AP Physics 1 - Understanding Circuit Diagrams

Circuitdiagram3

$R_1=1\Omega$

$R_2=2\Omega$

$R_3=3\Omega$

$R_4=4\Omega$

In the circuit above, what is the current passing through ?

Explanation

To approach this problem, note that there are no other resistors (or combinations or resistors) beyond the parallel arrangement shown, so the voltage drop across the top and the bottom is the same and equal to the voltage across the circuit, .

Furthermore, the current that passes through must be the same as the current that passes through .

Therefore, the current that passes through them can be found by rearranging Ohm's law, solving for current.

$I_{1,3}=\frac{V}{R_1+R_3}=\frac{10V}{1\Omega+3\Omega}$

$I_{1,3}=2.5A$

Circuit diagram

$R_1=6\Omega$

In the circuit above, what is the resistance of ?

$12\Omega$

$2.4\Omega$

$-2\Omega$

$10\Omega$

$2\Omega$

Explanation

Find the total resistance of the circuit, which can be determined using Ohm's law.

$R_T=\frac{V}{I}$

$R_T=\frac{5}{1.25}\Omega=4\Omega$

Now, the resistance of the second resistor can be found. Since the two resistors are in parallel, they're related to the total resistance as follows:

$\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}$

Rearrange and solve for

$\frac{1}{R_2}=\frac{1}{R_T}-\frac{1}{R_1}$

$\frac{1}{R_2}=\frac{1}{4\Omega }-\frac{1}{6\Omega }=\frac{6-4}{24\Omega }=\frac{1}{12\Omega }$

$R_2=12\Omega$

The following symbol represents what item in a circuit?

Capacitor

Battery

Resistor

Inductor

Explanation

The symbol for a capacitor is written as a break in the circuit separated by two parallel lines of equal length as shown below. This loosely resembles the most common type of capacitor, a parallel plate capacitor.

Capacitor

Basic circuit2

In the circuit above:

$V_{in}=12V$

$Z_1=11\Omega$

$Z_2=21\Omega$

$Z_3=16\Omega$

What is the current across ?

Explanation

Realize that the voltage drop across the combined resistances of and must be equal to the voltage of the circuit, since the parallel combination is the only presence of resistance in the circuit. This voltage drop must be the total voltage of the circuit, .

The current across and is the same, and is given as:

$I_1=I_2=\frac{12V}{Z_1+Z_2}$

$I_2=\frac{12V}{32\Omega}$

Basic circuit2

In the circuit above:

$V_{in}=12V$

$Z_1=11\Omega$

$Z_2=21\Omega$

$Z_3=16\Omega$

What is the total current in the circuit before it is encounters the parallel connection?

Explanation

Begin by finding the resistance of the parallel connection:

$\frac{1}{Z}=\frac{1}{Z_1+Z_2}+\frac{1}{Z_3}$

$\frac{1}{Z}=\frac{1}{32\Omega}+\frac{1}{16\Omega}$

$\frac{1}{Z}=\frac{3}{32\Omega}$

$Z=\frac{32}{3}\Omega$

The total current is then found using Ohm's law:

$I=\frac{V_{in}}{Z}$

$I=\frac{12V}{\frac{32}{3}\Omega}$

Circuitdiagram3

$R_1=1\Omega$

$R_2=2\Omega$

$R_3=3\Omega$

$R_4=4\Omega$

In the circuit above, what is the voltage drop across ?

Explanation

The voltage drop across can be found as:

$V_3=\frac{R_3}{R_1+R_3}V=\frac{3\Omega}{1\Omega+3\Omega}10V=\frac{3}{4}\cdot10V$

Circuitdiagram

What is the equivalent capacitance of the given circuit diagram?

$42\mu F$

$6.2\mu F$

$34\mu F$

$85\mu F$

$14\mu F$

Explanation

Capacitors add opposite of the way resistors add in a circuit. That is, for capacitors in series they add as such:

$C_{eq}^{-1} = C_{1}^{-1} + C_{2}^{-1} + ...$

Capacitors in parallel add as such:

$C_{eq} = C_{1} + C_{2} + ...$

Use this information to add all the capacitors in series together. The only branch this applies to is the right hand branch.

$C_{345} = (C_{3}^{-1} + C_{4}^{-1} + C_{5}^{-1})^{-1} = (\frac{1}{20\mu F}+\frac{1}{5\mu F}+\frac{1}{5\mu F})^{-1} = 2.2\mu F$

The equivalent circuit is shown below:

Circuitdiagram2

Add the capacitors in parallel.

$C_{eq} = C_{1} + C_{2} + C_{345} = 10\mu F + 30\mu F + 2.2\mu F = 42\mu F$

Circuitdiagram

What is the charge on capacitor $C_{3}$ in the given circuit diagram?

$30\mu C$

$210\mu C$

$180\mu C$

$50\mu C$

$3.3\mu C$

Explanation

The relationship between a capacitor's charge and the voltage drop across it is:

Since the voltage drop across both $C_{1}$ and $C_{2} + C_{3}$ are the same, we just have to worry about the right part of the circuit. Capacitors are the opposite of resistors when it comes to finding equivalent capacitance, so for capacitors in series the two capacitors on the right will add as such