AP Physics 1 - Equivalent Resistance

Consider the circuit:

Circuit_4

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

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

$0.94\Omega$

$1.88\Omega$

$6.72\Omega$

$12.46\Omega$

$21.09\Omega$

Explanation

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:

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

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

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

$R1 = 0.94\Omega$

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?

$0\Omega$

$4\Omega$

$8\Omega$

$2\Omega$

$6\Omega$

Explanation

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.

Consider the circuit:

Circuit_4

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

$R1 = 8\Omega,\ R2 = 20\Omega,\ R3 = 8\Omega,\ R4 = 30\Omega$

$R1 = 1\Omega,\ R2 = 0.25\Omega,\ R3 = 1.25\Omega,\ R4 = 0.5\Omega$

$R1 = 3\Omega,\ R2 = 7\Omega,\ R3 = 15\Omega,\ R4 = 5\Omega$

$R1 = 5\Omega,\ R2 = 5\Omega,\ R3 = 6\Omega,\ R4 = 8\Omega$

None of these

Explanation

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:

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

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

$R_{eq}= 3\Omega$

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

$25\Omega$

$10\Omega$

$15\Omega$

$5\Omega$

Explanation

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,

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

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

$6\Omega$

$25\Omega$

$5\Omega$

$10\Omega$

Explanation

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,

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

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

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

Consider the given circuit:

Circuit_2

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

$R1 = 3\Omega$

$R2 = 2\Omega$

Explanation

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:

$V=IR_{tot}$

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

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

$0.68 \Omega$

$1.47 \Omega$

$22.0 \Omega$

$11.0\Omega$

Explanation

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

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

Consider the given circuit:

Circuit_2

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

$R1 = 3\Omega$

$R2=2\Omega$

$2\Omega$

$1\Omega$

$3\Omega$

$4\Omega$

$5\Omega$

Explanation

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:

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

$R_{AB} = 2\Omega$

Consider the circuit:

Circuit_4

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

$40\Omega$

$2.5\Omega$

$25\Omega$

$4\Omega$

None of these

Explanation

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:

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

$R = 40\Omega$

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.

$\frac{1}{20}V$

Explanation

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 .

Equivalent Resistance

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