Question

Sketch the circuit labeling the meter and bulb as two separate resistors connected in parallel to the voltage source. Then 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 shows why measuring voltage with a multimeter does not affect the circuit.

Answer 1

Answer:

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Explanation:

Let's call V the voltage provided by the battery in the circuit. M is the multimeter (let's call $R_M$ its internal resistance) and R indicates the resistance of the light bulb.

We know that the meter's internal resistance is 1000 times higher than the bulb's resistance:

$R_M = 1000 R$ (1)

Both  the meter and the bulb are connected in parallel to the battery, so they both have same potential difference at their terminals:

$V_M = V_R$

Using Ohm's law, $V=RI$, we can rewrite the previous equation as:

$R_M I_M = R I_R$

where

$I_M$ is the current in the meter

$I_R$ is the current in the bulb

Using (1), this equation becomes

$(1000 R) I_M = R I_R \rightarrow I_M = \frac{I_R}{1000}$

so, the current in the meter is 1000 times less than through the bulb.