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2007-02-23 19:49:15 · 3 answers · asked by Dothy 1 in Science & Mathematics Physics

3 answers

Both the short-circuit current and the ½ voltage current are not practical to measure for a live source: try with, say, a car battery or the wall outlet and see the mess.

Do something fundamentally the same idea but crosses the source’s V vs. I curve tangentially rather than end-to-end (short circuit the source) or end-to-mid-point (short to ½ voltage drop).

Maybe, shunt the source with two known resistors, one at a time. Each case causes a sustainable current, and the two cases a detectable difference in the source’s voltage. The marginal source resistance is, then, (delta V) divided by (delta I), where each case's current I is the measured voltage divided by the known resistor value.

2007-02-23 21:31:14 · answer #1 · answered by sciquest 4 · 0 0

If you have a device you want to determine the internal resistance at the output, load the output with a variable resistor and measure the output voltage as you decrease the resistance. When the voltage drops to 1/2 the unloaded value, the external resistance equals the internal resistance.

You can measure input resistance using a variable resistor in series with a voltage source at the input. When the voltage at the input to th device is 1/2 the source voltage, the external resistance is equal to the internal resistance.

2007-02-24 03:59:55 · answer #2 · answered by gp4rts 7 · 0 0

If you want to know the internal resistance of, say, a battery, you need to measure its open circuit voltage and its short circuit current. R =E / I will give it to you.

2007-02-24 03:59:04 · answer #3 · answered by ZORCH 6 · 0 0

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