Resistivity - MCAT Chemical and Physical Foundations of Biological Systems

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Question

An electrician wishes to cut a copper wire $(\rho = 1.724 * $10^{-8}$\Omega m)$ that has no more than $10 \Omega$ of resistance. The wire has a radius of 0.725mm. Approximately what length of wire has a resistance equal to the maximum $10 \Omega$ ?

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Answer

To relate resistance R, resistivity $\rho$ , area A, and length L we use the equation $R = $\frac{\rho L}{A}$$ .

Rearranging to isolate the quantity we wish to solve for, L, gives the equation $L = $\frac{RA}{\rho }$$ . We must first solve for A using the radius, 0.725mm.

$A = \pi $r^{2}$ = \pi $(0.000725m)^{2}$=1.65 * $10^{-6}$$m^{2}$$

Plugging in our numbers gives the answer, 960m.

$L = $\frac{10\Omega * 1.65 * $10^{-6}$$$m^{2}$$}{1.724*10^{-8}$\Omega m} = 957m \approx960m$

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Question

Which of the following factors would decrease the resistance through an electrical cord?

Tap to reveal answer

Answer

The equation for resistance is given by $R = $\frac{\rho L}{A}$$ .

From this equation, we can see the best way to decrease resistance is by increasing the cross-sectional area, , of the cord. Increasing the length, , of the cord or the resistivity, $\rho$ , will increase the resistance.

← Didn't Know|Knew It →

Question

What is the current in a circuit with a $2\Omega$ resistor followed by a $3\Omega$ resistor that are both in parallel with a $5\Omega$ resistor? The voltage supplied to the circuit is 5V.

Tap to reveal answer

Answer

Resistors in serries add according to the formula:

Resistors in parallel add according to the formula:

We can find the total equivalent resistance:

$R_f=2.5\Omega$

Now we can use Ohm's law to find the current:

$I=\frac{V}{R}$$

Solve:

$I = $\frac{5 V}{2.5 \Omega }$=2 A$

← Didn't Know|Knew It →

Question

An electrician wishes to cut a copper wire $(\rho = 1.724 * $10^{-8}$\Omega m)$ that has no more than $10 \Omega$ of resistance. The wire has a radius of 0.725mm. Approximately what length of wire has a resistance equal to the maximum $10 \Omega$ ?

Tap to reveal answer

Answer

To relate resistance R, resistivity $\rho$ , area A, and length L we use the equation $R = $\frac{\rho L}{A}$$ .

Rearranging to isolate the quantity we wish to solve for, L, gives the equation $L = $\frac{RA}{\rho }$$ . We must first solve for A using the radius, 0.725mm.

$A = \pi $r^{2}$ = \pi $(0.000725m)^{2}$=1.65 * $10^{-6}$$m^{2}$$

Plugging in our numbers gives the answer, 960m.

$L = $\frac{10\Omega * 1.65 * $10^{-6}$$$m^{2}$$}{1.724*10^{-8}$\Omega m} = 957m \approx960m$

← Didn't Know|Knew It →

Question

Which of the following factors would decrease the resistance through an electrical cord?

Tap to reveal answer

Answer

The equation for resistance is given by $R = $\frac{\rho L}{A}$$ .

From this equation, we can see the best way to decrease resistance is by increasing the cross-sectional area, , of the cord. Increasing the length, , of the cord or the resistivity, $\rho$ , will increase the resistance.

← Didn't Know|Knew It →

Question

What is the current in a circuit with a $2\Omega$ resistor followed by a $3\Omega$ resistor that are both in parallel with a $5\Omega$ resistor? The voltage supplied to the circuit is 5V.

Tap to reveal answer

Answer

Resistors in serries add according to the formula:

Resistors in parallel add according to the formula:

We can find the total equivalent resistance:

$R_f=2.5\Omega$

Now we can use Ohm's law to find the current:

$I=\frac{V}{R}$$

Solve:

$I = $\frac{5 V}{2.5 \Omega }$=2 A$

← Didn't Know|Knew It →

Question

An electrician wishes to cut a copper wire $(\rho = 1.724 * $10^{-8}$\Omega m)$ that has no more than $10 \Omega$ of resistance. The wire has a radius of 0.725mm. Approximately what length of wire has a resistance equal to the maximum $10 \Omega$ ?

Tap to reveal answer

Answer

To relate resistance R, resistivity $\rho$ , area A, and length L we use the equation $R = $\frac{\rho L}{A}$$ .

