Understanding Circuit Diagrams - AP Physics 1

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Question

In the circuit above, what is the power dissipated by ?

Answer

The first step to solving a circuit problem would be to identify and calculate for all the currents that are unknown using Kirchoff's Laws. By drawing loops in the circuit and setting the voltage drop across a loop equal to zero we can calculate for the unknown currents and solve for the power. The two loops are indicated here in this diagram.

Designate the current flowing through the battery as $I_{1}$ , the current flowing through $R_{2}$ and $R_{3}$ as $I_{2}$ , and the current through $R_{1}$ as $I_{3}$ . The first loop on the left, when written out using the loop rule, gives:

$9V-30\Omega I_{3} = 0$

Solve for current.

$I_{3} = .3A$

The second loop gives us:

$-25\Omega I_{2}-15\Omega I_{2} + 30\Omega I_{3} = 0$

Plug in the value for $I_{3}$

$I_{2} = \frac{3}{4}I_{3} = 0.225A$

Find the power dissipated through $R_{2}$

$P=I^{2}R=I^{2}_{2}R_{2}=0.76W$

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Question

What is the equivalent resistance of the circuit shown above?

Answer

When resistors are in series, they add normally, such as

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

when in series, they add via their reciprocal

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

Using these rules, we can first combine all the resistors in series ( $R_{1}$ and $R_{2}$ , $R_{4}$ and $R_{5}\right)$ ), which can be diagrammed as such:

Using the parallel rule, find to total equivalent resistance.

$R_{eq} = \left(R_{12}^{-1} + R_{3}^{-1} + R_{45}^{-1}\right)^{-1} = \left(\frac{1}{35\Omega}+\frac{1}{15\Omega}+\frac{1}{10\Omega}\right )^{-1} = 5.1\Omega$

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Question

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

Answer

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

$C_{23} = (C_{2}^{-1}+C_{3}^{-1})^{-1} = 3.3\mu F$

Plugging into the first equation.

$Q_{23} = C_{23}V_{23} = 30\mu C$

Since the two capacitors are in series they must share the same charge as the equivalent capacitor.

$Q_{3} = Q_{23} = 30\mu C$

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Question

What is the equivalent capacitance of the given circuit diagram?

Answer

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:

Add the capacitors in parallel.

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

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Question

How much current runs through $R_{3}$ in the given circuit diagram?

Answer

The first step to solving this problem is to utilize Kirchoff's rules to write a set of equations to find the unknowns. Below two loops are diagrammed, and we assign 3 currents to the circuit. The first and second pass up through batteries 1 and 2 respectively, and the third current passes down through $R_{3}$ .

These initial conditions give us our first equation:

$I_{1}+I_{2} = I_{3}$

Now using the loop rule and setting all voltage changes across a loop equal to zero, we get these two equations for the two loops. The loop on the left gives us:

$-50I_{1} = 10V - 15V+35I_{2} = 0$

The loop on the right gives us:

$-35I_{2} + 15V-40I_{3} = 0$

Since we're looking to find the current that flows through $R_{3}$ we will need to solve for $I_{3}$ . The easiest way to do this would be to take the first equation and replace either $I_{1}$ or $I_{2}$ in one of the other two equations. The following solution will replace $I_{1}$ in the first equation to accomplish this.

$-50(I_{3}-I_{2}) + 35I_{2}-5V = 0$

Rearrange.

$17I_{2} - 10I_{3} = 1V$

Simplify the second equation.

$7I_{2}+8I_{3} = 3V$

Lastly, solve for $I_{2}$ and set the two equations equal to each other.

$\frac{1}{17}+\frac{10}{17}I_{3} = \frac{3}{7}-\frac{8}{7}I_{3}$

$I_{3} = \frac{22}{103} = 0.21A$

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Question

$R_1=1\Omega$

$R_2=4\Omega$

In the circuit above, what is the total current?

