Equivalent Resistance - AP Physics 1

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Question

Consider the following circuit:

What is the total equivalent resistance of the circuit?

$R1 = 5\Omega,\ R2 = 3\Omega,\ R3 = 4\Omega,\ R4=1\Omega,\ R5=2\Omega$

Answer

First we need to condense R3 and R4. They are in series, so we can simply add them to get:

$R_{34} = 5\Omega$

Now we can condense R2 and R34. They are in parallel, so we will use the following equation:

$\frac{1}{R_{eq}} = \sum\frac{1}{R} = \frac{1}{3}+\frac{1}{5} = \frac{8}{15}$

Therefore:

$R_{eq}=\frac{15}{8}$

The equivalent circuit now looks like:

Since everything is in series, we can simply add everything up:

$R_{total} = R1 + R_{eq}+ R5 = 5\Omega+\frac{15}{8}\Omega+2\Omega$

$R_{total} = 8\frac{7}{8}\Omega$

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Question

Consider the given circuit:

What is the current through the system if we attach a $5\Omega$ resistor from point A to B?

$R1 = 3\Omega$

$R2 = 2\Omega$

Answer

The new circuit has two resistors in parallel: R2 and the new one attached. To find the equivalent resistance of these two branches, we use the following expression:

$\frac{1}{R_{eq}}=\sum\frac{1}{R}=\frac{1}{R_2}+\frac{1}{5\Omega}=\frac{1}{2\Omega}+\frac{1}{5\Omega}=\frac{7}{10}$

$R_{eq}=\frac{10}{7}\Omega$

In this new equivalent circuit everything is in series, so we can simply add up the resistances:

$R_{tot}=R1+R_{eq}=3\Omega+\frac{10}{7}\Omega=4\frac{3}{7}\Omega$

Now we can use Ohm's law to calculate the total current through the circuit:

$V=IR_{tot}$

$I=\frac{V}{R_{tot}}=\frac{12V}{\frac{31}{7}\Omega} = 2.7A$

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Question

Consider the given circuit:

How much resistance must be applied between points A and B for the circuit to have a total current of 3A?

$R1 = 3\Omega$

$R2=2\Omega$

Answer

We will be working backwards on this problem, using the current to find the resistance. We know the voltage and desired current, so we can calculate the total necessary resistance:

$V=IR_{tot}$

$R_{tot}=\frac{12V}{3A} = 4\Omega$

Then we can calculate the equivalent resistance of the two resistors that are in parallel (R2 and our unknown):

$R_{tot}=R1+R_{eq}$

$R_{eq}=4\Omega-3\Omega=1\Omega$

Now we can calculate what the resistance between point A and B:

$\frac{1}{R_{eq}}=\sum\frac{1}{R}=\frac{1}{R2}+\frac{1}{R_{AB}}$

Rearranging for the desired resistance:

$\frac{1}{R_{AB}}=\frac{1}{R_{eq}}-\frac{1}{R2}=\frac{1}{1}-\frac{1}{2}=\frac{1}{2}$

$R_{AB} = 2\Omega$

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Question

Consider the circuit:

If the equivalent resistance of the circuit is $10 \Omega$ and each resistor is the same, what is the value of each resistor?

Answer

We can use the equation for equivalent resistance of parallel resistors to solve this equation:

$\frac{1}{R_{eq}}=\sum \frac{1}{R}$

We know the equivalent resistance, and we know that the resistance of each of the four resistors is equal:

$\frac{1}{10\Omega} = \frac{4}{R}$

$R = 40\Omega$

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Question

Consider the circuit:

If the power dissipated throughout the entire circuit is , what is the value of ?