Rearranging to isolate the quantity we wish to solve for, L, gives the equation $L = $\frac{RA}{\rho }$$ . We must first solve for A using the radius, 0.725mm.

$A = \pi $r^{2}$ = \pi $(0.000725m)^{2}$=1.65 * $10^{-6}$$m^{2}$$

Plugging in our numbers gives the answer, 960m.

$L = $\frac{10\Omega * 1.65 * $10^{-6}$$$m^{2}$$}{1.724*10^{-8}$\Omega m} = 957m \approx960m$

← Didn't Know|Knew It →

Question

Which of the following factors would decrease the resistance through an electrical cord?

Tap to reveal answer

Answer

The equation for resistance is given by $R = $\frac{\rho L}{A}$$ .

From this equation, we can see the best way to decrease resistance is by increasing the cross-sectional area, , of the cord. Increasing the length, , of the cord or the resistivity, $\rho$ , will increase the resistance.

← Didn't Know|Knew It →

Question

What is the current in a circuit with a $2\Omega$ resistor followed by a $3\Omega$ resistor that are both in parallel with a $5\Omega$ resistor? The voltage supplied to the circuit is 5V.

Tap to reveal answer

Answer

Resistors in serries add according to the formula:

Resistors in parallel add according to the formula:

We can find the total equivalent resistance:

$R_f=2.5\Omega$

Now we can use Ohm's law to find the current:

$I=\frac{V}{R}$$

Solve:

$I = $\frac{5 V}{2.5 \Omega }$=2 A$

← Didn't Know|Knew It →

Question

An electrician wishes to cut a copper wire $(\rho = 1.724 * $10^{-8}$\Omega m)$ that has no more than $10 \Omega$ of resistance. The wire has a radius of 0.725mm. Approximately what length of wire has a resistance equal to the maximum $10 \Omega$ ?

Tap to reveal answer

Answer

To relate resistance R, resistivity $\rho$ , area A, and length L we use the equation $R = $\frac{\rho L}{A}$$ .

Rearranging to isolate the quantity we wish to solve for, L, gives the equation $L = $\frac{RA}{\rho }$$ . We must first solve for A using the radius, 0.725mm.

$A = \pi $r^{2}$ = \pi $(0.000725m)^{2}$=1.65 * $10^{-6}$$m^{2}$$

Plugging in our numbers gives the answer, 960m.

$L = $\frac{10\Omega * 1.65 * $10^{-6}$$$m^{2}$$}{1.724*10^{-8}$\Omega m} = 957m \approx960m$

← Didn't Know|Knew It →

Question

Which of the following factors would decrease the resistance through an electrical cord?

Tap to reveal answer

Answer

The equation for resistance is given by $R = $\frac{\rho L}{A}$$ .

From this equation, we can see the best way to decrease resistance is by increasing the cross-sectional area, , of the cord. Increasing the length, , of the cord or the resistivity, $\rho$ , will increase the resistance.

← Didn't Know|Knew It →

Question

What is the current in a circuit with a $2\Omega$ resistor followed by a $3\Omega$ resistor that are both in parallel with a $5\Omega$ resistor? The voltage supplied to the circuit is 5V.

Tap to reveal answer

Answer

Resistors in serries add according to the formula:

Resistors in parallel add according to the formula:

We can find the total equivalent resistance:

$R_f=2.5\Omega$

Now we can use Ohm's law to find the current:

$I=\frac{V}{R}$$

Solve:

$I = $\frac{5 V}{2.5 \Omega }$=2 A$

← Didn't Know|Knew It →

Question

An electrician wishes to cut a copper wire $(\rho = 1.724 * $10^{-8}$\Omega m)$ that has no more than $10 \Omega$ of resistance. The wire has a radius of 0.725mm. Approximately what length of wire has a resistance equal to the maximum $10 \Omega$ ?

Tap to reveal answer

Answer

To relate resistance R, resistivity $\rho$ , area A, and length L we use the equation $R = $\frac{\rho L}{A}$$ .

Rearranging to isolate the quantity we wish to solve for, L, gives the equation $L = $\frac{RA}{\rho }$$ . We must first solve for A using the radius, 0.725mm.

$A = \pi $r^{2}$ = \pi $(0.000725m)^{2}$=1.65 * $10^{-6}$$m^{2}$$

Plugging in our numbers gives the answer, 960m.

