Calculating Total Resistance - High School Physics

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Question

Ten resistors, each with $300\Omega$ resistance, are set up in series. What is their equivalent resistance?

Answer

For resistors aligned in series, the equivalent resistance is the sum of the individual resistances.

$R_{eq}=R_1+R_2+R_3...$

Since all the resistors in this problem are equal, we can simplify with multiplication.

$R_{eq}=10 (300\Omega)$

$R_{eq}=3000\Omega$

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Question

Ten resistors, each with $300\Omega$ resistance, are set up in parallel. What is the equivalent resistance?

Answer

For resistors in parallel, the equivalent resistance can be found by summing the reciprocals of the individual resistances, then taking the reciprocal of the resultant sum.

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}...$

Since all the resistors in this problem are equal, we can simplify by multiplying.

$\frac{1}{R_{eq}}=10( \frac{1}{300\Omega})$

$\frac{1}{R_{eq}}=\frac{10}{300\Omega}$

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

$R_{eq}=30\Omega$

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Question

There are three resistors in parallel in a circuit with resistances of $10\Omega$ , $20\Omega$ , and $30\Omega$ .

What is the equivalent resistance?

Answer

The equation for resistors in parallel is:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}...$

We are given the values of the resistors. Using this formula, we can solve for the equivalent resistance.

Plug in the given values and solve.

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}...$

$\frac{1}{R_{eq}}=\frac{1}{10\Omega}+\frac{1}{20\Omega}+\frac{1}{30\Omega}...$

$\frac{1}{R_{eq}}=\frac{11}{60\Omega}$

$R_{eq}=\frac{60\Omega}{11}$

$R_{eq}=5.45\Omega$

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Question

Twelve resistors of equal resistance are set up in series. If the total resistance is $600\Omega$ , what is the resistance of each resistor?

Answer

The total resistance of resistors in series is the sum of the individual resistances.

$R_{eq}=R_1+R_2+R_3...$

In this case, we know the total resistance and the number of resistors, and we are told that all resistors are of equal strength. That means we can simplify this problem using multiplication.

$R_{eq}=12R_$

We can use the total resistance to solve for the individual value.

$600\Omega=12R$

$\frac{600\Omega}{12}=R$

$50\Omega=R$

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Question

Three resistors are in a series. They have $10\Omega$ , $18.3\Omega$ , and $21.22\Omega$ of resistance respectively. What is the total resistance?

Answer

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

$R_{eq}=R_1+R_2+R_3...$

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

$R_{eq}=10\Omega+18.3\Omega+21.22\Omega$

$R_{eq}=49.52\Omega$

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Question

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.

$R_{eq}=R_1+R_2+R_3...$

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

$R_{eq}=32.4\Omega + 8.7\Omega +0.77\Omega + 65.2\Omega + 10\Omega + 26.33\Omega$

$R_{eq}=143.4\Omega$

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Question

Three resistors are in a parallel circuit. They have resistances of $10\Omega$ , $18.3\Omega$ , and $21.22\Omega$ , respectively. What is their total resistance?

Answer

The formula for resistors in parallel is:

$\frac{1}{R_{eq}}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}...$

Using the given individual resistances, we can find the total resistance of the circuit.

$\frac{1}{R_{eq}}}=\frac{1}{10\Omega}+\frac{1}{18.3\Omega}+\frac{1}{21.22\Omega}$

$\frac{1}{R_{eq}}}=0.2017\Omega$

$R_{eq}=4.96\Omega$

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Question

Two resistors in series have $1.8\Omega$ and $32.4\Omega$ resistance, respectively. What is the total resistance?

Answer

For resistors in a series, total resistance is equal to the sum of each individual resistance.

$R_{eq}=R_1+R_2+R_3...$

We can use the individual resistances from the question to solve for the total resistance.

$R_{eq}=1.8\Omega+32.4\Omega$

$R_{eq}=34.2\Omega$

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Question

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

Answer

For a series circuit, the formula for total resistance is:

$R_{eq}=R_1+R_1+R_3...$

We are given the values of each resistance, allowing us to sum them to find the total resistance in the circuit.

$R_{eq}=2\Omega+0.55\Omega+4\Omega$

$R_{eq}=6.55\Omega$

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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:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}...$

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

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

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

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

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Question

Calculate the resistance of a copper wire with cross-sectional area of $\small 0.501cm^2$ and length of $\small 2m$ .

$\small rho_{Cu}=1.68*10^{-3}\Omega m$

Answer

The resistance of a wire is given by the following equation:

$R=\rho\frac{L}{A}$

We are given the resistivity ( $\small \rho$ ), cross-sectional area, and length. Using these values, we can solve for the resistance.

First, convert the cross-sectional area to square-meters.

$0.501cm^2\frac{1m^2}{10000cm^2}=5.0110^{-5}m^2$

Use the resistance equation to solve.

$R=(1.6810^{-3}\Omega m)\frac{2m}{5.0110^{-5}m^2}$

$R=67.1\Omega$

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Question

What is the total resistance of a $14\Omega$ , $2.2\Omega$ , and $13\Omega$ series circuit?

Answer

The formula for resistors in series is $R_{eq}=R_1+R_2+R_3...$ .

Plug in our given values.

$R_{eq}=R_1+R_2+R_3...$

$R_{eq}=14\Omega+2.2\Omega+13\Omega$

$R_{eq}=29.2\Omega$

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