$R1 = 10\Omega,\ R2 = 15\Omega,\ R3 = 20\Omega$

Answer

Since we know the power loss and voltage of the circuit, we can calculate the equivalent resistance of the circuit using the following equations:

Substituting Ohm's law into the equation for power, we get:

$P = \frac{V^2}{R}$

Rearranging for resistance, we get:

$R = \frac{V^2}{P}=\frac{(12V)^2}{40W} = 3.6\Omega$

This is the equivalent resistance of the entire circuit. Now we can calculate R4 using the expression for resistors in parallel:

$\frac{1}{R_{eq}}=\sum \frac{1}{R}$

$\frac{1}{3.6\Omega} = \frac{1}{10\Omega}+\frac{1}{15\Omega}+\frac{1}{20\Omega}+\frac{1}{R4}$

$R4 = 16.4\Omega$

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Question

Consider the circuit:

If the current flowing through the circuit is , what is the value of R1?

$R2 = 2 (R1),\ R3 =2(R2),\ R4 = 2 (R3)$

Answer

We can use Ohm's law to calculate the equivalent resistance of the circuit:

$V = IR_{eq}$

$R_{eq} = \frac{V}{I}=\frac{12V}{24A}= 0.5\Omega$

Now we can use the expression for combining parallel resistors to calculate R1:

$\frac{1}{R_{eq}}=\sum \frac{1}{R}=\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}+\frac{1}{R4}$

$\frac{1}{0.5\Omega}=\sum\frac{1}{R1}+\frac{1}{2R1}+\frac{1}{4R1}+\frac{1}{8R1}$

$\frac{1}{0.5\Omega}=\frac{15}{8R1}$

$R1 = 0.94\Omega$

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Question

Consider the circuit:

If the equivalent resistance of the circuit is $3\Omega$ , which of the following configuration of resistance values is possible?

Answer

We will need to test the values of each answer to find the one that generates an equivalent resistance of $3\Omega$ .

We know that when condensing parallel resistors, the equivalent resistance will never be larger than the largest single resistance, and will always be smaller than the smallest resistance. Therefore, two of the answer options cen be eliminated immediately.

After we have narrowed our choices down to the other options answers, we just have to test them with the following formula:

$\frac{1}{R_{eq}}=\sum\frac{1}{R}$

We will test the incorrect answer first:

$\frac{1}{R_{eq}} = \frac{1}{5\Omega}+\frac{1}{5\Omega}+\frac{1}{6\Omega}+\frac{1}{8\Omega}$

$R_{eq} = 1.4\Omega$

Now for the correct answer:

$\frac{1}{R_{eq}} = \frac{1}{8\Omega}+\frac{1}{20\Omega}+\frac{1}{8\Omega}+\frac{1}{30\Omega}$

$\frac{1}{R_{eq}}=\frac{40}{120}\Omega=\frac{1}{3}\Omega$

$R_{eq}= 3\Omega$

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Question

A circuit contains three resistors. Two of the resistors are in parallel with each other, and the third is connected in series with the parallel connection. If all the resistors' resistances must add to $8\Omega$ , what resistance should the resistor in series have to minimize the equivalent resistance?

Answer

The goal of this question is to realize that when two resistors are connected in parallel, the equivalent resistance is lower than either of the two original resistors. But when two resistors are connected in series, the equivalent resistance is the sum of the two original resistors. Therefore to minimize our equivalent resistance, we want all the resistance to be in the parallel resistors, leaving $0\Omega$ for the resistor in series.

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Question

What is the equivalent resistance from Point A to Point B?

Answer

Because this circuit is neither purely series or purely parallel, we must simplify it before we solve it. Replace the right branch, which is purely series, with its equivalent resistance:

$50\Omega+50\Omega=100\Omega$

Now we have a purely parallel circuit, each branch having a resistance of $100\Omega$ . Apply the parallel formula and solve:

$R_{tot}=\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}}$

$R_{tot}=\frac{1}{\frac{1}{100\Omega }+\frac{1}{100\Omega }}=50\Omega$

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Question

What is the equivalent resistance of the following resistors, all in series: $3 \Omega, 2 \Omega, 6\Omega$ ?

Answer

For resistors all in series, the equivalent resistance is equal to the sum of the resistances.

