Relevant equations: 

$\displaystyle V = IR + L\frac{dI}{dt}$

When the circuit is first connected, the current increases from zero to its final value. During this time as the current changes, the inductor has a voltage across it. After a long period, the current has built up to its maximum value.

After a long period, $\displaystyle \frac{dI}{dt}=0$.

$\displaystyle V = IR + L\frac{dI}{dt}\rightarrow V=IR$

Plugging in our values for voltage and resistance, we can solve for the final current.

$\displaystyle I=\frac{V}{R}=\frac{12V}{4000\Omega}=0.003A=3mA$

Relevant equations:

$\displaystyle \omega _o = \frac{1}{\sqrt{LC}}$

The current is maximized when the power supply frequency, $\displaystyle \omega$, equals the resonance frequency, $\displaystyle \omega _o$. Plugging in the given inductance and capacitance yields:

$\displaystyle \omega _o = \frac{1}{\sqrt{(40*10^{-3}H)(250*10^{-9}C)}}=\frac{1}{\sqrt{1*10^{-8}}}=\frac{1}{1*10^{-4}}= 10,000 \frac{rad}{s}$

Relevant equations:

$\displaystyle I_{max}=\frac{V_{max}}{Z}$

$\displaystyle Z = \sqrt{R^2+(X_L-X_C)}$

$\displaystyle X_L = X_C$ at resonance

$\displaystyle I_{rms}=\frac{I_{max}}{\sqrt2}$

Step 1: Find impedance, $\displaystyle Z$, at resonance:

$\displaystyle Z = \sqrt{R^2+0}=\sqrt{400}=20\Omega$

Step 2: Calculate $\displaystyle I_{max}$, using $\displaystyle V_{max}=200V$ and $\displaystyle Z = 20\Omega$:

$\displaystyle I_{max}=\frac{200V}{20\Omega}=10A$

Step 3: Use $\displaystyle I_{max}=10A$ to calculate $\displaystyle I_{rms}$:

$\displaystyle I_{rms}=\frac{10A}{\sqrt2}\approx7A$

Response 3 is the correct choice.  Electrons will space themselves as far apart as possible, because of charge repulsion; therefore, in a parallel arrangement, they will jump onto each capacitor and “load it up.” The parallel capacitance is calculated by simply adding the individual values. The value would be about 1.75 μF if the three were connected in series, where the formula is $\displaystyle \dpi{100} \small \frac{1}{C_{total}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}$ (reciprocal of total equals sum of reciprocal of each). The time constant of a circuit is given by the formula τ = RC, and it relates to how fast a capacitor can charge to full capacitance.

First find the equivalent capacitance of the two capacitors by adding their inverses.

$\displaystyle \frac{1}{C_{eq}}=\frac{1}{2}+\frac{1}{6}$ $\displaystyle \rightarrow C_{eq} = \frac{6}{4} = 1.5\mu F$

Then, we can find the charge stored on this equivalent capacitor.

$\displaystyle Q_{eq}=C_{eq}V = 1.5 * 12 = 18\mu C$

For capacitors in series the charges on each must be equal, and also equal to the charge on the equivalent capacitor. The answer is $\displaystyle 18\mu C$.

Charge is evenly distributed on the surface of the capacitor. If we think back to electric force, we know that positive charges repel other positive charges and negative charges repel other negative charges; thus, the charges are evenly distributed to minimize the force between them. We can see how this looks in diagrammatic form below.

First, we need to determine how these capacitors are being added. We can see that they are being added in in series. Remember that capacitors in series are added as reciprocals:

 $\displaystyle \frac{1}{C_{eq}}=\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{1\mu F}+\frac{1}{1\mu F}=2\mu F$

C_eq = 0.5μF

First, we need to determine how capacitors C₂ and C₃ are being added. We can see that they are being added in in series. Remember that capacitors in series are added as reciprocals.

$\displaystyle \frac{1}{C_{23}}=\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{1\mu F}+\frac{1}{1\mu F}=2\mu F$

C₂₃ = 0.5μF

Next, we need to determine how we can find the C_eq by simplifying C₂₃ and C₁. We can see that C_eq and C₁ are in parallel, thus we can directly add the individual capacitances.

C_eq = C₂₃ + C₁ = 0.5μF + 0.5μF = 1μF

In order to determine the time, we need to know the total charge stored on the capacitors. Remember that Q = CV, where Q is the total charge, C is the equivalent capacitance, and V is the voltage. We must first find the equivalent capacitance.

C₂ and C₃ are capacitors in series, while C₁ is in parallel.

$\displaystyle \frac{1}{C_{23}}=\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{1\mu F}+\frac{1}{1\mu F}=2\mu F$

C₂₃ = 0.5μF

C_eq = C₂₃ + C₁ = 0.5μF + 0.5μF = 1μF

Now we can plug in the C_eq and battery voltage to find the charge.

Q = (1μF)(10V) = 10μC

Additionally, we need to know the current the battery can provide (the charge per unit time). Knowing both the total charge and current will allow us to calculate the time. We can use V = IR to determine the current.

I = V/R = 10V/1Ω = 10A = 10C/sec

We can equate charge and current to determine time.

10μC = 10 C/t

t = 10 C/10 μC = 1 * 10⁶s or 11.6days

In this question, we are asked how the total charge stored on the surface of the capacitors would change if we inserted a dielectric between the parallel plates. As we can see in the equation for capacitance based on physical properties, $\displaystyle C=\frac{k\epsilon_0A}{d}$_, where k is the dielectric constant (k = 1 for air; k > 1 for all dielectric materials), ε₀ is the constant of the permeability of free space, A is the area of the plates, and d is the distance between the plates.

If we insert a dielectric material, k > 1, the value of C increases, thus the overall capacitance of the circuit increases.