Here is a more advanced experimental problem encountered in research labs. As yo

Question

Here is a more advanced experimental problem encountered in research labs. As you measured the voltage drop across the light bulb using a multimeter, you may have wondered why we ignored the current loop created by the multimeter leads. To be sure, a little bit of current does get diverted through the multimeter, but the meter is made to have a high internal resistance to greatly restrict the current flowing through it. Let’s do a little “back of the envelope calculation” to see how much current flows through the meter. To do this, we can treat it like a large resistor and follow the same methods we used for the parallel bulbs, except now the two resistance values are no longer identical. Show mathematically that if the meter’s internal resistance is 1000 times higher than the bulb’s resistance, the current in the meter is 1000 times less than through the bulb. This proves why the meter has no discernible effect on the circuit.

Explanation / Answer

According to Ohm's law, V = I R     or     I = V / R

To measure the voltage across the resistor, the multimeter is connected parallel to the resistor. That means, the voltage across both the resistor and multimeter is the same. Let it be V.

Let R be the test resistor. Then current across R:        IR = V / R

Let 1000 R be the multimeter resistance. Then current across multimeter:     IM = V / (1000 R)

Clearly:    IM = IR / 1000

IM < < IR and can be neglected for practical purpose.

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