Remember that convention dictates that current flows in the direction of positive charge (protons), thus, electrons flow in the opposite direction. Also remember that the larger length on the battery symbol on the circuit diagram indicates that current flows in this direction. In the diagram below, we can see that current flows clockwise, thus electrons flow counterclockwise.

Remember that convention dictates that current flows in the direction of positive charge (protons). Also remember that the larger length on the battery symbol on the circuit diagram indicates that current flows in this direction. In the diagram below, we can see that current flows counterclockwise.

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

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.

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.

We will need to calculate total equivalent resistance for each of the two scenarios to calculate power losses.

SCENARIO 1: Switch Closed

When the switch is closed, a resistance-free path is created. This effectively reduces the current flow through the parallel resistors to zero. Therefore, we have a much simpler circuit, consisting only of R1 and R5. Since they are in series, we can simply add them:

The expression for power is:

Substituting Ohm's law for current, we get:

Plugging in our values, we get:

SCENARIO 2: Switch Open

Now that we have resistors in parallel, it will take two steps to condense the circuit. For the parallel part, we get:

Now, since everything is in series in this equivalent circuit, we can simply add it all up:

Again using the formula for power, we get:

Now we can calculate the difference between the two scenarios:

Relevant equations: 

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, .

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

Relevant equations:

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

Relevant equations:

 at resonance

Step 1: Find impedance, , at resonance:

Step 2: Calculate , using  and :

Step 3: Use  to calculate :