$L = $\frac{10\Omega * 1.65 * $10^{-6}$$$m^{2}$$}{1.724*10^{-8}$\Omega m} = 957m \approx960m$

← Didn't Know|Knew It →

Question

Which of the following factors would decrease the resistance through an electrical cord?

Tap to reveal answer

Answer

The equation for resistance is given by $R = $\frac{\rho L}{A}$$ .

From this equation, we can see the best way to decrease resistance is by increasing the cross-sectional area, , of the cord. Increasing the length, , of the cord or the resistivity, $\rho$ , will increase the resistance.

← Didn't Know|Knew It →

Question

What is the current in a circuit with a $2\Omega$ resistor followed by a $3\Omega$ resistor that are both in parallel with a $5\Omega$ resistor? The voltage supplied to the circuit is 5V.

Tap to reveal answer

Answer

Resistors in serries add according to the formula:

Resistors in parallel add according to the formula:

We can find the total equivalent resistance:

$R_f=2.5\Omega$

Now we can use Ohm's law to find the current:

$I=\frac{V}{R}$$

Solve:

$I = $\frac{5 V}{2.5 \Omega }$=2 A$

← Didn't Know|Knew It →

Question

An electrician wishes to cut a copper wire $(\rho = 1.724 * $10^{-8}$\Omega m)$ that has no more than $10 \Omega$ of resistance. The wire has a radius of 0.725mm. Approximately what length of wire has a resistance equal to the maximum $10 \Omega$ ?

Tap to reveal answer

Answer

To relate resistance R, resistivity $\rho$ , area A, and length L we use the equation $R = $\frac{\rho L}{A}$$ .

Rearranging to isolate the quantity we wish to solve for, L, gives the equation $L = $\frac{RA}{\rho }$$ . We must first solve for A using the radius, 0.725mm.

$A = \pi $r^{2}$ = \pi $(0.000725m)^{2}$=1.65 * $10^{-6}$$m^{2}$$

Plugging in our numbers gives the answer, 960m.

$L = $\frac{10\Omega * 1.65 * $10^{-6}$$$m^{2}$$}{1.724*10^{-8}$\Omega m} = 957m \approx960m$

← Didn't Know|Knew It →

Question

Which of the following factors would decrease the resistance through an electrical cord?

Tap to reveal answer

Answer

The equation for resistance is given by $R = $\frac{\rho L}{A}$$ .

From this equation, we can see the best way to decrease resistance is by increasing the cross-sectional area, , of the cord. Increasing the length, , of the cord or the resistivity, $\rho$ , will increase the resistance.

← Didn't Know|Knew It →

Question

What is the current in a circuit with a $2\Omega$ resistor followed by a $3\Omega$ resistor that are both in parallel with a $5\Omega$ resistor? The voltage supplied to the circuit is 5V.

Tap to reveal answer

Answer

Resistors in serries add according to the formula:

Resistors in parallel add according to the formula:

We can find the total equivalent resistance:

$R_f=2.5\Omega$

Now we can use Ohm's law to find the current:

$I=\frac{V}{R}$$

Solve:

$I = $\frac{5 V}{2.5 \Omega }$=2 A$

← Didn't Know|Knew It →

Question

An electrician wishes to cut a copper wire $(\rho = 1.724 * $10^{-8}$\Omega m)$ that has no more than $10 \Omega$ of resistance. The wire has a radius of 0.725mm. Approximately what length of wire has a resistance equal to the maximum $10 \Omega$ ?

Tap to reveal answer

Answer

To relate resistance R, resistivity $\rho$ , area A, and length L we use the equation $R = $\frac{\rho L}{A}$$ .

Rearranging to isolate the quantity we wish to solve for, L, gives the equation $L = $\frac{RA}{\rho }$$ . We must first solve for A using the radius, 0.725mm.

$A = \pi $r^{2}$ = \pi $(0.000725m)^{2}$=1.65 * $10^{-6}$$m^{2}$$

Plugging in our numbers gives the answer, 960m.

$L = $\frac{10\Omega * 1.65 * $10^{-6}$$$m^{2}$$}{1.724*10^{-8}$\Omega m} = 957m \approx960m$

← Didn't Know|Knew It →

Question

Which of the following factors would decrease the resistance through an electrical cord?

Tap to reveal answer

Answer

The equation for resistance is given by $R = $\frac{\rho L}{A}$$ .

From this equation, we can see the best way to decrease resistance is by increasing the cross-sectional area, , of the cord. Increasing the length, , of the cord or the resistivity, $\rho$ , will increase the resistance.

