First, we need to determine the voltage drop across R₄. Given that R_A and R₄ are in parallel, we know that the voltage drop across each is the same.

The voltage drop across R₁ can be calculated using the I_circuit (all the current generated by the battery’s potential difference must pass through R₁ because it is in direct series with the battery), and the resistance of the resistor.

I can be calculated by V = IR because we have the voltage drop across the circuit (9 V) 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 allows us to find the current in the circuit.

V = IR

I = V/R = 9V/4Ω = 2.25 A

Plugging this value in, we can find the voltage drop across R₁.

V_R1 = IR = (2.25 A)(2Ω) = 4.5 V

Now we can determine the voltage drop across the parallel resistors by subtracting the voltage drop across R₁ from the battery.

V_A = 9V – 4.5V = 4.5V

this value will be the same for either R_A or R₄. We can now use V = IR to determine the amount of current that flows through R₄.

V = IR

I = V/R = 4.5V/3Ω = 1.5A

Additionally, due to Kirchhoff’s first law, we know that current in must equal current out in a junction. Thus, because R₄ has 1.5 A of current, R_A must have 0.75A. Thus makes logical sense because electrons take the path of least resistance, meaning the resistor with the lowest resistance will have the greater current.

Because we are being asked the current that flows through the circuit, 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 find the current.

V = IR

I = V/R = 9V/4Ω = 2.25A

For this question we need to know that the voltage drop is equal through all paths of a parallel circuit; therefore we can realize that the voltage drop through both resistors are independent of each other. The current through resistor A will decrease with the removal of resistor B. Electrons want to travel through the path of least resistance.

Given that resistor R = 2 Ω, we can see that the parallel aspect of the circuit has a resistance of 2 Ω as a whole, which means the total effective resistance of the circuit is 4 Ω. The potential difference drops 2V over the first resistor, then 2 V through the parallel aspect of the circuit. Potential difference, or voltage, drops through an equal amount in all branches of resistors in parallel. Therefore the remaining 2 V of potential difference is dropped by electrons moving through both branches of the parallel resistor, and the voltage drop through the 4 Ω resistor is 2V.

Given that

V(q) = U

When V is voltage, U is electrical potential energy and q is charge, we can solve by plugging in 4 for V and -2 for q. Also, we must understand that the electric potential energy of a particle decreases as it moves from an area of higher energy to one of lower energy. For an electron, it has higher energy in the negative terminal of the battery than it does in when getting to the positive terminal.

This question asks us directly about one of Kirchhoff’s Loop Laws—more specifically the second law that states that the sum of the voltage around any loop must be equal to zero.

ΣV_circuit = 0V

Because the battery provides 9V, the voltage drop across the circuit must be 9V.

As an aside, the other Kirchhoff Law, the first law, states that current into a junction must equal the current that exits the junction, due to conservation of charge.

As a common rule to all circuits, all resistors in series share the same current (due to conservation of charge) and all resistors in parallel share the same voltage. This information is very helpful in determining the voltage drops across sets of resistors and the current that flows through them.

First, we need to determine the voltage drop across R₁, and then we can subtract that from the 9V battery to determine the voltage drop across R_A. Given that R_A and R₄ are in parallel, we know that the voltage drop across each is the same.

The voltage drop across R₁ can be calculated using the I_circuit (all the current generated by the battery’s potential difference must pass through R₁ because it is in direct series with the battery), and the resistance of the resistor.

I can be calculated by 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 allows us to find the current in the circuit.

V = IR

I = V/R = 9V/4Ω = 2.25 A

Plugging this value in, we can find the voltage drop across R₁.

V_R1 = IR = (2.25 A)(2Ω) = 4.5 V

Now we can determine the voltage drop across the parallel resistors by subtracting the voltage drop across R₁ from the battery.

V_A = 9V – 4.5V = 4.5V

Because R_A and R₄ are in parallel, the voltage drop across R₄ is also 4.5V

Regardless of whether the  resistor is present, we can simply focus on the current loop containing the battery and  resistor.

For resistors in parallel, each component/branch will have the same voltage drop, but the current may vary. Initially, the voltage through the  resistor will be:

After the  resistor is removed, the voltage will remain unchanged.

Recall that , where . The shape, material characteristics, and resistance do not appear in this equation. So only the size (area) of the loop influences the emf.