Calculating Circuit Power - AP Physics C Electricity & Magnetism

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Question

A straight copper wire has a fixed voltage applied across its length. Which of the following changes would increase the power dissipated by this wire?

Answer

Relevant equations:

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

Current and resistance are inversely proportional to one another, assuming voltage is fixed. Since , changes in current effect the power more than changes in resistance do. Thus, we need current to increase, meaning that resistance must decrease.

To decrease resistance, we could:

1. Change the material of the wire to one of lesser resistivity

2. Decrease the length of the wire

3. Increase the cross-sectional area of the wire

4. Decrease the temperature of the wire (very slight effect on resistance)

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Question

A battery is measured to have a potential of 5V. When connected to a wire with no resistors or other components, the voltage measured is 4.9V.

If the current through the wire is measured to be 2A, how much thermal energy is being lost per second as soon as the wire is connected to the battery?

Answer

First, we must know that the wire has some internal resistance . To calculate this, we need to know the potential drop through the wire, which must be the difference we saw from the initial voltage reading to the second. This value, 0.1V, we plug into Ohm's law to calculate the resistance of the wire.

$R=0.05\Omega$

The question asks for energy lost per second; this value is equivalent to the power.

Use our values to solve.

$P=(2A)^2(0.05\Omega)=0.2W$

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Question

A simple circuit contains two $12\Omega$ resistors in parallel, connected to a 20V source. What power is being provided by the source to the circuit?

Answer

The power supplied to the circuit can be calculated using the equation:

$P=IV=\frac{V^2}{R}$

To use this equation, we need to find the equivalent resistance of the circuit. Use the equation for equivalent resistance in parallel:

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

$\frac{1}{R_{eq}}=\frac{1}{12\Omega}+\frac{1}{12\Omega}=\frac{1}{6\Omega}$

$R_{eq}=6\Omega$

Now that we have the resistance and the voltage, we can solve for the power.

$P=\frac{V^2}{R}=\frac{(20V)^2}{6\Omega}\approx 67W$

Compare your answer with the correct one above

Question

A straight copper wire has a fixed voltage applied across its length. Which of the following changes would increase the power dissipated by this wire?

Answer

Relevant equations:

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

Current and resistance are inversely proportional to one another, assuming voltage is fixed. Since , changes in current effect the power more than changes in resistance do. Thus, we need current to increase, meaning that resistance must decrease.

To decrease resistance, we could:

1. Change the material of the wire to one of lesser resistivity

2. Decrease the length of the wire

3. Increase the cross-sectional area of the wire

4. Decrease the temperature of the wire (very slight effect on resistance)

Compare your answer with the correct one above

Question

A battery is measured to have a potential of 5V. When connected to a wire with no resistors or other components, the voltage measured is 4.9V.

If the current through the wire is measured to be 2A, how much thermal energy is being lost per second as soon as the wire is connected to the battery?

Answer

First, we must know that the wire has some internal resistance . To calculate this, we need to know the potential drop through the wire, which must be the difference we saw from the initial voltage reading to the second. This value, 0.1V, we plug into Ohm's law to calculate the resistance of the wire.

$R=0.05\Omega$

The question asks for energy lost per second; this value is equivalent to the power.

Use our values to solve.

$P=(2A)^2(0.05\Omega)=0.2W$

Compare your answer with the correct one above

Question

A simple circuit contains two $12\Omega$ resistors in parallel, connected to a 20V source. What power is being provided by the source to the circuit?

Answer

The power supplied to the circuit can be calculated using the equation:

$P=IV=\frac{V^2}{R}$

To use this equation, we need to find the equivalent resistance of the circuit. Use the equation for equivalent resistance in parallel:

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

$\frac{1}{R_{eq}}=\frac{1}{12\Omega}+\frac{1}{12\Omega}=\frac{1}{6\Omega}$

$R_{eq}=6\Omega$

Now that we have the resistance and the voltage, we can solve for the power.

$P=\frac{V^2}{R}=\frac{(20V)^2}{6\Omega}\approx 67W$

Compare your answer with the correct one above

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