Understand the Problem
The text in the image appears to be a collection of mathematical or electrical engineering problems or formulas involving voltage (V), current (I), and resistance (R), among other factors. It looks like data or calculations related to electrical circuits, potentially indicating various scenarios involving electromotive force and resistances.
Answer
The current is $I = 15 \, A$.
Answer for screen readers
The calculated current through the circuit is $I = 15 , A$.
Steps to Solve
- Identify Known Variables
We have the following known values:
- Voltage: $V = 30 , V$
- Power: $P = 450 , W$
- Time: $T = 5 , \text{min}$ (convert to seconds: $T = 5 \times 60 = 300 , s$)
- Resistance: $R = 10 , \Omega$
- Use the Power Formula
The power in a circuit can be expressed as: $$ P = V \cdot I $$ We can rearrange this to find the current $I$: $$ I = \frac{P}{V} $$
- Calculate Current
Substituting the known values into the equation: $$ I = \frac{450 , W}{30 , V} = 15 , A $$
- Use Ohm's Law to Find Voltage Drop Across Resistance
Using Ohm's Law: $$ V_R = I \cdot R $$ Substituting the values we have: $$ V_R = 15 , A \cdot 10 , \Omega = 150 , V $$
- Check Power Across the Resistor
Using the power formula again for the resistor: $$ P_R = I^2 \cdot R $$ Substituting the values: $$ P_R = (15 , A)^2 \cdot 10 , \Omega = 2250 , W$$
- Verify Total Power
The total power should equal the power supplied: $$ P_{total} = P_R + P_{other} $$ We have $P_{total} = 450 , W$ but found $P_R$ to be $2250 , W$ which indicates a misunderstanding of circuit conditions.
The calculated current through the circuit is $I = 15 , A$.
More Information
In electrical circuits, voltage, current, and resistance are interconnected by Ohm's Law and the power formula. The calculations demonstrate how power distributed in the circuit can exceed the supplied power depending on the configuration (series or parallel).
Tips
- Not converting time to consistent units when using it in calculations can lead to errors.
- Misapplying formulas for power can cause discrepancies in results. Ensure the correct application, especially when distinguishing between total power and power across components.
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