The electrical conductivity is the reciprocal of electrical resistivity; therefore, an increase in electrical resistivity will lead to a decrease in electrical conductivity, and vice versa.

Recall the definition of electrical resistivity:

Here,  is resistance,  is cross-sectional area, and  is the length of the rod. This equation reveals that an increase in resistance and area will increase resistivity, whereas an increase in length will decrease resistivity. Since conductivity is the reciprocal of resistivity, increasing resistance and area will decrease conductivity, whereas increasing the length will increase conductivity.

The question is asking about the rates of charge flow. Recall that the current is defined as the amount of charge flowing through a point in a given time; therefore, we are looking for the amount of current flowing through the two rods. Current flow is higher for a material with higher electrical conductivity. This means that we need to find the rod with the higher electrical conductivity.

Since there are no easy equations to find conductivity, we need to find the resistivity first. Electrical conductivity is the reciprocal of electrical resistivity (measure of the ability of a material to resist current flow); therefore, an increase in resistivity leads to a decrease in conductivity, and vice versa. Resistivity is defined as:

Here,  is the resistance,  is the cross-sectional area, and  is the length of the rod. The question states that the two rods have the same geometrical properties; therefore, the area and the length of the rods are the same. However, the resistance of rod A is higher. This means that the resistivity is higher and conductivity, consequently, is lower for rod A. Since it has a lower conductivity, rod A has a lower charge flow rate. 

Batteries are a series of electrochemical cells that use the electrical energy associated with the movement of electrons to perform other processes. The electrical energy is not itself usable; instead, batteries convert the electrical energy into other forms of energy, such as chemical energy, that can be used to drive a process. The electrical energy is derived from the movement of electrons associated with redox reactions. 

Batteries contain one or more electrochemical cells. Each electrochemical cell houses a cathode and an anode, which facilitate a redox reaction. The cathode is the site of the reduction half-reaction, whereas the anode is the site of the oxidation half-reaction. Reduction is the process by which reactants gain electrons and become more negative; oxidation is the process by which reactants lose electrons and become more positive. As such, electrons are products at an anode and reactants at a cathode. The redox reaction proceeds in the spontaneous reaction direction when the battery is being used, but proceeds in the nonspontaneous reaction direction when the battery is being recharged after use. Recall that nonspontaneous reactions require energy. When you are charging a battery, you are supplying energy in the form of voltage to drive the nonspontaneous reaction. On the other hand, when you are discharging (or using) the battery, the spontaneous redox reaction is supplying energy to drive another process.

An electrochemical cell in a battery contains two electrodes: the anode and the cathode. Anodes are the site of oxidation half-reactions and cathodes are the site of reduction half-reactions. Recall that oxidation involves loss of electrons from reactants. The reactants at the anode become more positive, and there is an increase in their oxidation number. Reduction, on the other hand, involves a gain of electrons and a decrease in the reactant oxidation number at the cathode. 

Electric power is expressed in a number of ways. We will need to use the relationship that involves current and resistance:

We are given the current and resistance, allowing us to calculate the power.

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

The formula we can use here is the power formula that involved both resistance and voltage:

We are given the voltage and power, allowing us to solve for the resistance.

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.