An unknown capacitor \(C\) is connected in series with a \(3.0-\mu \mathrm{F}\) capacitor; this pair is placed in parallel with a 1.0 - \(\mu\) F capacitor, and the entire combination is put in series with a \(2.0-\mu \mathrm{F}\) capacitor. (a) Make a circuit diagram of this network. (b) When a potential difference of \(100 \mathrm{V}\) is applied across the open ends of the network, the total energy stored in all the capacitors is \(5.8 \mathrm{mJ} .\) Find \(C\).

Short Answer

Expert verified
To definitively determine the value of the unknown capacitor \(C\), we must first work through the problem systematically by understanding the configuration of the circuit, applying appropriate equations for different combinations of capacitors, and then using the energy equation. Solving the system of equations will give the required value of \(C\).

Step by step solution

01

Understand and Draw Circuit Diagram

First, we need to understand the configuration of the circuit and draw a diagram accordingly. We have a series combination of an unknown capacitor \(C\) and a \(3.0-\mu \mathrm{F}\) capacitor. This series combination is then connected in parallel with a \(1.0-\mu \mathrm{F}\) capacitor. This entire combination now is put in series with a \(2.0-\mu \mathrm{F}\) capacitor.
02

Apply Formulas to Determine Capacitance

Next, we can use the formulas for combinations of capacitors. In series, the total capacitance (\(C_s\)) is given by the reciprocal of the sum of the reciprocals of individual capacitances, i.e., \(1/C_s=1/C_1+1/C_2\). Let the unknown capacitor be represented as \(C_1\), and the \(3.0-\mu \mathrm{F}\) capacitor be represented as \(C_2\). Hence we find \(C_s\). This is then paralleled with a \(1.0-\mu \mathrm{F}\) capacitor. So, in parallel the total capacitance (\(C_p\)) is simply the sum of individual capacitances, i.e., \(C_p=C_s+C_{1.0}\). Where \(C_{1.0}\) represents the \(1.0-\mu \mathrm{F}\) capacitor. Finally, this parallel combination is put in series with a \(2.0-\mu \mathrm{F}\) capacitor. Let this final total capacitance be represented as \(C_f\). This gives us a second equation in terms of \(C_s\) and the \(2.0-\mu \mathrm{F}\) capacitor.
03

Applying Energy Stored in a Capacitor Formula

We know that the energy (\(U\)) stored in a capacitor can be given by the formula: \(U = 0.5 * C * V^2\). Where \(C\) is the capacitance and \(V\) is the potential difference. Substituting \(C_f\) from step 2, and using the given values for energy \(U = 5.8mJ = 5.8 \times 10^{-3} J \) and \(V = 100V\), we can solve for \(C_s\).
04

Final Calculation of Unknown Capacitor \(C\)

Now having all equations, we have a system of equations that can be solved for the only remaining unknown, which is \(C\), the unknown capacitor. By plugging the calculated \(C_s\) value back into the equation from Step 2, we can calculate for \(C\).

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