Question

75. The capacitor in FIGURE P 30.75 is initially charged to $100\text{V}$, the $1200\mu \text{F}$capacitor is uncharged, and the switches are both open.

a. What is the maximum voltage to which you can charge the $1200\mu \text{F}$capacitor by the proper closing and opening of the two switches?

b. How would you do it? Describe the sequence in which you would close and open switches and the times at which you would do so. The first switch is closed at $t=0\text{s}$.

Short answer

Part (a) The maximum voltage has $1200\mu F$is 50 V

Step-by-step solution

Step 1. Given information

The capacitance of two different capacitor is

$\begin{array}{l}{C}_1=300\mu \text{F}\\ C_2=1200\mu \text{F}\end{array}$

The amount of energy stored in the capacitor is

$U_{\text{c}}=\frac{1}{2}C{V}^2$

Part (a)

The maximum energy stored in the first capacitor is equal to the maximum energy stored in the second capacitor.

$\begin{array}{l}{U}_1=U_2\\ \frac{1}{2}{C}_1{V}_1^2=\frac{1}{2}{C}_2{V}_2^2\\ V_2=V_1\times \sqrt{\frac{C_1}{C_2}}\\ V_2=100\times \sqrt{\frac{300\mu F}{1200\mu F}}\\ \Rightarrow V_2=50\text{V}\end{array}$

The maximum voltage $1200\mu F$has is $50\text{V}$

Part (b)

$C_2$Possible Sequence is:

1). Close $S_1$and hence tank circuit is formed between $C_1$and $L$leading to sinusoidal oscillations of energy moving between $C_1$and L.

2). Close $S_2$ open $S_1$at an instant where the potential on $C_1=0V$. Thus the energy present in $C_1$becomes zero and consequently transferred to the and stored in it as magnetic energy. Now a new current direction for the inductor current is through$C_2$ , making a new tank circuit which contains all the initial energy of $C_1$

3). Open $S_2$when current drops to zero in new tank circuit, parting $C_2$with all the energy and hence the maximum charge is achieved.

Conclusion:

The possible steps for switching strategies to keep maximum voltage of 1200 micro are Close , close open and Open .