Question

An ac generator with emf amplitude ${\mathit{\epsilon }}_{\mathbf{m}}\mathbf{=}\mathbf{220}\mathbf{}\mathbf{\text{V}}$ and operating at frequency ${\mathit{f}}_{\mathbf{d}}\mathbf{=}\mathbf{400}\mathbf{}\mathbf{\text{Hz}}$causes oscillations in a series$\mathit{R}\mathit{L}\mathit{C}$ circuit having $\mathit{R}\mathbf{}\mathbf{=}\mathbf{}\mathbf{220}\mathbf{}\mathbf{\text{Ω}}$,$\mathit{L}\mathbf{=}\mathbf{150}\mathbf{}\mathbf{\text{mH}}$ , and$\mathit{C}\mathbf{=24.0}\mathbf{\text{μF}}$ . Find (a) the capacitive reactance $\mathbf{}{\mathit{X}}_{\mathbf{C}}$, (b) the impedance Z, and (c) the current amplitude I. A second capacitor of the same capacitance is then connected in series with the other components. Determine whether the values of (d)${\mathit{X}}_{\mathbf{C}}$ , (e) Z, and (f) Iincrease, decrease, or remain the same.

Short answer

1. The capacitive reactance is $16.6\text{Ω}$.
2. Impedance of series RLC circuit is $422\text{Ω}$.
3. Current amplitude is$0.52\text{A}$.
4. Thecapacitive reactance increases.
5. The impedance decreases.
6. The current increases.

Step-by-step solution

The given data

1. Amplitude of emf, ${\epsilon }_m=220\text{V}$
2. Driving frequency,$f_d=400\text{Hz}$
3. Resistance,$R=220\text{Ω}$
4. Inductance,$L=150\text{mH}$
5. Capacitance, $C=24.0\text{μF}$

Understanding the concept of oscillations of LC circuit

In the given circuit, the resistor, capacitor, and inductor are in series. To find the impedance in the circuit, we have to find out the capacitive and inductive reactance. Then by using impedance, we can find the amplitude of current and phase angle. Then a second capacitor is connected in series with the other components. We can find the equivalence capacitance for the modified circuit and corresponding capacitive reactance. We can also find the impedance and the current of the modified circuit.

The impedance of theLCRcircuit for the driving frequency$\left({\omega }_d\right)$,

$\mathbf{Z}\mathbf{=}\sqrt{{\mathbf{R}}^{\mathbf{2}}\mathbf{+}{\left({\mathbf{X}}_{\mathbf{L}}\mathbf{-}{\mathbf{X}}_{\mathbf{c}}\right)}^{\mathbf{2}}}$ ...(1)

The capacitive reactance of the capacitor,

${\mathit{X}}_{\mathit{C}}\mathbf{=}\frac{\mathbf{1}}{{\mathit{\omega }}_{\mathit{d}}\mathit{C}}$ ...(2)

The inductive reactance of the inductor,

${\mathit{X}}_{\mathit{L}}\mathbf{=}{\mathit{\omega }}_{\mathit{d}}\mathit{L}$ …(3)

The angular frequency of the LC oscillation,
${\mathit{\omega }}_{\mathit{d}}\mathbf{=}\mathbf{2}\mathit{\pi }{\mathit{f}}_{\mathit{d}}$ …(4)

The current equation using Ohm’s law,

$\mathit{I}\mathbf{=}\frac{{\mathit{\epsilon }}_{\mathit{m}}}{\mathit{Z}}$ …(5)

The equivalent capacitance of a series combination,

$\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{eq}}}\mathbf{=}\sum _{\mathbf{i}}^{\mathbf{n}}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{i}}}$ …(6)

Here, R is the resistance of the resistor, Cis the capacitance of the capacitor, L is the inductance of the inductor and${\mathit{f}}_{\mathit{d}}$is the frequency of the LC oscillation.

a) Calculation of the capacitive reactance

For the given value of frequency, equation (5) gives-

$$\begin{gathered} {\omega }_d=2\pi \times 400\text{Hz} \\ =2513.28\text{rad/s} \end{gathered}$$

Capacitive reactance is given using equation (1) as:

$$\begin{gathered} X_c=\frac{1}{2513.28\text{rad/s}\times 24.0\times {10}^{-6}\text{F}} \\ =16.57\text{Ω} \\ \approx 16.6\text{Ω} \end{gathered}$$

Hence, the capacitive reactance is role="math" localid="1662813034707" $16.6\text{Ω}$.

b) Calculation of the impedance of the circuit

Inductive reactance is given using equation (2) as:

$$\begin{gathered} X_L=2513.28\text{rad/s}\times 150\times {10}^{-3}\text{H} \\ =376.99\text{Ω} \\ \approx 377\text{Ω} \end{gathered}$$

Now, the impedance of the RLC circuit can be given using equation (3) as:

$$\begin{gathered} Z=\sqrt{{\left(220\text{Ω}\right)}^2+{\left\{377\text{Ω}-16.6\text{Ω}\right\}}^2} \\ =422.24\text{Ω} \\ \approx 422\text{Ω} \end{gathered}$$

Hence, the impedance is $422\text{Ω}$.

c) Calculation of the current amplitude

Amplitude of current is given using the data in equation (4) as follows:

$$\begin{gathered} I=\frac{220V}{422\text{Ω}} \\ =0.521\text{A} \end{gathered}$$

Hence, the current amplitude is $0.521\text{A}$.

d) Calculation of the modified capacitance reactance

Now, the second capacitor ofthesame capacitance is connected in series.

Then, the equivalent capacitance is given using equation (6) as follows:

$$\begin{gathered} \frac{1}{C_{eq}}=\frac{1}{C}+\frac{1}{C} \\ {C}_{eq}=\frac{C}{2} \\ =\frac{24.0\times {10}^{-6}\text{F}}{2} \\ =12.0\times {10}^{-6}\text{F} \end{gathered}$$

Thus, the modified capacitive reactance is given using equation (1) as follows:

$$\begin{gathered} X_c=\frac{1}{2513.28\text{rad/s}\times 12.0\times {10}^{-6}\text{F}} \\ =33.2\text{Ω} \end{gathered}$$

Thus, the modified capacitive reactance increases

e) Calculation of the modified impedance

Modified impedance is given using the above values in equation (iii) as follows:

$$\begin{gathered} Z=\sqrt{{\left(220\text{Ω}\right)}^2+{\left\{377\text{Ω}-33.2\text{Ω}\right\}}^2} \\ =408.16\text{Ω} \end{gathered}$$

Thus, the modified impedance decreases.

f) Calculation of the modified current amplitude

Modified amplitude of currentis given using the above values in equation (iv) as follows:

$$\begin{gathered} I=\frac{220\text{V}}{408.16\text{Ω}} \\ =0.539\text{A} \end{gathered}$$

Therefore, the modified current increases.