← Didn't Know|Knew It →

Knew It

Resistivity - MCAT Chemical and Physical Foundations of Biological Systems

An electrician wishes to cut a copper wire that has no more than of resistance. The wire has a radius of 0.725mm. Approximately what length of wire has a resistance equal to the maximum ?

To relate resistance R, resistivity , area A, and length L we use the equation. Rearranging to isolate the quantity we wish to solve for, L, gives the equation . We must first solve for A using the radius, 0.725mm. Plugging in our numbers gives the answer, 960m.

Which of the following factors would decrease the resistance through an electrical cord?

The equation for resistance is given by . From this equation, we can see the best way to decrease resistance is by increasing the cross-sectional area, , of the cord. Increasing the length, , of the cord or the resistivity, , will increase the resistance.

What is the current in a circuit with a resistor followed by a resistor that are both in parallel with a resistor? The voltage supplied to the circuit is 5V.

Resistors in serries add according to the formula: Resistors in parallel add according to the formula: We can find the total equivalent resistance: Now we can use Ohm's law to find the current: Solve:

An electrician wishes to cut a copper wire that has no more than of resistance. The wire has a radius of 0.725mm. Approximately what length of wire has a resistance equal to the maximum ?

To relate resistance R, resistivity , area A, and length L we use the equation. Rearranging to isolate the quantity we wish to solve for, L, gives the equation . We must first solve for A using the radius, 0.725mm. Plugging in our numbers gives the answer, 960m.

Which of the following factors would decrease the resistance through an electrical cord?

The equation for resistance is given by . From this equation, we can see the best way to decrease resistance is by increasing the cross-sectional area, , of the cord. Increasing the length, , of the cord or the resistivity, , will increase the resistance.

What is the current in a circuit with a resistor followed by a resistor that are both in parallel with a resistor? The voltage supplied to the circuit is 5V.

Resistors in serries add according to the formula: Resistors in parallel add according to the formula: We can find the total equivalent resistance: Now we can use Ohm's law to find the current: Solve:

An electrician wishes to cut a copper wire that has no more than of resistance. The wire has a radius of 0.725mm. Approximately what length of wire has a resistance equal to the maximum ?

To relate resistance R, resistivity , area A, and length L we use the equation. Rearranging to isolate the quantity we wish to solve for, L, gives the equation . We must first solve for A using the radius, 0.725mm. Plugging in our numbers gives the answer, 960m.

Which of the following factors would decrease the resistance through an electrical cord?

The equation for resistance is given by . From this equation, we can see the best way to decrease resistance is by increasing the cross-sectional area, , of the cord. Increasing the length, , of the cord or the resistivity, , will increase the resistance.

What is the current in a circuit with a resistor followed by a resistor that are both in parallel with a resistor? The voltage supplied to the circuit is 5V.

Resistors in serries add according to the formula: Resistors in parallel add according to the formula: We can find the total equivalent resistance: Now we can use Ohm's law to find the current: Solve:

An electrician wishes to cut a copper wire that has no more than of resistance. The wire has a radius of 0.725mm. Approximately what length of wire has a resistance equal to the maximum ?

To relate resistance R, resistivity , area A, and length L we use the equation. Rearranging to isolate the quantity we wish to solve for, L, gives the equation . We must first solve for A using the radius, 0.725mm. Plugging in our numbers gives the answer, 960m.

Which of the following factors would decrease the resistance through an electrical cord?

The equation for resistance is given by . From this equation, we can see the best way to decrease resistance is by increasing the cross-sectional area, , of the cord. Increasing the length, , of the cord or the resistivity, , will increase the resistance.

What is the current in a circuit with a resistor followed by a resistor that are both in parallel with a resistor? The voltage supplied to the circuit is 5V.

Resistors in serries add according to the formula: Resistors in parallel add according to the formula: We can find the total equivalent resistance: Now we can use Ohm's law to find the current: Solve:

An electrician wishes to cut a copper wire that has no more than of resistance. The wire has a radius of 0.725mm. Approximately what length of wire has a resistance equal to the maximum ?

To relate resistance R, resistivity , area A, and length L we use the equation. Rearranging to isolate the quantity we wish to solve for, L, gives the equation . We must first solve for A using the radius, 0.725mm. Plugging in our numbers gives the answer, 960m.

Which of the following factors would decrease the resistance through an electrical cord?

