Question 1.1: RLC Circuit Calculations

Question

ScanSolve AI Tutor · Accepted Answer

Here are the solutions to your questions.

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 $(V_L - V_C)$ .

Question 1.1: RLC Circuit Calculations

Question 1.1: RLC Circuit Calculations

Still stuck on this one?

More Physics Questions