Other Circuit Principles - AP Physics 2

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Question

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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Answer

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 $6\Omega$ .

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Question

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

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Answer

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 $\textup{m}^2$})$

$\textup R= 0.0068\Omega$

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Question

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the total resistance of the given circuit.

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Answer

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:

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Question

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 reveal answer

Answer

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 $6\Omega$ .

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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 $\textup{m}^2$})$

$\textup R= 0.0068\Omega$

← Didn't Know|Knew It →

Question

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the total resistance of the given circuit.

Tap to reveal answer

Answer

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:

← Didn't Know|Knew It →

Question

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 reveal answer

Answer

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 $6\Omega$ .

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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 $\textup{m}^2$})$

$\textup R= 0.0068\Omega$

← Didn't Know|Knew It →

Question

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the total resistance of the given circuit.

Tap to reveal answer

Answer

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:

← Didn't Know|Knew It →

Question

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 reveal answer

Answer

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 $6\Omega$ .

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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 $\textup{m}^2$})$

$\textup R= 0.0068\Omega$

← Didn't Know|Knew It →

Question

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the total resistance of the given circuit.

Tap to reveal answer

Answer

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:

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