Question

The figure below shows 3 charged capacitors connected to a long solenoid with the switch initially open. The equivalent capacitance of the three capacitors is Ceq = 8 mF where G = 18.0 mF, C3 = 5.0 mF and C is unknown. The inductance of the solenoid is L = 32.0 pH. Initially, the potential of point a is 14.0 V greater than at point b. cs llllllll b Part A Determine the capacitance C2 μΑ Value Units

Answer 1

Part A

To calculate the capacitance for we could start with the expression for capcitance for capacitors in series. In this case we would have:

C:

+ Ceq C1 C2+3

where $C_{2+3}$ is the equivalent capacitance of the parallel capacitors

Since this capacitors are in Parallel the equivalent capacitance is:

C2+3 C2 + C3

Applying this into the firs equation we get :

$\frac{1}{C_{eq}}=\frac{1}{C_{1}}+\frac{1}{C_{2}+C_{3}}$

and solving for $C_{2}$ we get:

$\frac{1}{C_{2}+C_{3}}=\frac{1}{C_{1}}-\frac{1}{C_{eq}} ; \\ \frac{1}{C_{2}+C_{3}}=\frac{C_{1}-C_{eq}}{C_{eq}C_{1}} ; \\ C_{2}+C_{3}=\frac{C_{eq}C_{1}}{C_{1}-C_{eq}} ; \\ C_{2}=\frac{C_{eq}C_{1}}{C_{1}-C_{eq}}-C_{3} ; \\$

Now using all the known values we get:

$C_{2}=\frac{C_{eq}C_{1}}{C_{1}-C_{eq}}-C_{3} =\frac{8*18}{18-8}-5=9.4mF$

Part B

To calculate the voltage in capacitor 3 we need first to find the total charge in the system and then the voltage in capacitor 1.

The total charge in the system is :

$Q=V_{T}*C_{eq}=14*8=112mC$

Since we have a system where capacitor 1 is in series with the equivalent capacitance of $C_{2+3}$ then the charge is the same in both parts. Now lets calculate the Voltage in capacitor 1.

$V_{1}=\frac{Q_{1}}{C_{1}}=\frac{112}{18}=6.22V$

For capacitors in series the toal voltage is just the addition of the voltages.

$V_{T}=V_{1}+V_{2+3}$

We need to find what is the voltage in the system of capacitors in parallel $C_{2+3}$ to find the final voltage in capacitor 3.

$V_{2+3}=V_{T}-V_{1}=14-6.22=7.78V$

Now since C3 and C2 are in parallel we know that its voltage is the same.

$V_{2+3}=V_{2}=V_{3}=7.78V$

Part C

To know the maximum we need first to calculate the Period of the circuit

7

$\tau = 2\pi\sqrt {LC_{eq}}=2\pi\sqrt {8*10^{-3*}32*10^{-6}}=0.0031s=3.1ms$

The first maximum value of the current is when the capacitor is completely discharged, this is $t=\frac {\tau}{4}=\frac{3.1}{4}=0.77ms$

Part D

To find the number of turns in the selenoid we can use the equation:

$L=\mu_{0}\frac{N^{2}A}{l}$

Here $\mu_{0}=4\pi*10^{-7}$ is the permeability of free space, N is the number of turns, L is the inductance, A is the core area of the inductor and l is the lenght of the coil in meters. Then solving for N in the equatio above we have:

$N=\sqrt{\frac{L*l}{\mu _{0}*A}}=\sqrt{\frac{L*l}{4*\pi*10^{-7}*\pi*r^{2}}}=\frac{1}{2\pi}\sqrt{\frac{L*l*10^{7}}{r^{2}}} \\ = \frac{1}{2\pi}\sqrt{\frac{300*10^{-7}*.4*10^{7}}{(.07)^{2}}}=\frac{1}{2\pi}\sqrt{\frac{120}{.0049}}=\frac{156.5}{2\pi}\approx 25turns$

Part E

The answer is the last one, the same magnetic field B, since the maginitc field inside the selenoid is constant. The magnetic field inside a solenoid is proportional to both the applied current and the number of turns. The field strength doesn't depend on the position inside the solenoid.

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