To solve this problem, we need to analyze the behavior of the voltage source with different load resistances. We can start by applying Ohm's Law, which states that voltage (V) is equal to the current (I) multiplied by the resistance (R), or V = I × R. Let's break down the information given and find the voltage across the source when a 15-ohm load is connected.
Understanding the Circuit Behavior
Initially, we have a 10-ohm load connected to the voltage source, and a current of 2A flows. Using Ohm's Law, we can calculate the voltage across the source:
- For the 10-ohm load: V = I × R = 2A × 10Ω = 20V
This means the voltage source provides 20 volts when a 10-ohm load is connected. Now, when we increase the load to 20 ohms, the current drops to 1A. Let's verify this:
- For the 20-ohm load: V = I × R = 1A × 20Ω = 20V
Again, we find that the voltage across the source remains 20 volts. This indicates that the voltage source is likely a constant voltage source, maintaining 20 volts regardless of the load resistance, as long as the current does not exceed its limits.
Calculating Voltage Across a 15-ohm Load
Now, let's determine the voltage when a 15-ohm load is connected. Since we established that the voltage source maintains a constant voltage of 20 volts, we can directly state:
- Voltage across 15-ohm load: 20V
Power Dissipation in the Circuit
Next, we need to calculate the power dissipated in the circuit when the 15-ohm load is connected. Power (P) can be calculated using the formula:
Substituting the values we have:
- P = 20V² / 15Ω = 400V² / 15Ω ≈ 26.67W
Thus, the power dissipated in the circuit with a 15-ohm load is approximately 26.67 watts.
Summary of Findings
To summarize, when a 15-ohm load is connected to the voltage source, the voltage across it is 20 volts, and the power dissipated in the circuit is around 26.67 watts. This analysis shows how the voltage source behaves consistently under varying load conditions, allowing us to predict the outcomes accurately.