The equation for resistance is given by . From this equation, we can see the best way to decrease resistance is by increasing the cross-sectional area, , of the cord. Increasing the length, , of the cord or the resistivity, , will increase the resistance.

What is the current in a circuit with a resistor followed by a resistor that are both in parallel with a resistor? The voltage supplied to the circuit is 5V.

Resistors in serries add according to the formula: Resistors in parallel add according to the formula: We can find the total equivalent resistance: Now we can use Ohm's law to find the current: Solve:

An electrician wishes to cut a copper wire that has no more than of resistance. The wire has a radius of 0.725mm. Approximately what length of wire has a resistance equal to the maximum ?

To relate resistance R, resistivity , area A, and length L we use the equation. Rearranging to isolate the quantity we wish to solve for, L, gives the equation . We must first solve for A using the radius, 0.725mm. Plugging in our numbers gives the answer, 960m.

Which of the following factors would decrease the resistance through an electrical cord?

The equation for resistance is given by . From this equation, we can see the best way to decrease resistance is by increasing the cross-sectional area, , of the cord. Increasing the length, , of the cord or the resistivity, , will increase the resistance.

What is the current in a circuit with a resistor followed by a resistor that are both in parallel with a resistor? The voltage supplied to the circuit is 5V.

Resistors in serries add according to the formula: Resistors in parallel add according to the formula: We can find the total equivalent resistance: Now we can use Ohm's law to find the current: Solve:

An electrician wishes to cut a copper wire that has no more than of resistance. The wire has a radius of 0.725mm. Approximately what length of wire has a resistance equal to the maximum ?

To relate resistance R, resistivity , area A, and length L we use the equation. Rearranging to isolate the quantity we wish to solve for, L, gives the equation . We must first solve for A using the radius, 0.725mm. Plugging in our numbers gives the answer, 960m.

Which of the following factors would decrease the resistance through an electrical cord?

The equation for resistance is given by . From this equation, we can see the best way to decrease resistance is by increasing the cross-sectional area, , of the cord. Increasing the length, , of the cord or the resistivity, , will increase the resistance.

An electrician wishes to cut a copper wire $(\rho = 1.724 * $10^{-8}$\Omega m)$ that has no more than $10 \Omega$ of resistance. The wire has a radius of 0.725mm. Approximately what length of wire has a resistance equal to the maximum $10 \Omega$ ?

The equation for resistance is given by $R = $\frac{\rho L}{A}$$ .

From this equation, we can see the best way to decrease resistance is by increasing the cross-sectional area, , of the cord. Increasing the length, , of the cord or the resistivity, $\rho$ , will increase the resistance.

What is the current in a circuit with a $2\Omega$ resistor followed by a $3\Omega$ resistor that are both in parallel with a $5\Omega$ resistor? The voltage supplied to the circuit is 5V.

Resistors in serries add according to the formula:

Resistors in parallel add according to the formula:

We can find the total equivalent resistance:

$R_f=2.5\Omega$

Now we can use Ohm's law to find the current:

$I=\frac{V}{R}$$

Solve:

$I = $\frac{5 V}{2.5 \Omega }$=2 A$

An electrician wishes to cut a copper wire $(\rho = 1.724 * $10^{-8}$\Omega m)$ that has no more than $10 \Omega$ of resistance. The wire has a radius of 0.725mm. Approximately what length of wire has a resistance equal to the maximum $10 \Omega$ ?

The equation for resistance is given by $R = $\frac{\rho L}{A}$$ .

From this equation, we can see the best way to decrease resistance is by increasing the cross-sectional area, , of the cord. Increasing the length, , of the cord or the resistivity, $\rho$ , will increase the resistance.

What is the current in a circuit with a $2\Omega$ resistor followed by a $3\Omega$ resistor that are both in parallel with a $5\Omega$ resistor? The voltage supplied to the circuit is 5V.

Resistors in serries add according to the formula:

Resistors in parallel add according to the formula:

We can find the total equivalent resistance:

$R_f=2.5\Omega$

Now we can use Ohm's law to find the current:

$I=\frac{V}{R}$$

Solve:

$I = $\frac{5 V}{2.5 \Omega }$=2 A$

An electrician wishes to cut a copper wire $(\rho = 1.724 * $10^{-8}$\Omega m)$ that has no more than $10 \Omega$ of resistance. The wire has a radius of 0.725mm. Approximately what length of wire has a resistance equal to the maximum $10 \Omega$ ?

