if that’s the case how does that happen? because if we say p=I^2R and the value of I decreases by the same proportion then why is the impact greater on the internal resistance? the value will be smaller ofc because r is usually smaller than R but I’m referring to the decrease
Question
if that’s the case how does that happen? because if we say p=I^2R and the value of I decreases by the same proportion then why is the impact greater on the internal resistance? the value will be smaller ofc because r is usually smaller than R but I’m referring to the decrease
Solution
The question seems to be about the relationship between power (P), current (I), and resistance (R) in an electrical circuit, as described by the formula P=I^2R.
The question is asking why a decrease in current (I) has a greater impact on the internal resistance (r) than on the total resistance (R).
Here's a step-by-step explanation:
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The formula P=I^2R tells us that power is directly proportional to the square of the current and the resistance. This means that if the current decreases, the power will decrease, but not linearly. The decrease will be squared because of the I^2 term in the formula.
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Now, let's consider the internal resistance (r). If the current decreases, the power dissipated across the internal resistance will also decrease. However, because of the I^2 term, this decrease will be more significant than the decrease in power across the total resistance (R).
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This is because the power dissipated across the internal resistance is a smaller proportion of the total power to begin with (since r is usually smaller than R). So, when the current decreases, this smaller proportion decreases even more significantly.
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Therefore, a decrease in current has a greater impact on the internal resistance than on the total resistance.
I hope this helps! Let me know if you have any other questions.
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