Related questions with answers

(b). For low frequencies, the reactance of the capacitor is large so that very little current exists in the capacitor branch. The reactance of the inductor is small so that current exists in the inductor branch and the lightbulb glows. As the frequency increases, the inductive reactance increases and the capacitive reactance decreases. At high frequencies, more current exists in the capacitor branch than the inductor branch and the lightbulb glows more dimly

The total power delivered to a balanced threephase load when operating at a line voltage of $2400 \sqrt{3} \mathrm{~V}$ is $720 \mathrm{~kW}$ at a lagging power factor of $0.8$ . The impedance of the distribution line supplying the load is $0.8+j 6.4 \Omega / \phi$ . Under these operating conditions, the drop in the magnitude of the line voltage between the sending end and the load end of the line is excessive. To compensate, a bank of $\Delta$ -connected capacitors is placed in parallel with the load. The capacitor bank is designed to furnish $576 \mathrm{kVAR}$ of magnetizing reactive power when operated at a line voltage of $2400 \sqrt{3} \mathrm{~V}$ . a) What is the magnitude of the voltage at the sending end of the line when the load is operating at a line voltage of $2400 \sqrt{3} \mathrm{~V}$ and the capacitor bank is disconnected? b) Repeat (a) with the capacitor bank connected. c) What is the average power efficiency of the line in (a)? d) What is the average power efficiency in (b)? e) If the system is operating at a frequency of $60 \mathrm{~Hz}$ , what is the size of each capacitor in microfarads?

A $500 \; \mu\text{F}$ capacitor provides a load current of $200 \text{ mA}$ at $8\%$ ripple. Calculate the peak rectified voltage obtained from the $60-\text{Hz}$ supply and the dc voltage across the filter capacitor.

Question

In Fig. we saw earlier, $C_1=$ $6.00 \mu \mathrm{F}, \quad C_2=3.00 \mu \mathrm{F}$ . and $C_3=5.00 \mu \mathrm{F}$ . The capacitor network is connected to an applied potential $V_{a b}$ . After the charges on the capacitors have reached their final values, the charge on $C_2$ is $30.0 \mu \mathrm{C}$ . (b) What is the applied voltage $V_{a b}$ ?

Solution

Verified

Answered 11 months ago

Step 1

1 of 2

(b) Now we want to determine the applied voltage across $ab$ . It is given by equation 24.1 in the next form

V_{ab} = \dfrac{Q}{C_{\text{eq}}} \tag{2}

Where $Q$ represents the charge on the third capacitor $Q = 90.0 \mathrm{~\mu C}$ and $C_{\text{eq}}$ is the equivalent capacitance on the system (between $C_{3}$ and ${C}'_{\text{eq}}$ ) in series. In series the equivalent capacitance $C_{\text{eq}}$ could be calculated by equation 24.5 in the next form

\begin{gathered} \dfrac{1}{C_{\text{eq}}} = \dfrac{1}{{C}'_{\text{eq}}} + \dfrac{1}{C_{3}} \tag{3} \end{gathered}

Where ${C}'_{\text{eq}}$ represents the equivalent capacitance between $C_{1}$ and $C_{2}$ and as both capacitors are in parallel, hence ${C}'_{\text{eq}}$ is given by equation 24.7