The equation for resistance is given by $R = $\frac{\rho L}{A}$$ .

From this equation, we can see the best way to decrease resistance is by increasing the cross-sectional area, , of the cord. Increasing the length, , of the cord or the resistivity, $\rho$ , will increase the resistance.

What is the current in a circuit with a $2\Omega$ resistor followed by a $3\Omega$ resistor that are both in parallel with a $5\Omega$ resistor? The voltage supplied to the circuit is 5V.

Resistors in serries add according to the formula:

Resistors in parallel add according to the formula:

We can find the total equivalent resistance:

$R_f=2.5\Omega$

Now we can use Ohm's law to find the current:

$I=\frac{V}{R}$$

Solve:

$I = $\frac{5 V}{2.5 \Omega }$=2 A$

An electrician wishes to cut a copper wire $(\rho = 1.724 * $10^{-8}$\Omega m)$ that has no more than $10 \Omega$ of resistance. The wire has a radius of 0.725mm. Approximately what length of wire has a resistance equal to the maximum $10 \Omega$ ?

The equation for resistance is given by $R = $\frac{\rho L}{A}$$ .

From this equation, we can see the best way to decrease resistance is by increasing the cross-sectional area, , of the cord. Increasing the length, , of the cord or the resistivity, $\rho$ , will increase the resistance.

What is the current in a circuit with a $2\Omega$ resistor followed by a $3\Omega$ resistor that are both in parallel with a $5\Omega$ resistor? The voltage supplied to the circuit is 5V.

Resistors in serries add according to the formula:

Resistors in parallel add according to the formula:

We can find the total equivalent resistance:

$R_f=2.5\Omega$

Now we can use Ohm's law to find the current:

$I=\frac{V}{R}$$

Solve:

$I = $\frac{5 V}{2.5 \Omega }$=2 A$

An electrician wishes to cut a copper wire $(\rho = 1.724 * $10^{-8}$\Omega m)$ that has no more than $10 \Omega$ of resistance. The wire has a radius of 0.725mm. Approximately what length of wire has a resistance equal to the maximum $10 \Omega$ ?

The equation for resistance is given by $R = $\frac{\rho L}{A}$$ .

From this equation, we can see the best way to decrease resistance is by increasing the cross-sectional area, , of the cord. Increasing the length, , of the cord or the resistivity, $\rho$ , will increase the resistance.

What is the current in a circuit with a $2\Omega$ resistor followed by a $3\Omega$ resistor that are both in parallel with a $5\Omega$ resistor? The voltage supplied to the circuit is 5V.

Resistors in serries add according to the formula:

Resistors in parallel add according to the formula:

We can find the total equivalent resistance:

$R_f=2.5\Omega$

Now we can use Ohm's law to find the current:

$I=\frac{V}{R}$$

Solve:

$I = $\frac{5 V}{2.5 \Omega }$=2 A$

An electrician wishes to cut a copper wire $(\rho = 1.724 * $10^{-8}$\Omega m)$ that has no more than $10 \Omega$ of resistance. The wire has a radius of 0.725mm. Approximately what length of wire has a resistance equal to the maximum $10 \Omega$ ?

The equation for resistance is given by $R = $\frac{\rho L}{A}$$ .

From this equation, we can see the best way to decrease resistance is by increasing the cross-sectional area, , of the cord. Increasing the length, , of the cord or the resistivity, $\rho$ , will increase the resistance.

What is the current in a circuit with a $2\Omega$ resistor followed by a $3\Omega$ resistor that are both in parallel with a $5\Omega$ resistor? The voltage supplied to the circuit is 5V.

Resistors in serries add according to the formula:

Resistors in parallel add according to the formula:

We can find the total equivalent resistance:

$R_f=2.5\Omega$

Now we can use Ohm's law to find the current:

$I=\frac{V}{R}$$

Solve:

$I = $\frac{5 V}{2.5 \Omega }$=2 A$

An electrician wishes to cut a copper wire $(\rho = 1.724 * $10^{-8}$\Omega m)$ that has no more than $10 \Omega$ of resistance. The wire has a radius of 0.725mm. Approximately what length of wire has a resistance equal to the maximum $10 \Omega$ ?

The equation for resistance is given by $R = $\frac{\rho L}{A}$$ .

From this equation, we can see the best way to decrease resistance is by increasing the cross-sectional area, , of the cord. Increasing the length, , of the cord or the resistivity, $\rho$ , will increase the resistance.