Solve for V P=(V^2)/R
The question presented relates to electric circuits and requires the application of Ohm's law, and specifically, it involves restructuring the equation to solve for voltage (V). Ohm's law states that the power (P) dissipated in an electrical resistor is proportional to the square of the voltage (V) across the resistor, divided by the resistance (R) of the resistor. The question asks to rearrange the provided formula to isolate and solve for the voltage (V) in terms of the power (P) and resistance (R).
$P = \frac{V^{2}}{R}$
Answer
Solution:
Step 1:
Express the given equation in the form $\frac{V^2}{R} = P$.
Step 2:
To isolate $V^2$, multiply both sides by $R$ to get $V^2 = PR$.
Step 3:
Proceed to simplify the equation.
Step 3.1:
Eliminate the $R$ on the left side by canceling it out.
Step 3.1.1:
Perform the cancellation: $V^2 = PR$.
Step 3.1.2:
State the simplified equation: $V^2 = PR$.
Step 4:
Now, we need to find the value of $V$.
Step 4.1:
Apply the square root to both sides of the equation to solve for $V$: $V = \pm\sqrt{PR}$.
Step 4.2:
Consider both the positive and negative square roots for the complete solution.
Step 4.2.1:
For the positive root: $V = \sqrt{PR}$.
Step 4.2.2:
For the negative root: $V = -\sqrt{PR}$.
Step 4.2.3:
Combine both solutions to present the full set of possible values for $V$: $V = \sqrt{PR}$ or $V = -\sqrt{PR}$.
Knowledge Notes:
The problem involves solving for the variable $V$ in the equation $P = \frac{V^2}{R}$, which represents the power dissipated by a resistor in an electrical circuit, where $P$ is the power in watts, $V$ is the voltage in volts, and $R$ is the resistance in ohms.
The relevant knowledge points for solving this problem include:
Algebraic manipulation: The ability to rearrange equations and solve for a particular variable is fundamental in algebra. This includes multiplying both sides of an equation by a term to eliminate fractions and taking roots to solve for variables raised to a power.
Square roots: The square root function is the inverse of squaring a number. When we have $V^2 = PR$, we take the square root of both sides to solve for $V$. Since squaring a positive or negative number yields the same result, we consider both the positive and negative square roots.
Physical interpretation: In the context of electrical circuits, only the positive square root is typically considered for voltage, as voltage is usually defined as a scalar quantity. However, in mathematics, both roots are valid solutions to the equation.
Units and dimensional analysis: When solving equations in physics or engineering, it's important to ensure that the units on both sides of the equation match and that the final answer is given in the correct units. In this case, the units for $V$ should be volts.
By understanding these concepts, one can solve a variety of problems involving algebraic equations in both pure mathematics and applied sciences.
