Problem 30
Question
Solve each formula for the quantity given. $$ \frac{V_{P}}{V_{S}}=\frac{N_{P}}{N_{S}} \text { for } N_{S} $$
Step-by-Step Solution
Verified Answer
\( N_{S} = \frac{V_{S} \cdot N_{P}}{V_{P}} \)
1Step 1: Identify the Equation
The equation given is \( \frac{V_{P}}{V_{S}} = \frac{N_{P}}{N_{S}} \). We need to solve for \( N_{S} \).
2Step 2: Cross-Multiply the Fractions
To eliminate the fractions, cross-multiply: \( V_{P} \cdot N_{S} = V_{S} \cdot N_{P} \). This step rearranges the equation to facilitate solving for \( N_{S} \).
3Step 3: Isolate \( N_{S} \)
Solve for \( N_{S} \) by dividing both sides of the equation by \( V_{P} \): \( N_{S} = \frac{V_{S} \cdot N_{P}}{V_{P}} \). This isolates the variable we are solving for.
Key Concepts
Voltage RatioTurns RatioCross-MultiplicationIsolating a Variable
Voltage Ratio
When dealing with transformers, understanding the voltage ratio is essential. The voltage ratio compares the voltages on the primary side (\( V_{P} \)) and the secondary side (\( V_{S} \)) of a transformer. This ratio is vital because it shows how voltages are transformed between the two sides.
Knowing the voltage ratio helps predict how the voltage changes from input to output.
Knowing the voltage ratio helps predict how the voltage changes from input to output.
- Primary voltage (\( V_{P} \)) is the voltage applied to the input winding.
- Secondary voltage (\( V_{S} \)) is the voltage detected in the output winding.
Turns Ratio
The turns ratio is another critical concept in transformer equations. It involves the number of turns in the primary and secondary windings of a transformer, denoted as \( N_{P} \) and \( N_{S} \) respectively. In calculations, the turns ratio is given by \( \frac{N_{P}}{N_{S}} \).
This ratio directly affects the voltage ratio of a transformer since they are proportional.
This ratio directly affects the voltage ratio of a transformer since they are proportional.
- A higher turns ratio indicates a greater difference in the voltage transformation between primary and secondary sides.
- It also determines how much a transformer can "step up" or "step down" the voltage.
Cross-Multiplication
Cross-multiplication is an algebraic method used to solve equations involving two fractions set equal to one another. This technique eliminates the fractions by multiplying across the equals sign, which simplifies the equation.
Using cross-multiplication, the equation \( \frac{V_{P}}{V_{S}} = \frac{N_{P}}{N_{S}} \) becomes \( V_{P} \cdot N_{S} = V_{S} \cdot N_{P} \).This rearrangement is essential because:
Using cross-multiplication, the equation \( \frac{V_{P}}{V_{S}} = \frac{N_{P}}{N_{S}} \) becomes \( V_{P} \cdot N_{S} = V_{S} \cdot N_{P} \).This rearrangement is essential because:
- It transforms the equation into a more straightforward form to manipulate.
- It lays the foundation for isolating a specific variable.
Isolating a Variable
Isolating a variable means rearranging an equation so that one variable stands alone on one side of the equality. This process is crucial when solving equations to find the specific value of a variable you're interested in.
In our example, we need to solve for \( N_{S} \). After cross-multiplying, we get \( V_{P} \cdot N_{S} = V_{S} \cdot N_{P} \).To isolate \( N_{S} \), divide both sides of the equation by \( V_{P} \):\[ N_{S} = \frac{V_{S} \cdot N_{P}}{V_{P}}.\]
In our example, we need to solve for \( N_{S} \). After cross-multiplying, we get \( V_{P} \cdot N_{S} = V_{S} \cdot N_{P} \).To isolate \( N_{S} \), divide both sides of the equation by \( V_{P} \):\[ N_{S} = \frac{V_{S} \cdot N_{P}}{V_{P}}.\]
- This step ensures that \( N_{S} \) is alone on one side, making it straightforward to calculate.
- Isolating a variable is often the final step when solving for a specific unknown in algebra.
Other exercises in this chapter
Problem 29
Solve each formula for the quantity given. $$ \frac{I_{S}}{I_{P}}=\frac{N_{P}}{N_{S}} \text { for } I_{P} $$
View solution Problem 30
What length of sidewalk \(4.00\) in. thick and \(4.00 \mathrm{ft}\) wide could be poured with \(2.00 \mathrm{yd}^{3}\) of concrete?
View solution Problem 31
Solve each formula for the quantity given. $$ v_{\text {avg }}=\frac{1}{2}\left(v_{f}+v_{i}\right) \text { for } v_{i} $$
View solution Problem 32
Solve each formula for the quantity given. $$ 2 a\left(s-s_{i}\right)=v^{2}-v_{i}^{2} \text { for } a $$
View solution