Question
A
5.0
Ω
resistance is connected in series with a 0.2-H pure inductance and a 40.0-nF pure capacitance. The combination is connected across a 30-V, 1780-Hz power supply. Calculate the impedance of the combination.
Question 1Select one:
5.0
Ω
20.0
Ω
25.0
Ω
10.0
Ω
5.0
Ω
resistance is connected in series with a 0.2-H pure inductance and a 40.0-nF pure capacitance. The combination is connected across a 30-V, 1780-Hz power supply. Calculate the impedance of the combination.
Question 1Select one:
5.0
Ω
20.0
Ω
25.0
Ω
10.0
Ω
Answers
GPT-4o mini
To calculate the impedance of the combination of the resistor (R), inductor (L), and capacitor (C) connected in series, we will use the following formulas:
1. The inductive reactance (\(X_L\)) is given by:
\[
X_L = 2 \pi f L
\]
where \(f\) is the frequency in Hz and \(L\) is the inductance in Henries.
2. The capacitive reactance (\(X_C\)) is given by:
\[
X_C = \frac{1}{2 \pi f C}
\]
where \(C\) is the capacitance in Farads.
3. The total impedance (\(Z\)) in a series circuit containing resistance (R), inductive reactance (\(X_L\)), and capacitive reactance (\(X_C\)) is given by:
\[
Z = \sqrt{R^2 + (X_L - X_C)^2}
\]
Given:
- \(R = 5.0 \, \Omega\)
- \(L = 0.2 \, H\)
- \(C = 40.0 \, nF = 40.0 \times 10^{-9} \, F\)
- \(f = 1780 \, Hz\)
**Step 1: Calculate \(X_L\)**
\[
X_L = 2 \pi (1780) (0.2) \\
= 2 \pi \cdot 1780 \cdot 0.2 \\
= 2 \cdot 3.14159 \cdot 1780 \cdot 0.2 \\
\approx 223.64 \, \Omega
\]
**Step 2: Calculate \(X_C\)**
\[
X_C = \frac{1}{2 \pi (1780) (40.0 \times 10^{-9})} \\
= \frac{1}{2 \cdot 3.14159 \cdot 1780 \cdot 40.0 \times 10^{-9}} \\
\approx \frac{1}{0.0004481} \\
\approx 2232.62 \, \Omega
\]
**Step 3: Calculate the total impedance \(Z\)**
Using the formula for impedance:
\[
Z = \sqrt{R^2 + (X_L - X_C)^2}
\]
\[
X_L - X_C = 223.64 - 2232.62 \approx -2008.98 \, \Omega
\]
Now, calculate the impedance:
\[
Z = \sqrt{5.0^2 + (-2008.98)^2} \\
= \sqrt{25 + 4035964.6} \\
= \sqrt{4035989.6} \\
\approx 2008.98 \, \Omega
\]
None of the options given (5.0 Ω, 20.0 Ω, 25.0 Ω, 10.0 Ω) match, which likely indicates the need to check the option choices versus the calculated value.
### Answer
If your aim was deferring from guiding towards an impedance calculation, please let me know for clarity on specification or detail in dimensions.
Would you like help with something else?
1. The inductive reactance (\(X_L\)) is given by:
\[
X_L = 2 \pi f L
\]
where \(f\) is the frequency in Hz and \(L\) is the inductance in Henries.
2. The capacitive reactance (\(X_C\)) is given by:
\[
X_C = \frac{1}{2 \pi f C}
\]
where \(C\) is the capacitance in Farads.
3. The total impedance (\(Z\)) in a series circuit containing resistance (R), inductive reactance (\(X_L\)), and capacitive reactance (\(X_C\)) is given by:
\[
Z = \sqrt{R^2 + (X_L - X_C)^2}
\]
Given:
- \(R = 5.0 \, \Omega\)
- \(L = 0.2 \, H\)
- \(C = 40.0 \, nF = 40.0 \times 10^{-9} \, F\)
- \(f = 1780 \, Hz\)
**Step 1: Calculate \(X_L\)**
\[
X_L = 2 \pi (1780) (0.2) \\
= 2 \pi \cdot 1780 \cdot 0.2 \\
= 2 \cdot 3.14159 \cdot 1780 \cdot 0.2 \\
\approx 223.64 \, \Omega
\]
**Step 2: Calculate \(X_C\)**
\[
X_C = \frac{1}{2 \pi (1780) (40.0 \times 10^{-9})} \\
= \frac{1}{2 \cdot 3.14159 \cdot 1780 \cdot 40.0 \times 10^{-9}} \\
\approx \frac{1}{0.0004481} \\
\approx 2232.62 \, \Omega
\]
**Step 3: Calculate the total impedance \(Z\)**
Using the formula for impedance:
\[
Z = \sqrt{R^2 + (X_L - X_C)^2}
\]
\[
X_L - X_C = 223.64 - 2232.62 \approx -2008.98 \, \Omega
\]
Now, calculate the impedance:
\[
Z = \sqrt{5.0^2 + (-2008.98)^2} \\
= \sqrt{25 + 4035964.6} \\
= \sqrt{4035989.6} \\
\approx 2008.98 \, \Omega
\]
None of the options given (5.0 Ω, 20.0 Ω, 25.0 Ω, 10.0 Ω) match, which likely indicates the need to check the option choices versus the calculated value.
### Answer
If your aim was deferring from guiding towards an impedance calculation, please let me know for clarity on specification or detail in dimensions.
Would you like help with something else?