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.

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, _, 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.

Dielectrics are put between two parallel plates of a capacitor to increase capacitance. This is done by holding voltage constant and increasing charge, or decreasing voltage and holding charge constant.

This can be understood using the equation . Either an increase in charge or a decrease in voltage will increase the capacitance.

First, we need to find the equivalent capacitance for the capacitors in series.

Now we can find the charge using the equation .

The power consumed by a capacitor is given by

After finding power, we can calculate the energy used as we are told that the capacitor took 5 seconds to charge. The units of power are watts which can be further broken down into joules per second. By multiplying the time taken to charge the capacitor by the power, we can find the total energy required to charge the capacitor.

The brightness of each bulb depends on its power, or how much energy it dissipates per unit time. This can be calculated using the equation:

If the bulbs are in parallel with each other, the bulb with lower resistance will receive more current, since current will tend toward the path of least resistance. Current is squared in the power equation, so it is of greater importance in determining the power. The bulb with greater current and lower resistance will thus have more power and will shine more brightly.

This question asks us about the power of the circuit, meaning the amount of energy per unit time. The power equation is P = IV. V = IR can be substituted in to allow P to be calculated from a number of parameters.

P = IV = I²R = V²/R

To solve for P, we first need the current supplied by the battery.

We can use the formula V = IR because we have the voltage drop across the circuit (9V) and can calculate the equivalent resistance.

By taking the inverse of the equation, we can see that R_A4 is equal to 2Ω.

R_eq = R_A4 + R₁ = 2Ω + 2Ω = 4Ω

Now, using V=IR, we can solve for the current.

V = IR

I = V/R = 9V/4Ω = 2.25A

Now, we can use the current we calculated and the voltage that is dissipated across the circuit to calculate the power.

P = IV = (2.25 A)(9V) = 20.25W

First, find the rms voltage across the resistor with the equation:

The pure voltage is , allowing us to solve for the rms voltage.

Now we can use Ohm's law to find the rms current.

Alternative current (AC) voltage is governed by the RMS (root-mean square) voltage law. This means that the RMS voltage of the circuit will be , but the actual voltage will fluctuate between two values. These values are determined by the equation:

Using the values from our question, we can find the maximum voltage value.

The highest energy of any visible light belongs to violet. The greater the wavelength, the lower the energy of the light. The greater the frequency, the higher the energy of the light. This is why ultraviolet light ("ultra" meaning "beyond" violet) is so damaging to DNA. Out of the answer choices, blue light has the lowest wavelength and greatest frequency, making it the highest energy.