Question

The circuit shown has a source voltage of \(440 \mathrm{V}\) ms, resistance \(R=250 \Omega\), inductance \(L=0.800 \mathrm{H},\) and capacitance $C=2.22 \mu \mathrm{F} .\( (a) Find the angular frequency \)\omega_{0}$ for resonance in this circuit. (b) Draw a phasor diagram for the circuit at resonance. (c) Find these rms voltages measured between various points in the circuit: \(V_{a b}, V_{b c}, V_{c d}, V_{b d},\) and \(V_{c d},\) (d) The resistor is replaced with one of \(R=125 \Omega\). Now what is the angular frequency for resonance? (e) What is the rms current in the circuit operated at resonance with the new resistor?

Short answer

Answer: The new rms current at resonance is $$I' = \frac{440V}{125\Omega}$$.

Step-by-step solution

Find the angular frequency for resonance

At resonance, the inductive reactance equals the capacitive reactance: \(X_L = X_C\). We know that the inductive and capacitive reactance are given by: \(X_L = \omega_0 L\) and \(X_C = \frac{1}{\omega_0 C}\), where \(\omega_0\) is the angular frequency. Therefore, \(\omega_0 L = \frac{1}{\omega_0 C}\). Rearranging for \(\omega_0\), we get: $$\omega_0 = \sqrt{\frac{1}{LC}}$$ Substitute the given values to find the angular frequency for resonance: $$\omega_0 = \sqrt{\frac{1}{0.800*2.22*10^{-6}}}$$

Calculate the impedance at resonance

At resonance, the total impedance \(Z\) is given by the resistance, since \(X_L = X_C\), and therefore, the impedance of each component cancels out. So, \(Z = R = 250\Omega\).

Calculate the rms current at resonance

Using Ohm's law, the rms current at resonance is given by: $$I = \frac{V}{Z} = \frac{440V}{250\Omega}$$

Calculate the rms voltages between various points in the circuit

a) \(V_{ab} = V_R = IR = 250\Omega * I\) b) \(V_{bc} = V_L = I X_L = I\omega_0 L\) c) \(V_{cd} = V_C = I X_C = \frac{I}{\omega_0 C}\) d) \(V_{bd} = V_{bc} + V_{cd}\) e) \(V_{ac} = V_{ab} + V_{bc}\) Calculate the given rms voltages using the previously found values for \(I\) and \(\omega_0\).

Replace the resistor with a new value and find the new angular frequency for resonance

The new resistance value is \(R' = 125 \Omega\). Since the capacitance and inductance are unchanged, the resonance frequency remains the same. So, the new resonance frequency is equal to the previous angular frequency, \(\omega_0\).

Calculate the new rms current in the circuit at resonance

With the new resistance, the impedance at resonance is also equal to the new resistance \(R' = 125\Omega\). Using Ohm's law, calculate the new rms current at resonance: $$I' = \frac{V}{R'} = \frac{440V}{125\Omega}$$