Understanding Series and Parallel - Physics

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Question

Each component of a series circuit carries an equal __________.

Answer

A series circuit has the same current running through each component, though the voltage across each component may vary. In contrast, a parallel circuit will carry a constant voltage in each branch and the current may be split in inequal amounts between branches.

Resistance and capacitance are dependent on the elements present in a given circuit, so no claims can be made. Power is similarly dependent on the elements present in a specific circuit.

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Question

Each branch of a parallel circuit carries an equal __________.

Answer

By definition, each branch of a parallel circuit will have the same voltage. The current, however, may be split between branches in inequal amounts. In contrast, a series circuit will carry a constant current throughout the circuit, but voltage may differ between elements.

Resistance and capacitance are dependent on the elements present in a given circuit, so no claims can be made. Power is similarly dependent on the elements present in a specific circuit.

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Question

Two $10\Omega$ resistors are placed in a circuit. What is the difference in equivalent resistance when the resistors in series versus when they are in parallel?

Answer

In series, the equivalent resistance is equal to the sum of each individual resistance. In parallel, the inverse of equivalent resistance is equal to the sum of the inverses of each individual resistance.

$R_{eq_{s}}=R_1+R_2$

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

We can calculate these two values, and then find the difference between them.

$R_{eq_s}=10\Omega+10\Omega=20\Omega$

$\frac{1}{R_{eq_p}}=\frac{1}{10\Omega}+\frac{1}{10\Omega}=\frac{1}{5\Omega}\rightarrow R_{eq_p}=5\Omega$

Find the difference:

$R_{eq_s}-R_{eq_p}=20\Omega-5\Omega=15\Omega$

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Question

What is the net resistance of the circuit connected to the battery?

Answer

To find the net resistance of the circuit, we must summarize all of the different resistors, one piece at a time. To begin, let’s start with the two resistors in the upper left had corner which are in series. In series resistance can be added together. We can replace these two resistors with a single resistor that is valued at .

Next, this new resistor is in parallel with the resistor that is on the diagonal line from point A to point B. In parallel we can add resistors through the equation $\frac{1}{R_{eq}} = \frac{1}{R_1} +\frac{1}{R_2}$

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

$\frac{1}{R_{eq}} = \frac{3}{2R}$

$R_{eq} = \frac{2}{3}R$

This new resistor is now in series with the resistor that is between points B & C. In series we can just add these resistors up.

$R_{eq} = \frac{2}{3}R + R$

$R_{eq} = \frac{5}{3}R$

This new resistor is now in parallel with the resistor that is between points A & C. In parallel we can add resistors through the equation $\frac{1}{R_{eq}} = \frac{1}{R_1} +\frac{1}{R_2}$

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

$\frac{1}{R_{eq}} = \frac{8}{5R}$

$R_{eq} = \frac{5}{8}R$

This new resistor is now in series with the resistor that is between the battery and point C. In series we can just add these resistors up.

$R_{eq} = \frac{5}{8}R + R$

$R_{eq} = \frac{13}{8}R$

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Question

What is the net resistance of the circuit connected to the battery?

Answer

To begin, let us start with the $10\Omega$ resistor and the $15\Omega$ resistor that are in parallel. In parallel we can add resistors through the equation $\frac{1}{R_{eq}} = \frac{1}{R_1} +\frac{1}{R_2}$

$\frac{1}{R_{eq}} = \frac{1}{10} + \frac {1}{15}$

$\frac{1}{R_{eq}} = \frac{3}{30} + \frac{2}{30}$

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

$R_eq = \frac{30}{5}$

$R_{eq} = 6$

This new resistor is now in series with the two $4\Omega$ resistors. In series we can just add these resistors up.

$R_{eq} = 4 + 4 + 6$

$R_{eq} = 14$

This new resistor is now in parallel with the $14\Omega$ resistor. In parallel we can add resistors through the equation $\frac{1}{R_{eq}} = \frac{1}{R_1} +\frac{1}{R_2}$

$\frac{1}{R_eq} = \frac{1}{14} +\frac{1}{14}$

$\frac{1}{R_eq} =\frac{2}{14}$

$R_{eq} = \frac{14}{2}$

$R_{eq} = 7$

This new resistor is now in series with the $3\Omega$ and $5\Omega$ resistor. In series we can just add these resistors up.

$R_{eq} = 7 + 3 + 5$

$R_{eq} = 15$

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Question

Basic Computational

Six resistors are arranged in a series. Their resistances are $32.4\Omega$ , $8.7\Omega$ , $0.77\Omega$ , $65.2\Omega$ , $10\Omega$ , and $26.33\Omega$ . What is their total resistance?

Answer

When working in a series, the total resistance is the sum of the individual resistances.

Use the given values for each individual resistor to solve for the total resistance.

