Using Ohm's Law - AP Physics C Electricity & Magnetism

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Question

A simple circuit consists of a $2\ \text{M}\Omega$ resistor that's connected to a battery. How much power is dissipated by the resistor?

Answer

One of the formulas for power is $P=\frac{V^2}{R}$ . We are given the values of the voltage and resistance.

$V=9\ \text{V}$

$R=2\ \text{M}\Omega=210^6\ \Omega$

Using these values, we can solve for power.

$P=\frac{(9\ \text{V})^2}{210^6\ \Omega}$

$P=4*10^{-5}\ \text{W}$

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Question

$R_1=15\Omega,\ R_2=20\Omega,\ R_3=20\Omega$

What is the current delivered to the curcuit by the battery?

Answer

Using Ohm's law to solve for the value of the current from the battery requires calculation of the equivalent resistance of the circuit.

$V=IR_{eq}$

The resistors R2 and R3 are in parallel with one another. Once combined, their requivalent resistor (R23) is in series with R1.

The equivalent resistance is thus given by:

$R_{eq}=\left ( \frac{1}{R_{2}} + \frac{1}{R_{3}} \right )^{-1} + R_{1}$

$R_{eq}=\left ( \frac{1}{20\Omega} + \frac{1}{20\Omega} \right )^{-1} + 15\Omega$

$R_{eq}=10\Omega + 15\Omega=25\Omega$

Use this value and the given voltage to solve for the current in the circuit:

$I = \frac{V}{R_{eq}}$

$I=\frac{50V}{25\Omega}=2.0A$

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Question

A cicuit with a voltage source of 40V has a $15\Omega$ resistor and a $5\Omega$ resistor in series, followed by two $5\Omega$ resistors in parallel. After the parallel branches rejoin, there is a $10\Omega$ resistor before the circuit is closed at the voltage source.

What is the current supplied by the source?

Answer

The current supplied by the source can be calculated using a derivation of Ohm's law:

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

Start by finding the equivalent resistance of the circuit.

Sum the first two resistors in series:

$R_{12}=15\Omega+5\Omega=20\Omega$

Calculate the equivalent resistance of the two resistors in parallel:

$\frac{1}{R_{34}}=\frac{1}{5\Omega}+\frac{1}{5\Omega}=\frac{2}{5\Omega}$

$R_{34}=\frac{5}{2}\Omega=2.5\Omega$

Now, all of the reistances can be viewed as being in series.

$R_{eq}=R_{12}+R_{34}+R_{5}$

$R_{eq}=20\Omega+2.5\Omega+10\Omega=32.5\Omega$

Returning to our current calulation, we can find our final answer:

$I=\frac{V}{R}=\frac{40V}{32.5\Omega}=1.2A$

Compare your answer with the correct one above

Question

A simple circuit consists of a $2\ \text{M}\Omega$ resistor that's connected to a battery. How much power is dissipated by the resistor?

Answer

One of the formulas for power is $P=\frac{V^2}{R}$ . We are given the values of the voltage and resistance.

$V=9\ \text{V}$

$R=2\ \text{M}\Omega=210^6\ \Omega$

Using these values, we can solve for power.

$P=\frac{(9\ \text{V})^2}{210^6\ \Omega}$

$P=4*10^{-5}\ \text{W}$

Compare your answer with the correct one above

Question

$R_1=15\Omega,\ R_2=20\Omega,\ R_3=20\Omega$

What is the current delivered to the curcuit by the battery?

Answer

Using Ohm's law to solve for the value of the current from the battery requires calculation of the equivalent resistance of the circuit.

$V=IR_{eq}$

The resistors R2 and R3 are in parallel with one another. Once combined, their requivalent resistor (R23) is in series with R1.

The equivalent resistance is thus given by:

$R_{eq}=\left ( \frac{1}{R_{2}} + \frac{1}{R_{3}} \right )^{-1} + R_{1}$

$R_{eq}=\left ( \frac{1}{20\Omega} + \frac{1}{20\Omega} \right )^{-1} + 15\Omega$

$R_{eq}=10\Omega + 15\Omega=25\Omega$

Use this value and the given voltage to solve for the current in the circuit:

$I = \frac{V}{R_{eq}}$

$I=\frac{50V}{25\Omega}=2.0A$

Compare your answer with the correct one above

Question

A cicuit with a voltage source of 40V has a $15\Omega$ resistor and a $5\Omega$ resistor in series, followed by two $5\Omega$ resistors in parallel. After the parallel branches rejoin, there is a $10\Omega$ resistor before the circuit is closed at the voltage source.

What is the current supplied by the source?

Answer

The current supplied by the source can be calculated using a derivation of Ohm's law:

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

Start by finding the equivalent resistance of the circuit.

Sum the first two resistors in series:

$R_{12}=15\Omega+5\Omega=20\Omega$

Calculate the equivalent resistance of the two resistors in parallel:

$\frac{1}{R_{34}}=\frac{1}{5\Omega}+\frac{1}{5\Omega}=\frac{2}{5\Omega}$

$R_{34}=\frac{5}{2}\Omega=2.5\Omega$

Now, all of the reistances can be viewed as being in series.

$R_{eq}=R_{12}+R_{34}+R_{5}$

$R_{eq}=20\Omega+2.5\Omega+10\Omega=32.5\Omega$

Returning to our current calulation, we can find our final answer:

$I=\frac{V}{R}=\frac{40V}{32.5\Omega}=1.2A$

Compare your answer with the correct one above

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