$R_{eq}=3+2+6=11\Omega$

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Question

What is the equivalent resistance of a circuit consisting of a group of resistors (all in parallel), with the following resistances: $5\Omega, 5\Omega, 5\Omega, 5\Omega, 5\Omega$ ?

Answer

The reciprocal of the equivalent resistance for resistors in parallel is equal to the sum of the reciprocals of the resistances:

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

$\frac{1}{R_{eq}}=\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=1\Omega$

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Question

Three identical resistors connected in parallel have an equivalent resistance equal to $2: \Omega$ . What is the resistance of each of the individual resistors in this circuit?

Answer

In the question stem, we're told that a circuit containing three identical resistors connected in parallel has an equivalent resistance equal to $2:\Omega$ . We are then asked to solve for the resistance of each individual resistor.

To start with, it's important to remember that resistors in parallel add inversely. Thus, the inverse of the equivalent resistance is equal to the sum of the inverse of each individual resistor. Put another way:

$\frac{1}{R_{eq}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+....\frac{1}{R_{n}}$

Since we know the three resistors we're dealing with are identical, we can assign each of them a value of .

$\frac{1}{2: \Omega }=\frac{1}{x}+\frac{1}{x}+\frac{1}{x}=\frac{3}{x}$

And rearranging, we obtain:

$x=(2: \Omega)(3)=6: \Omega$

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Question

Two resistors, $R_{1}=15\Omega$ and $R_{2}=10\Omega$ , are connected in series. What is the equivalent resistance of this setup?

Answer

The equivalent resistance of resistors connected in series is the sum of the resistance values of each resistor, or

$R_{s}=\sum R_{n}=R_{1}+R_{2}+...$

In our problem,

$R_{s}=R_{1}+R_{2}=15+10=25\Omega$

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Question

Two resistors, $R_{1}=15\Omega$ and $R_{2}=10\Omega$ , are connected in parallel. What is the equivalent resistance of this setup?

Answer

The equivalent resistance of resistors connected in parallel is given by the following equation,

$\frac{1}{R_{p}}=\sum \left ( \frac{1}{R_{n}} \right )=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+...$

In our problem,

$\frac{1}{R_{p}}=\frac{1}{15}+\frac{1}{10}=\frac{2}{30}+\frac{3}{30}$

$\frac{1}{R_{p}}=\frac{5}{30}$

$R_{p}=\frac{30}{5}=6\Omega$

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Question

Three resistors, $R_{1}=15\Omega$ , $R_{2}=10\Omega$ , and $R_{3}=5\Omega$ , are connected in series. What is the equivalent resistance of this setup?

Answer

The equivalent resistance of resistors connected in series is the sum of the resistance values of each resistor, or

$R_{s}=\sum R_{n}=R_{1}+R_{2}+...$

In our problem,

$R_{s}=R_{1}+R_{2}+R_{5}=15+10+5=30\Omega$

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Question

Three resistors, $R_{1}=15\Omega$ , $R_{2}=10\Omega$ , and $R_{3}=5\Omega$ , are connected in parallel. What is the equivalent resistance of this setup?

Answer

The equivalent resistance of resistors connected in parallel is given by the following equation,

$\frac{1}{R_{p}}=\sum \left ( \frac{1}{R_{n}} \right )=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+...$

In our problem,

$\frac{1}{R_{p}}=\frac{1}{15}+\frac{1}{10}+\frac{1}{5}=\frac{2}{30}+\frac{3}{30}+\frac{6}{30}$

$\frac{1}{R_{p}}=\frac{11}{30}$

$R_{p}=\frac{30}{11}=2.73\Omega$

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Question

$R_{1}=10\Omega, R_{2}=15\Omega, R_{3}=25\Omega.$

Three resistors are connected to a battery. What is the equivalent resistance of the given circuit?

