Other Circuit Principles - AP Physics 2

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You have 4 resistors and an ammeter arranged as shown in the diagram below.

An ammeter measures current. The ammeter in this setup reads 0A. What is the resistance of ?

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This setup is called a Wheatstone bridge. It's used to find the resistance of a resistor with an unknown value. When the ammeter reads 0, the two sides are at equipotential, so

$\frac{1}{3}$=\frac{2}{R}$ .

Therefore, the resistance of R is $6\Omega$ .

What is the resistance of a $\textup {1.5m}$ long copper wire with a diameter of $\textup {2.2mm}$ ?

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Write the following formula to find the resistance of the copper wire.

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

where $\rho$ is the resistivity in $\Omega \textup{m}$ , $\textup L$ is the length of the wire in $\textup m$ , and $\textup A$ is the cross sectional area of the wire in $\textup{m}^\textup {2}$ .

The resistivity of copper is: $1.72 * $10^{-8}$ \Omega \textup{m}$ .

$\textup L=1.5 \textup{m}$

$\textup A=\pi $r^2$$=\pi(0.0011m)^2$ = $1.2110^{-6}$ \pi $m^2$$

Substitute the givens to the resistance formula and solve.

$\textup R=(1.72 $10^{-8}$\Omega \textup{m})($\frac{1.5 \textup{m}$}{1.21*10^-6\pi $\textup{m}^2$})$

$\textup R= 0.0068\Omega$

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the total resistance of the given circuit.

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In order to find the total resistance of the circuit, we need to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with :

$$\frac{1}{R_1}$+\frac{1}{R_$2}$=\frac{1}{R_{1,2}$$}$

Combine with :

$$\frac{1}{R_3}$+\frac{1}{R_$4}$=\frac{1}{R_{3,4}$$}$

Combine with :

$$\frac{1}{R_5}$+\frac{1}{R_$6}$=\frac{1}{R_{5,6}$$}$

Then, add the combined resistors, which are now all in series:

You have 4 resistors and an ammeter arranged as shown in the diagram below.

An ammeter measures current. The ammeter in this setup reads 0A. What is the resistance of ?

Tap to see back →

This setup is called a Wheatstone bridge. It's used to find the resistance of a resistor with an unknown value. When the ammeter reads 0, the two sides are at equipotential, so

$\frac{1}{3}$=\frac{2}{R}$ .

Therefore, the resistance of R is $6\Omega$ .

What is the resistance of a $\textup {1.5m}$ long copper wire with a diameter of $\textup {2.2mm}$ ?

Tap to see back →

Write the following formula to find the resistance of the copper wire.

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

where $\rho$ is the resistivity in $\Omega \textup{m}$ , $\textup L$ is the length of the wire in $\textup m$ , and $\textup A$ is the cross sectional area of the wire in $\textup{m}^\textup {2}$ .

The resistivity of copper is: $1.72 * $10^{-8}$ \Omega \textup{m}$ .

$\textup L=1.5 \textup{m}$

$\textup A=\pi $r^2$$=\pi(0.0011m)^2$ = $1.2110^{-6}$ \pi $m^2$$

Substitute the givens to the resistance formula and solve.

$\textup R=(1.72 $10^{-8}$\Omega \textup{m})($\frac{1.5 \textup{m}$}{1.21*10^-6\pi $\textup{m}^2$})$

$\textup R= 0.0068\Omega$

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the total resistance of the given circuit.

Tap to see back →

In order to find the total resistance of the circuit, we need to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with :

$$\frac{1}{R_1}$+\frac{1}{R_$2}$=\frac{1}{R_{1,2}$$}$

Combine with :

$$\frac{1}{R_3}$+\frac{1}{R_$4}$=\frac{1}{R_{3,4}$$}$

Combine with :

$$\frac{1}{R_5}$+\frac{1}{R_$6}$=\frac{1}{R_{5,6}$$}$

Then, add the combined resistors, which are now all in series:

You have 4 resistors and an ammeter arranged as shown in the diagram below.