$Req=32.4\Omega +8.7\Omega +0.77\Omega +65.2\Omega +10\Omega +26.33\Omega$

$Req=143.4\Omega$

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Question

What is the total resistance of a parallel circuit with resistors of $2\Omega$ , $0.55\Omega$ , and $4\Omega$ ?

Answer

The formula for resistance in parallel is:

$\frac{1}{R_{eq}} = \frac{1}{R_1} +\frac{1}{R_2} ...$

We are given the values for each individual resistor, allowing us to solve for the total resistance.

$\frac{1}{R_{eq}} = \frac{1}{2} + \frac{1}{0.55} + {1}{4}$

$\frac{1}{R_{eq}} = \frac{113}{44}$

$R_{eq}= \frac{44}{113} = 0.389$

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Question

Conceptual

When resistors are connected in series

Answer

In series, there are no branches for the current to travel down. Therefore it is like a one lane highway so the current is the same throughout the entire circuit.

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Question

When resistors are connected in parallel, we can be certain that

Answer

Consider a set of parallel resistors that are connected to the battery. Each resistor is on its own loop to the battery. Therefore each resistor has its own potential difference that is equal to the battery. That is why if one bulb goes out in a parallel circuit, the other stays lit. Each resistor has its own connection independent of the others in the circuit.

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Question

An electric circuit is set up in series with five resistors. If the resistors remain the same, but the circuit is now set up with the resistors in parallel, how would this affect the total resistance?

Answer

Think of resistors as doors, preventing the flow of people (electrons). Imagine the following scenarios: a large group of people are in a room and all try to leave at once. If the five resistors are in series, that's like having all of these people trying to go through all five doors before they can leave. In a circuit, all the electrons in the current must pass through every resistor.

If the resistors are in parallel, it's like having five separate doors from the room. All of a sudden, the group can leave MUCH faster, encountering less resistance to their flow out of the room. The path of the electrons can split, allowing each particle to pass through only one resistor.

From a formula perspective, the resistors in series are simply summed together to find the equivalent resistance.

$R_{eq} = R_1 +R_2 + R3$

In parallel, however, the reciprocals are summed to find the reciprocal equivalent resistance.

$\frac{1}{R_{eq}} = \frac{1}{R_1} +\frac{1}{R_2} + ...$

Adding whole numbers will always give you a much greater result than adding fractions. For the exact same set of resistors, arrangement in series will have a greater total resistance than arrangement in parallel.

Compare your answer with the correct one above

Question

Each component of a series circuit carries an equal __________.

Answer

A series circuit has the same current running through each component, though the voltage across each component may vary. In contrast, a parallel circuit will carry a constant voltage in each branch and the current may be split in inequal amounts between branches.

Resistance and capacitance are dependent on the elements present in a given circuit, so no claims can be made. Power is similarly dependent on the elements present in a specific circuit.

Compare your answer with the correct one above

Question

Each branch of a parallel circuit carries an equal __________.

Answer

By definition, each branch of a parallel circuit will have the same voltage. The current, however, may be split between branches in inequal amounts. In contrast, a series circuit will carry a constant current throughout the circuit, but voltage may differ between elements.

Resistance and capacitance are dependent on the elements present in a given circuit, so no claims can be made. Power is similarly dependent on the elements present in a specific circuit.

Compare your answer with the correct one above

Question

Two $10\Omega$ resistors are placed in a circuit. What is the difference in equivalent resistance when the resistors in series versus when they are in parallel?

Answer

In series, the equivalent resistance is equal to the sum of each individual resistance. In parallel, the inverse of equivalent resistance is equal to the sum of the inverses of each individual resistance.

$R_{eq_{s}}=R_1+R_2$

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

We can calculate these two values, and then find the difference between them.

$R_{eq_s}=10\Omega+10\Omega=20\Omega$

$\frac{1}{R_{eq_p}}=\frac{1}{10\Omega}+\frac{1}{10\Omega}=\frac{1}{5\Omega}\rightarrow R_{eq_p}=5\Omega$

Find the difference:

$R_{eq_s}-R_{eq_p}=20\Omega-5\Omega=15\Omega$

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Question

What is the net resistance of the circuit connected to the battery?

Answer

To find the net resistance of the circuit, we must summarize all of the different resistors, one piece at a time. To begin, let’s start with the two resistors in the upper left had corner which are in series. In series resistance can be added together. We can replace these two resistors with a single resistor that is valued at .

Next, this new resistor is in parallel with the resistor that is on the diagonal line from point A to point B. In parallel we can add resistors through the equation $\frac{1}{R_{eq}} = \frac{1}{R_1} +\frac{1}{R_2}$

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

$\frac{1}{R_{eq}} = \frac{3}{2R}$

$R_{eq} = \frac{2}{3}R$

This new resistor is now in series with the resistor that is between points B & C. In series we can just add these resistors up.

