Questions: XC = ΩΩ XL = -Ω Z = Ω

Transcript text: ``` X_{C}=\Omega \Omega \\ X_{L}=-\Omega \\ Z=\Omega ```

Solution

Solution Steps

Here's the solution for the first three circuits:

Step 1: Calculate Reactances for Circuit 1

Given \(R_1 = 100 \Omega\), \(L_1 = 5\)mH, \(C_1 = 10\mu\)F, and \(f = 1\)kHz. \(X_C = \frac{1}{2\pi fC} = \frac{1}{2\pi(1000)(10 \times 10^{-6})} \approx 15.92\Omega\) \(X_L = 2\pi fL = 2\pi(1000)(5 \times 10^{-3}) \approx 31.42\Omega\)

Step 2: Calculate Impedance for Circuit 1

Since the inductor and capacitor are in series, the total reactance is \(X = X_L - X_C \approx 31.42 - 15.92 = 15.5\Omega\). Impedance \(Z = \sqrt{R^2 + X^2} = \sqrt{100^2 + 15.5^2} \approx \sqrt{10000 + 240.25} \approx \sqrt{10240.25} \approx 101.2\Omega\).

Step 3: Calculate Reactances for Circuit 2

Given \(R_1 = 100 \Omega\), \(L_1 = 20\)mH, \(C_1 = 500\mu\)F, and \(f = 1\)kHz. \(X_C = \frac{1}{2\pi fC} = \frac{1}{2\pi(1000)(500 \times 10^{-6})} \approx 0.318\Omega\) \(X_L = 2\pi fL = 2\pi(1000)(20 \times 10^{-3}) \approx 125.66\Omega\)

Step 4: Calculate Impedance for Circuit 2

Since the inductor and capacitor are in parallel, their combined reactance is given by: \(X_{parallel} = \frac{X_L X_C}{X_L - X_C} \approx \frac{(125.66)(0.318)}{125.66 - 0.318} \approx 0.32\Omega\) Then the inductor-capacitor parallel combination is in series with the resistor. The magnitude of the total impedance is \(Z \approx \sqrt{100^2 + 0.32^2} \approx \sqrt{10000 + 0.1024} \approx \sqrt{10000.1024} \approx 100 \Omega\). For practical purpose, we can approximate the total impedance to \(100 \Omega\). Since the combined reactance is dominated by the capacitor, the total impedance is \(Z \approx \sqrt{100^2 + 0.318^2} \approx 100\Omega\).

Step 5: Calculate Reactances for Circuit 3

Given \(R_1 = 100 \Omega\), \(L_1 = 85\)mH, \(C_1 = 25\mu\)F, and \(f = 1\)kHz. \(X_C = \frac{1}{2\pi fC} = \frac{1}{2\pi(1000)(25 \times 10^{-6})} \approx 6.37\Omega\) \(X_L = 2\pi fL = 2\pi(1000)(85 \times 10^{-3}) \approx 534.07\Omega\)

Step 6: Calculate Impedance for Circuit 3

Impedance \(Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{100^2 + (534.07 - 6.37)^2} \approx \sqrt{10000 + 527.7^2} \approx \sqrt{10000 + 278487.29} \approx \sqrt{288487.29} \approx 537.11\Omega\)

Final Answer

Circuit 1: \(X_C \approx \boxed{15.92 \Omega}\), \(X_L \approx \boxed{31.42 \Omega}\), \(Z \approx \boxed{101.2 \Omega}\) Circuit 2: \(X_C \approx \boxed{0.32 \Omega}\), \(X_L \approx \boxed{125.66 \Omega}\), \(Z \approx \boxed{100 \Omega}\) Circuit 3: \(X_C \approx \boxed{6.37 \Omega}\), \(X_L \approx \boxed{534.07 \Omega}\), \(Z \approx \boxed{537.11 \Omega}\)

Was this solution helpful?

Unhelpful

Helpful

Questions asked by other users

< Previous Next >