Answer

To find the amperage, first find the combined resistances of the resistors in parallel:

$\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{1\Omega }+\frac{1}{4\Omega }$

$\frac{1}{R_T}=\frac{5}{4\Omega }$

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

After that, calculate the current using Ohm's Law:

$I=\frac{V}{R}=\frac{5V}{\frac{4}{5}\Omega}$

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Question

$R_1=2\Omega$

$R_2=3\Omega$

In the circuit above, what is the total voltage?

Answer

To find the voltage, first find the combined resistances of the resistors in parallel:

$\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{2\Omega }+\frac{1}{3\Omega }$

$\frac{1}{R_T}=\frac{5}{6\Omega }$

$R_T=\frac{6}{5}\Omega$

Use Ohm's law to find the voltage.

$V=30A\left(\frac{6}{5}\Omega\right)=36V$

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Question

$R_1=6\Omega$

In the circuit above, what is the resistance of ?

Answer

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

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

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

$R_2=12\Omega$

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Question

$R_1=3\Omega$

$R_2=6\Omega$

$R_3=2\Omega$

$R_4=16\Omega$

In the circuit above, what is the total resistance?

Answer

Find the combined resistances for the resistors in parallel:

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

$\frac{1}{R_A}=\frac{1}{3\Omega}+\frac{1}{6\Omega}=\frac{9}{18\Omega}$

$R_A=2\Omega$

$\frac{1}{R_B}=\frac{1}{R_3}+\frac{1}{R_4}$

$\frac{1}{R_B}=\frac{1}{2\Omega}+\frac{1}{16\Omega}=\frac{18}{32\Omega}$

$R_B=\frac{16}{9}\Omega$

Combine these two combined series resistors to find the total resistance:

$R_T=R_A+R_B=2\Omega+\frac{16}{9}\Omega=\frac{34}{9}\Omega$

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Question

$R_1=1\Omega$

$R_2=2\Omega$

$R_3=3\Omega$

$R_4=4\Omega$

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

Answer

Find the total resistance of the circuit. First, calculate the values of the combined resistances of the resistors in parallel:

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

$\frac{1}{R_A}=\frac{1}{1\Omega}+\frac{1}{2\Omega}=\frac{3}{2\Omega}$

$R_A=\frac{2}{3}\Omega$

$\frac{1}{R_B}=\frac{1}{R_3}+\frac{1}{R_4}$

$\frac{1}{R_B}=\frac{1}{3\Omega}+\frac{1}{4\Omega}=\frac{7}{12\Omega}$

$R_B=\frac{12}{7}\Omega$

Therefore, the total resistance is:

$R_T=R_A+R_B=\frac{2}{3}\Omega+\frac{12}{7}\Omega=\frac{50}{21}\Omega$

Now, note that since and are in parallel, the voltage drop across them is the same. Use Ohm's law to relate current in terms of voltage and resistance.

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

Substitute into Ohm's law for the resistance across :

$V_{3,4}=\frac{V}{R_T}R_B$

$V_{3,4}=10V\left(\frac{21}{50\Omega}\right)\left(\frac{12\Omega}{7}\right)$

$V_{3,4}=7.2V$

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Question

$R_1=1\Omega$

$R_2=2\Omega$

$R_3=3\Omega$

$R_4=4\Omega$

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

Answer

Find the total resistance of the circuit. First, calculate the values of the combined resistances of the resistors in parallel:

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

$\frac{1}{R_A}=\frac{1}{1\Omega}+\frac{1}{2\Omega}=\frac{3}{2\Omega}$

$R_A=\frac{2}{3}\Omega$

$\frac{1}{R_B}=\frac{1}{R_3}+\frac{1}{R_4}$

$\frac{1}{R_B}=\frac{1}{3\Omega}+\frac{1}{4\Omega}=\frac{7}{12\Omega}$

$R_B=\frac{12}{7}\Omega$

Therefore, the total resistance is:

$R_T=R_A+R_B=\frac{2}{3}\Omega+\frac{12}{7}\Omega=\frac{50}{21}\Omega$

From Ohm's law, we know that $\frac{V}{R_T}$ is the current traveling through the circuit.

$I=10V\left(\frac{21}{50\Omega}\right)=4.2A$

This current will be divided between and , with more current taking the path of lower resistance.