Answer

Remember the rules of adding resistors. For resistors in series,

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

and for resistors in parallel,

$R_{eq}=\left ( \frac{1}{R_{1}}+\frac{1}{R_{2}}+...+\frac{1}{R_{n}}\right )^{-1}$

The idea is to start from the side furthest away from the battery and work back toward it. Notice that $R_{2}$ and $R_{3}$ are in parallel. We can add them in parallel so that they have an equivalent resistance $R_{23}$ ,

This can be calculated, but for 3 resistors we can leave it in equation form until the end. This looks like:

$R_{23}=\left ( \frac{1}{15\Omega}+\frac{1}{25\Omega}\right )^{-1}=9.375\Omega$

Notice that $R_{1}$ is now in series with $R_{23}$ ,

$R_{eq}=R_{1}+R_{23}=R_{1}+\left ( \frac{1}{15\Omega}+\frac{1}{25\Omega}\right )^{-1}=19.375\Omega$

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Question

You are given the following circuit:

The resistor values are as follows:

$R1=R5=R6=2k\Omega$

$R2=R3=R4=R10=8k\Omega$

$R7=R8=R9=R11=6k\Omega$

Find the equivalent resistance.

Answer

In order to find the equivalent resistance, you must take small steps and slowly work towards finding the total equivalent resistance.You can combine resistors if they are in parallel. The expression to find parallel equivalent resistance is:

$R_1\left | \right |R_2=\frac{R_1R_2}{R_1+R_2}$

You can also combine resistors in series, or one after the other. The expression to find resistors in series is:

$R_1\rightarrow R_2=R_1+R_2$

To start off, you can combine R2 and R3, since they are both in parallel:

$R_{2_3}=\frac{R2R3}{R2+R3}=\frac{8k\Omega8k\Omega}{8k\Omega+8k\Omega}=4k\Omega$

Next, you can combine some of the series resistors together.

$R_{1_2_3}=2k\Omega+4k\Omega=6k\Omega$

$R_{5_6}=2k\Omega+2k\Omega=4k\Omega$

$R_{8_9}=6k\Omega+6k\Omega=12k\Omega$

$R_{10_11}=8k\Omega+6k\Omega=14k\Omega$

After this, you can combine the following resistors in parallel:

$R_{7_8_9}=R_7\left | \right |R_{8_9}=\frac{6k\Omega12k\Omega}{6k\Omega+12k\Omega}=4k\Omega$

Then, combine and $R_{7_8_9}}$ in series:

$R_{4_7_8_9}=4k\Omega+8k\Omega=12k\Omega$

Now get rid of the last parallel by combining $R_{5_6}$ and $R_{4_7_8_9}$ in parallel:

$R_{4_5_6_7_8_9}=\frac{4k\Omega*12k\Omega}{4k\Omega+12k\Omega}=3k\Omega$

Now finally, add the remaining resistors in series to find the equivalent resistance:

$R_{1_2_3_4_5_6_7_8_9_10_11}=6k\Omega+14k\Omega+3k\Omega=23k\Omega$

$R_{equivalent}=23k\Omega$

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Question

A circuit has 10 identical resistors in parallel with a battery of , and a total resistance of $\frac{1}{20} \Omega$ . Determine the voltage drop across one resistor.

Answer

Since the resistors are in parallel to a battery, the voltage drop across each resistor has to be equal to the voltage gain across the battery. Therefore, the voltage drop for any of the resistors will be .

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Question

Calculate the equivalent resistance, of four resistors in parallel, which values $R_1 = 5 \Omega, R_2 = 10 \Omega, R_3 = 1 \Omega, R_4 = 6 \Omega$ .

Answer

In order to find the equivalent resistance of resistors in parallel, we add the inverses of their values, as shown below

$\frac{1}{R_{tot}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{5 \Omega}+\frac{1}{10 \Omega}+\frac{1}{1\Omega}+\frac{1}{6 \Omega}=1.467 \Omega ^{-1}$

Finally

$\frac{1}{R_{tot}}=1.467 \Omega ^{-1} \Rightarrow R_{tot}=\frac{1}{1.467 \Omega ^{-1}}=0.68 \Omega$

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