$R_{eq} = \frac{2}{3}R + R$

$R_{eq} = \frac{5}{3}R$

This new resistor is now in parallel with the resistor that is between points A & C. In parallel we can add resistors through the equation $\frac{1}{R_{eq}} = \frac{1}{R_1} +\frac{1}{R_2}$

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

$\frac{1}{R_{eq}} = \frac{8}{5R}$

$R_{eq} = \frac{5}{8}R$

This new resistor is now in series with the resistor that is between the battery and point C. In series we can just add these resistors up.

$R_{eq} = \frac{5}{8}R + R$

$R_{eq} = \frac{13}{8}R$

Compare your answer with the correct one above

Question

What is the net resistance of the circuit connected to the battery?

Answer

To begin, let us start with the $10\Omega$ resistor and the $15\Omega$ resistor that are in parallel. In parallel we can add resistors through the equation $\frac{1}{R_{eq}} = \frac{1}{R_1} +\frac{1}{R_2}$

$\frac{1}{R_{eq}} = \frac{1}{10} + \frac {1}{15}$

$\frac{1}{R_{eq}} = \frac{3}{30} + \frac{2}{30}$

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

$R_eq = \frac{30}{5}$

$R_{eq} = 6$

This new resistor is now in series with the two $4\Omega$ resistors. In series we can just add these resistors up.

$R_{eq} = 4 + 4 + 6$

$R_{eq} = 14$

This new resistor is now in parallel with the $14\Omega$ resistor. In parallel we can add resistors through the equation $\frac{1}{R_{eq}} = \frac{1}{R_1} +\frac{1}{R_2}$

$\frac{1}{R_eq} = \frac{1}{14} +\frac{1}{14}$

$\frac{1}{R_eq} =\frac{2}{14}$

$R_{eq} = \frac{14}{2}$

$R_{eq} = 7$

This new resistor is now in series with the $3\Omega$ and $5\Omega$ resistor. In series we can just add these resistors up.

$R_{eq} = 7 + 3 + 5$

$R_{eq} = 15$

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Question

Basic Computational

Six resistors are arranged in a series. Their resistances are $32.4\Omega$ , $8.7\Omega$ , $0.77\Omega$ , $65.2\Omega$ , $10\Omega$ , and $26.33\Omega$ . What is their total resistance?

Answer

When working in a series, the total resistance is the sum of the individual resistances.

Use the given values for each individual resistor to solve for the total resistance.

$Req=32.4\Omega +8.7\Omega +0.77\Omega +65.2\Omega +10\Omega +26.33\Omega$

$Req=143.4\Omega$

Compare your answer with the correct one above

Question

What is the total resistance of a parallel circuit with resistors of $2\Omega$ , $0.55\Omega$ , and $4\Omega$ ?

Answer

The formula for resistance in parallel is:

$\frac{1}{R_{eq}} = \frac{1}{R_1} +\frac{1}{R_2} ...$

We are given the values for each individual resistor, allowing us to solve for the total resistance.

$\frac{1}{R_{eq}} = \frac{1}{2} + \frac{1}{0.55} + {1}{4}$

$\frac{1}{R_{eq}} = \frac{113}{44}$

$R_{eq}= \frac{44}{113} = 0.389$

Compare your answer with the correct one above

Question

Conceptual

When resistors are connected in series

Answer

In series, there are no branches for the current to travel down. Therefore it is like a one lane highway so the current is the same throughout the entire circuit.

Compare your answer with the correct one above

Question

When resistors are connected in parallel, we can be certain that

Answer

Consider a set of parallel resistors that are connected to the battery. Each resistor is on its own loop to the battery. Therefore each resistor has its own potential difference that is equal to the battery. That is why if one bulb goes out in a parallel circuit, the other stays lit. Each resistor has its own connection independent of the others in the circuit.

Compare your answer with the correct one above

Question

An electric circuit is set up in series with five resistors. If the resistors remain the same, but the circuit is now set up with the resistors in parallel, how would this affect the total resistance?

Answer

Think of resistors as doors, preventing the flow of people (electrons). Imagine the following scenarios: a large group of people are in a room and all try to leave at once. If the five resistors are in series, that's like having all of these people trying to go through all five doors before they can leave. In a circuit, all the electrons in the current must pass through every resistor.

If the resistors are in parallel, it's like having five separate doors from the room. All of a sudden, the group can leave MUCH faster, encountering less resistance to their flow out of the room. The path of the electrons can split, allowing each particle to pass through only one resistor.

From a formula perspective, the resistors in series are simply summed together to find the equivalent resistance.

$R_{eq} = R_1 +R_2 + R3$

In parallel, however, the reciprocals are summed to find the reciprocal equivalent resistance.

$\frac{1}{R_{eq}} = \frac{1}{R_1} +\frac{1}{R_2} + ...$

Adding whole numbers will always give you a much greater result than adding fractions. For the exact same set of resistors, arrangement in series will have a greater total resistance than arrangement in parallel.

Compare your answer with the correct one above

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