Total voltage drop across $R_{3,4}$ :

$V_{3,4}=I_TR_{3,4}$

$V_{3,4}=4.2A\cdot\frac{12}{7}\Omega$

$V_{3,4}=7.2V$

The current through is given by:

$I_3=\frac{V_3}{R_3}$

$I_3=\frac{7.2V}{3\Omega}=2.4A$

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Question

$R_1=3\Omega$

$R_2=6\Omega$

$R_3=2\Omega$

$R_4=10\Omega$

In the circuit above, what is the total resistance?

Answer

Begin by combining the resistors that are immediately in series:

$R_A=R_1+R_3=3\Omega+2\Omega=5\Omega$

$R_B=R_2+R_4=6\Omega+10\Omega=16\Omega$

Now to find the total resistance, combine these two new resistance values, which are in parallel:

$\frac{1}{R_T}=\frac{1}{5\Omega}+\frac{1}{16\Omega}=\frac{21}{80\Omega}$

$R_T=\frac{80}{21}\Omega$

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Question

$R_1=1\Omega$

$R_2=2\Omega$

$R_3=3\Omega$

$R_4=4\Omega$

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

Answer

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, .

The voltage drop across can be found as:

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

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Question

$R_1=1\Omega$

$R_2=2\Omega$

$R_3=3\Omega$

$R_4=4\Omega$

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

Answer

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.

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

$I_{1,3}=2.5A$

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Question

In the circuit above:

$V_{in}=12V$

$Z_1=11\Omega$

$Z_2=21\Omega$

$Z_3=16\Omega$

What is the current across ?

Answer

The quickest way to approach this problem is to realize that the voltage drop across is the same as the voltage drop across the combined resistances of and . Since this parallel combination is the only presence of resistance in the circuit, this voltage drop must be the total voltage of the circuit, .

Therefore, the current across is:

$I_3=\frac{V_3}{Z_3}$

$I_3=\frac{12V}{16\Omega}$

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Question

In the circuit above:

$V_{in}=12V$

$Z_1=11\Omega$

$Z_2=21\Omega$

$Z_3=16\Omega$

What is the current across ?

Answer

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:

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

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

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Question

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?

Answer

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:

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

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

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Question

The following symbol represents what item in a circuit?

Answer

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.

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Question

In the circuit above, what is the power dissipated by ?

Answer

The first step to solving a circuit problem would be to identify and calculate for all the currents that are unknown using Kirchoff's Laws. By drawing loops in the circuit and setting the voltage drop across a loop equal to zero we can calculate for the unknown currents and solve for the power. The two loops are indicated here in this diagram.

Designate the current flowing through the battery as $I_{1}$ , the current flowing through $R_{2}$ and $R_{3}$ as $I_{2}$ , and the current through $R_{1}$ as $I_{3}$ . The first loop on the left, when written out using the loop rule, gives:

$9V-30\Omega I_{3} = 0$

Solve for current.

$I_{3} = .3A$

The second loop gives us:

$-25\Omega I_{2}-15\Omega I_{2} + 30\Omega I_{3} = 0$

Plug in the value for $I_{3}$

$I_{2} = \frac{3}{4}I_{3} = 0.225A$

Find the power dissipated through $R_{2}$

$P=I^{2}R=I^{2}_{2}R_{2}=0.76W$

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Question

What is the equivalent resistance of the circuit shown above?

Answer

When resistors are in series, they add normally, such as

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

when in series, they add via their reciprocal

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

Using these rules, we can first combine all the resistors in series ( $R_{1}$ and $R_{2}$ , $R_{4}$ and $R_{5}\right)$ ), which can be diagrammed as such:

Using the parallel rule, find to total equivalent resistance.

$R_{eq} = \left(R_{12}^{-1} + R_{3}^{-1} + R_{45}^{-1}\right)^{-1} = \left(\frac{1}{35\Omega}+\frac{1}{15\Omega}+\frac{1}{10\Omega}\right )^{-1} = 5.1\Omega$

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