Question 1.1: RLC Circuit Calculations
This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Question 1.1: RLC Circuit Calculations
Given:
Resistance, $R = 13 \, \Omega$
Inductive Reactance, $X_L = 47.13 \, \Omega$ (given in $\Omega$, so it's reactance)
Capacitance, $C = 100 \, \mu F = 100 \times 10^{-6} \, F$
Supply Voltage, $V_{supply} = 100 \, V$
Frequency, $f = 50 \, Hz$
Step 1: Calculate Capacitive Reactance ($X_C$)
$$X_C = \frac{1}{2 \pi f C}$$
$$X_C = \frac{1}{2 \pi (50 \, \text{Hz}) (100 \times 10^{-6} \, \text{F})}$$
$$X_C = \frac{1}{0.0314159}$$
$$X_C \approx 31.83 \, \Omega$$
Step 2: Calculate Total Impedance ($Z$)
$$Z = \sqrt{R^2 + (X_L - X_C)^2}$$
$$Z = \sqrt{(13 \, \Omega)^2 + (47.13 \, \Omega - 31.83 \, \Omega)^2}$$
$$Z = \sqrt{169 \, \Omega^2 + (15.3 \, \Omega)^2}$$
$$Z = \sqrt{169 \, \Omega^2 + 234.09 \, \Omega^2}$$
$$Z = \sqrt{403.09 \, \Omega^2}$$
$$Z \approx 20.08 \, \Omega$$
1.1.1 Draw the circuit diagram.
A series RLC circuit consists of a resistor, an inductor, and a capacitor connected in series to an AC voltage source.
`
R L C
---[ R ]---[ L ]---[ C ]---
| |
+---( ~ )-----------+
AC Source
`
1.1.2 Calculate the total current flowing in the circuit.
$$I = \frac{V_{supply}}{Z}$$
$$I = \frac{100 \, \text{V}}{20.08 \, \Omega}$$
$$I \approx 4.98 \, \text{A}$$
The total current flowing in the circuit is $\boxed{\text{4.98 A}}$.
1.1.3 Calculate the voltage drop across the resistor.
$$V_R = I \cdot R$$
$$V_R = (4.98 \, \text{A}) \cdot (13 \, \Omega)$$
$$V_R \approx 64.74 \, \text{V}$$
The voltage drop across the resistor is $\boxed{\text{64.74 V}}$.
1.1.4 Calculate the voltage drop across the inductor.
$$V_L = I \cdot X_L$$
$$V_L = (4.98 \, \text{A}) \cdot (47.13 \, \Omega)$$
$$V_L \approx 234.71 \, \text{V}$$
The voltage drop across the inductor is $\boxed{\text{234.71 V}}$.
1.1.5 Calculate the voltage drop across the capacitor.
$$V_C = I \cdot X_C$$
$$V_C = (4.98 \, \text{A}) \cdot (31.83 \, \Omega)$$
$$V_C \approx 158.52 \, \text{V}$$
The voltage drop across the capacitor is $\boxed{\text{158.52 V}}$.
1.1.6 Calculate the phase angle.
$$\phi = \arctan\left(\frac{X_L - X_C}{R}\right)$$
$$\phi = \arctan\left(\frac{47.13 \, \Omega - 31.83 \, \Omega}{13 \, \Omega}\right)$$
$$\phi = \arctan\left(\frac{15.3 \, \Omega}{13 \, \Omega}\right)$$
$$\phi = \arctan(1.1769)$$
$$\phi \approx 49.64^\circ$$
The phase angle is $\boxed{\text{49.64}^\circ}$.
1.1.7 Draw the phase diagram of the above circuit.
The phase diagram (phasor diagram) shows the voltage phasors relative to the current. Since $X_L > X_C$, the circuit is inductive, and the total voltage leads the current.
$V_R$ is in phase with the current $I$.
$V_L$ leads the current $I$ by $90^\circ$.
$V_C$ lags the current $I$ by $90^\circ$.
$V_{supply}$ is the vector sum of $V_R$ and $(V_L - V_C)$.
`
^ V_L
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