The power in a circuit is determined by the equation \(\displaystyle P=I^{2}R\), where \(\displaystyle P\) is the power of the circuit, \(\displaystyle I\) is the current in the circuit, and \(\displaystyle R\) is the equivalent  resistance of the circuit.

Since we are given resistance and voltage, we will also need Ohm's law, \(\displaystyle V=IR\), where \(\displaystyle V\) is the voltage, \(\displaystyle I\) is the current in the circuit, and \(\displaystyle R\) is the equivalent  resistance of the circuit.

Solving Ohm's law for current gives us \(\displaystyle I=\frac{V}{R}\).

Substituting this form of Ohm's law into the power equation gives us

\(\displaystyle P=I^{2}R=\left ( \frac{V}{R} \right )^{2}R=\frac{V^{2}}{R} \right\)

The power equation is now in a form that we can solve with the information we are given.

\(\displaystyle P=\frac{V^{2}}{R} \right=\frac{3^{2}}{10} \right=0.9W\)

The power in a circuit is determined by the equation \(\displaystyle P=I^{2}R\), where \(\displaystyle P\) is the power of the circuit, \(\displaystyle I\) is the current in the circuit, and \(\displaystyle R\) is the equivalent  resistance of the circuit.

To relate power to voltage, we will also need Ohm's law, \(\displaystyle V=IR\), where \(\displaystyle V\) is the voltage, \(\displaystyle I\) is the current in the circuit, and \(\displaystyle R\) is the equivalent  resistance of the circuit.

Solving Ohm's law for current gives us \(\displaystyle I=\frac{V}{R}\).

Substituting this form of Ohm's law into the power equation gives us

\(\displaystyle P=I^{2}R=\left ( \frac{V}{R} \right )^{2}R=\frac{V^{2}}{R} \right\)

\(\displaystyle P=\frac{V^{2}}{R} \right\)

Assuming the current stays the same, if the voltage is doubled, the power will be four times larger.

Expressed mathematically,

If \(\displaystyle V=2V\)

\(\displaystyle P=\frac{(2V)^{2}}{R} \right=\frac{4(V)^{2}}{R} \right=4P\)

The power in a circuit is determined by the equation \(\displaystyle P=I^{2}R\), where \(\displaystyle P\) is the power of the circuit, \(\displaystyle I\) is the current in the circuit, and \(\displaystyle R\) is the equivalent  resistance of the circuit.

To relate power to voltage, we will also need Ohm's law, \(\displaystyle V=IR\), where \(\displaystyle V\) is the voltage, \(\displaystyle I\) is the current in the circuit, and \(\displaystyle R\) is the equivalent  resistance of the circuit.

Solving Ohm's law for current gives us \(\displaystyle I=\frac{V}{R}\).

Substituting this form of Ohm's law into the power equation gives us

\(\displaystyle P=I^{2}R=\left ( \frac{V}{R} \right )^{2}R=\frac{V^{2}}{R} \right\)

\(\displaystyle P=\frac{V^{2}}{R} \right\)

Energy is related to power by the equation \(\displaystyle E=P\cdot t\), where \(\displaystyle E\) is energy, \(\displaystyle P\) is the power of the circuit, and \(\displaystyle t\) is time.

Substituting theequation for power into the equation for energy gives

\(\displaystyle E=P\cdot t=\frac{V^{2}}{R} t\)

For our problem

\(\displaystyle E=P\cdot t=\frac{V^{2}}{R} t=\frac{(110)^{2}}{10} (5)=6050 J\)

The power in a circuit is determined by the equation \(\displaystyle P=I^{2}R\), where \(\displaystyle P\) is the power of the circuit, \(\displaystyle I\) is the current in the circuit, and \(\displaystyle R\) is the equivalent  resistance of the circuit.

Solving the power equation for current gives 

\(\displaystyle I=\sqrt{\frac{P}{R}}\)

In this problem, \(\displaystyle I=\sqrt{\frac{P}{R}}=\sqrt{\frac{60}{15}}=\sqrt{4}=2Amps\)

Energy is related to power by the equation \(\displaystyle E=P\cdot t\), where \(\displaystyle E\) is energy, \(\displaystyle P\) is the power of the circuit, and \(\displaystyle t\) is time.

Time needs to be in seconds.

\(\displaystyle 5min\times \frac{60sec}{1min}=300 sec\)

For our problem, energy is

\(\displaystyle E=75\cdot(300)=22500J\)

Recall the formula for circuit power:

\(\displaystyle P=\frac{V^2}{R}\) 

We know that \(\displaystyle V=50V\). To determine resistance of 3 resistors in parallel we set up the equation for total resistance for resistors in parallel:

\(\displaystyle \frac{1}{R_{total}}=\frac{1}{3 \Omega}+\frac{1}{3 \Omega}+\frac{1}{3 \Omega}\)

\(\displaystyle \frac{1}{R_{total}}=3*(\frac{1}{3 \Omega})\)

\(\displaystyle R_{total}=1\Omega\)

From our formula we know that the circuit power is:

\(\displaystyle P=\frac{(50V)^2}{1\Omega}=2500 W\)

We can write the power of a circuit as: 

\(\displaystyle P=IV\)

This however, does not help much because we do not know the current in the circuit. But by Ohm's law: 

\(\displaystyle V=IR\Rightarrow I= \frac{V}{R}\)

Substitution of this expression into the expression for power gives us:

\(\displaystyle P = \frac{V^2}{R}\)

Where \(\displaystyle R= R_{tot}\), the total resistance of the circuit. For a circuit in series, the total resistance is simply the sum of the resistors:

\(\displaystyle R_{tot}=R_1+R_2+R_3 = (10+3+7) \Omega=20 \Omega\) 

Therefore, the power of the circuit is:

\(\displaystyle P=\frac{(18V)^2}{20 \Omega}=16.2W\)

The power \(\displaystyle P\) of a circuit is given by:

\(\displaystyle P=IV\).

By substitution of Ohm's law, we can show that:

\(\displaystyle P=\frac{V^2}{R}\)

Where \(\displaystyle R=R_{tot}\), or the total resistance of the circuit. For resistors in parallel, we add the inverses of their values. This is shown more plainly as: 

\(\displaystyle \frac{1}{R_{tot}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\) 

From the given values, we can calculate the total resistance as \(\displaystyle R_{tot}=2.86 \Omega\). This leads us to the final answer:

\(\displaystyle P=\frac{(12V)^2}{2.86 \Omega}=50.35 W\)

The power provided by the battery is equal to the product of the current that passes through it and the voltage it provides. In order to find the current, you must find the equivalent resistance of the circuit.

Since R1 is in series with R2 and R3, who are in parallel, the equivalent resistance is given by: 

\(\displaystyle R_{equivalent}=R1+\frac{R2*R3}{R2+R3}=5k\Omega+8k\Omega+12k\Omega=25k\Omega\)

Then, to find power, you can use the following formula:

\(\displaystyle P=IV=\frac{V^2}{R}=\frac{10^2}{25k\Omega}=4mW\)

The battery supplies \(\displaystyle 4mW\) of power.

The equation for power in a circuit is given as:

\(\displaystyle P = I^{2}R\)

\(\displaystyle P\) is the power in Watts, \(\displaystyle I\) is the current in amps, and \(\displaystyle R\) is the resistance in ohms. By solving the equation and substituting our values, we obtain:

\(\displaystyle P = (2 A)^{2}(10\Omega ) = 40 W\)