An ammeter measures current. The ammeter in this setup reads 0A. What is the resistance of ?

Tap to see back →

This setup is called a Wheatstone bridge. It's used to find the resistance of a resistor with an unknown value. When the ammeter reads 0, the two sides are at equipotential, so

$\frac{1}{3}$=\frac{2}{R}$ .

Therefore, the resistance of R is $6\Omega$ .

What is the resistance of a $\textup {1.5m}$ long copper wire with a diameter of $\textup {2.2mm}$ ?

Tap to see back →

Write the following formula to find the resistance of the copper wire.

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

where $\rho$ is the resistivity in $\Omega \textup{m}$ , $\textup L$ is the length of the wire in $\textup m$ , and $\textup A$ is the cross sectional area of the wire in $\textup{m}^\textup {2}$ .

The resistivity of copper is: $1.72 * $10^{-8}$ \Omega \textup{m}$ .

$\textup L=1.5 \textup{m}$

$\textup A=\pi $r^2$$=\pi(0.0011m)^2$ = $1.2110^{-6}$ \pi $m^2$$

Substitute the givens to the resistance formula and solve.

$\textup R=(1.72 $10^{-8}$\Omega \textup{m})($\frac{1.5 \textup{m}$}{1.21*10^-6\pi $\textup{m}^2$})$

$\textup R= 0.0068\Omega$

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the total resistance of the given circuit.

Tap to see back →

In order to find the total resistance of the circuit, we need to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with :

$$\frac{1}{R_1}$+\frac{1}{R_$2}$=\frac{1}{R_{1,2}$$}$

Combine with :

$$\frac{1}{R_3}$+\frac{1}{R_$4}$=\frac{1}{R_{3,4}$$}$

Combine with :

$$\frac{1}{R_5}$+\frac{1}{R_$6}$=\frac{1}{R_{5,6}$$}$

Then, add the combined resistors, which are now all in series:

You have 4 resistors and an ammeter arranged as shown in the diagram below.

An ammeter measures current. The ammeter in this setup reads 0A. What is the resistance of ?

Tap to see back →

This setup is called a Wheatstone bridge. It's used to find the resistance of a resistor with an unknown value. When the ammeter reads 0, the two sides are at equipotential, so

$\frac{1}{3}$=\frac{2}{R}$ .

Therefore, the resistance of R is $6\Omega$ .

What is the resistance of a $\textup {1.5m}$ long copper wire with a diameter of $\textup {2.2mm}$ ?

Tap to see back →

Write the following formula to find the resistance of the copper wire.

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

where $\rho$ is the resistivity in $\Omega \textup{m}$ , $\textup L$ is the length of the wire in $\textup m$ , and $\textup A$ is the cross sectional area of the wire in $\textup{m}^\textup {2}$ .

The resistivity of copper is: $1.72 * $10^{-8}$ \Omega \textup{m}$ .

$\textup L=1.5 \textup{m}$

$\textup A=\pi $r^2$$=\pi(0.0011m)^2$ = $1.2110^{-6}$ \pi $m^2$$

Substitute the givens to the resistance formula and solve.

$\textup R=(1.72 $10^{-8}$\Omega \textup{m})($\frac{1.5 \textup{m}$}{1.21*10^-6\pi $\textup{m}^2$})$

$\textup R= 0.0068\Omega$

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the total resistance of the given circuit.

Tap to see back →

In order to find the total resistance of the circuit, we need to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with :

$$\frac{1}{R_1}$+\frac{1}{R_$2}$=\frac{1}{R_{1,2}$$}$

Combine with :

$$\frac{1}{R_3}$+\frac{1}{R_$4}$=\frac{1}{R_{3,4}$$}$

Combine with :

$$\frac{1}{R_5}$+\frac{1}{R_$6}$=\frac{1}{R_{5,6}$$}$

Then, add the combined resistors, which are now all in series: