Since the resistors  and  form a parallel network, removing  from the circuit increases the resistance of that part of the circuit. Because the new circuit is the series combination of  and , the increased resistance leads to lower current in each of these resistors.

To determine total charge stored, we need to add up the capacitance of each capacitor(because they are capacitors in parallel) and multiply by the voltage difference. Recall that for capacitors,

For parallel plate capacitors:

Here, , which is the permittivity of empty space,  is the dielectric constant, which is  since there is only vacuum present, , which is the area of the parallel plates, and , which is the distant between the plates. 

Plug in known values to solve for capacitance.

Each of the two capacitors has capacitance 

Since the capacitors are in parallel, the total capacitance is the sum of each individual capacitance. The total capacitance in the circuit  is given by:

Plug this value into our first equation and solve for the total charge stored.

, where  is the total charge stored in the capacitor. Since 

First we need to determine the overall resistance of the circuit before we know how much current is flowing through. Since the resistors are in parallel, their resistances will add reciprocally:

where  is the total resistance of the circuit. 

Now that we've solved for , we know that the current  flowing through the circuit can be found using Ohm's law:

The total resistance of a circuit in series can be described by the equation:

The series circuit in ths problem therefore has a resistance of:

The resistance of a circuit wired in parallel has a total resistance of:

We are assuming that the two circuits have the same total resistance, so to find the resistance of the unknown resistor, we set up the following equation:

This, when solved, gives us a resitance of 200 ohms for our unknown element.

The equivalent voltage of batteries connected in series is the sum of the voltage of each battery, or

In our problem,

The equivalent voltage of batteries connected in parallel is equal to the voltage of 1 battery.

In this problem, 

Note: Connecting batteries in parallel increases the capacity of the battery.

To find the equivalent resistance of this system, we must first find the equivalent resistance of the resistors in parallel, then evaluate the resistors in parallel.

The parallel resistor equivalence is given by the following equation,

In our problem,

The parallel resistors can now be treated as one resistor with the resistance .  To find the total resistance, we add the resistances of ,  and .

To find the equivalent resistance of this system, we must first find the equivalent resistance of the resistors in parallel, then evaluate the resistors in parallel.

The parallel resistor equivalence is given by the following equation,

In our problem,

The parallel resistors can now be treated as one resistor with the resistance .  To find the total resistance, we add the resistance of  and .

Let's go through and figure out what the equivalent resistances are.

(A) This is a resistor in parallel with two series resistors. This looks like:

(B) These are just two resistors in series,

(C) This is just one resistor,

(D) All three resistors are in parallel,

Ranking them we see that the largest to smallest values are 

The key hallmark of parallel circuits is that the elements connected in parallel have the same voltage drop across them. It doesn’t matter if the circuit element is a resistor, capacitor, or an inductor; the voltage drop across all elements is the same. This means that the voltage across both resistors A and B is same. On the other hand, circuit elements connected in series have the same current flowing through them; however, they have different voltage drops.