Solved Examples of LCR Circuit
Here are two solved examples of a LCR circuit:
Q1. A resistor of 200 Ω and a capacitor of 15.0 μF are connected in series to a 220 V, 50 Hz ac source.
- Calculate the current in the circuit;
- Calculate the voltage (rms) across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage? If yes, resolve the paradox.
A1. Given,
R = 200 Ω,
\( {C} = {15.0 μF} = {15.0} × {10^{-6}} \)
V = 220 V,
ν = 50 Hz
In order to calculate the current, we need the impedance of \( {Z} = \sqrt {R^2+ X_C^2} = \sqrt {R^2 + (2πνC)^{-2}}
\( = 291.5Ω \)
Therefore, current in the circuit is
\( {I} = {V\cdot Z} \)
\( = {220 V\cdot 291.5Ω} \)
\( = {0.755 A} \)
Since the current is the same throughout the circuit, we have
\( {V_R} = {I}×{R} = (0.755 A)×(200Ω) = 151 V \)
\( {V_C} = {I}×{X_C} = (0.755 A)×(212.3Ω) = 160.3 V \)
The algebraic sum of the two voltages, \(V_R\) and \(V_C\) is 311.3 V which is more than the source voltage of 220 V. To resolve this paradox we’ve to keep in mind that the two voltages are not in the same phase. Therefore, they cannot be added like ordinary numbers. The two voltages are out of phase by ninety degrees. Therefore, the total of these voltages must be obtained using the Pythagorean theorem:
\( {V_R+C} = \sqrt{V_R^2 + V_C^2} \)
\( = 220V \)
Thus if the phase difference between two voltages is properly taken into account, the total voltage across the resistor and the capacitor is equal to the voltage of the source.
Q.2 A series LCR circuit with R = 20Ω, L = 1.5 H and C = 35μF is connected to a variable-frequency 200 V AC supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?
A2. Given,
R = 20Ω,
L = 1.5 H,
\( {C} = {35 μF} = {35} × {10^{-6}} \),
V = 200 V
Therefore, \( {Z} = \sqrt{R^2 + (ωL – 1 \cdot ωC)^{2}} \)
At resonance, \( {ωL} = {1 \cdot ωC} \)
Therefore, \( {Z} = {R} = {20Ω} \)
So, the current present in the circuit can be calculated as:
\( {I} = {V\cdot Z} \)
\( = {200\cdot20} \)
\( = 10 A \)
Therefore, the average power transferred to the circuit = V×I
= 200 × 10 = 2000 W.
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