][/tex]
Given [tex]\( Z_{\text{eq}} = 5 \, \Omega \)[/tex], we can write:
[tex]\[ \frac{1}{5} = \frac{1}{8 + j(6 - X)} + \frac{1}{R} \][/tex]
This equation involves complex numbers, so we must separate it into real and imaginary parts. The first term on the right-hand side is the reciprocal of a complex number:
[tex]\[ \frac{1}{8 + j(6 - X)} = \frac{8 - j(6 - X)}{(8)^2 + (6 - X)^2} = \frac{8 - j(6 - X)}{64 + 36 - 12X + X^2} = \frac{8 - j(6 - X)}{100 - 12X + X^2} \][/tex]
Multiplying both numerator and denominator by the conjugate:
[tex]\[ \frac{1}{8 + j(6 - X)} = \frac{8 - j(6 - X)}{100 - 12X + X^2} \][/tex]
Now we separate into real and imaginary parts:
[tex]\[ \text{Real part: } \frac{8}{100 - 12X + X^2} \][/tex]
[tex]\[ \text{Imaginary part: } \frac{-(6 - X)}{100 - 12X + X^2} \][/tex]
We come back to our equation:
[tex]\[ \frac{1}{5} = \frac{8}{100 - 12X + X^2} + \frac{1}{R} \][/tex]
Because [tex]\( Z_{\text{eq}} \)[/tex] is purely real, the imaginary part must be zero:
[tex]\[ \frac{-(6 - X)}{100 - 12X + X^2} = 0 \][/tex]
[tex]\[ -(6 - X) = 0 \][/tex]
[tex]\[ X = 6 \][/tex]
Now substituting [tex]\( X = 6 \)[/tex] into the real part:
[tex]\[ \frac{1}{5} = \frac{8}{100 - 12(6) + 6^2} + \frac{1}{R} \][/tex]
[tex]\[ \frac{1}{5} = \frac{8}{100 - 72 + 36} + \frac{1}{R} \][/tex]
[tex]\[ \frac{1}{5} = \frac{8}{64} + \frac{1}{R} \][/tex]
[tex]\[ \frac{1}{5} = \frac{1}{8} + \frac{1}{R} \][/tex]
Solving for [tex]\( R \)[/tex]:
[tex]\[ \frac{1}{R} = \frac{1}{5} - \frac{1}{8} \][/tex]
[tex]\[ \frac{1}{R} = \frac{8 - 5}{40} \][/tex]
[tex]\[ \frac{1}{R} = \frac{3}{40} \][/tex]
[tex]\[ R = \frac{40}{3} = 13.33 \, \Omega \][/tex]
Thus, the values are:
- The resistor [tex]\( R = 13.33 \, \Omega \)[/tex]
- The capacitive reactance [tex]\( X = 6 \, \Omega \)[/tex]
