Calculating Current in a Circuit without an Ammeter

[saw176]

Homework Statement

An ammeter whose internal resistance is 63 ohm reads 5.25 mA when connected in a circuit comtaining a battery and two resistors in series whose values are 750 ohm and 480 ohm. What is the actual current when the ammeter is absent?

Homework Equations

V = IR
R_series = R₁ + R₂

The Attempt at a Solution

First I found the voltage using the internal resistance of the ammeter and the current:
V = IR
V = 5.25 x 10^-3 (63) = 0.33 V

Then I added the two resistances in series to find the total resistance
R = 750 + 480 = 1230 ohm

Then I tried to apply V = IR with the new resistance to find the current
0.33 V = I (1230 ohm)
And get a value of 2.68 x 10^-4A, which is not correct (correct answer: 5.52 x 10^-3)

I'm not very good at these problems, so I have a feeling I'm doing something silly wrong, thanks for the input!

 

[gneill]

The voltage that you calculated is the voltage that appears across the ammeter when it's in the circuit. It's not helpful when the ammeter is removed...

Instead, can you find the voltage of the battery?

 

[saw176]

I think I got it! Okay, I went V = 5.25 x 10^-3A (480 +750)ohm to get a battery voltage of 6.45 volts. I then added the 0.33 volts that the ammeter was reading to this, to get 6.78 volts. Then, I applied V=IR, 6.78 V = I (1230 ohm) to get the answer 5.52 x 10^-3A. This is the correct answer, so I assume my thinking of adding the 0.33 volts the ammeter "took" to the battery voltage was correct? Thanks for your help!

 

[gneill]

I think I got it! Okay, I went V = 5.25 x 10^-3A (480 +750)ohm to get a battery voltage of 6.45 volts. I then added the 0.33 volts that the ammeter was reading to this, to get 6.78 volts. Then, I applied V=IR, 6.78 V = I (1230 ohm) to get the answer 5.52 x 10^-3A. This is the correct answer, so I assume my thinking of adding the 0.33 volts the ammeter "took" to the battery voltage was correct? Thanks for your help!

Well, you got to the answer, but by way of some shifty moves

An ammeter doesn't read voltage, it reads current. Any voltage that appears across the ammeter is due to that current flowing through the meter's internal resistance (63 Ohms in this case). The battery voltage will be the sum of the voltage drops across all three resistances.

Since all the resistances (including the ammeter) are in series, you can just add them up to find a total resistance. Since the same current flows through all of them you can find the battery voltage by Ohm's law: V = I*R.

Once you know the battery voltage you can always find the current through any resistive load, again using Ohm's law. So you can remove one resistor (the ammeter for example ) and calculate the current that the battery will cause through the remaining resistance.

 

[Ouabache]

I think I got it! Okay, I went V = 5.25 x 10^-3A (480 +750)ohm to get a battery voltage of 6.45 volts. I then added the 0.33 volts that the ammeter was reading to this, to get 6.78 volts. Then, I applied V=IR, 6.78 V = I (1230 ohm) to get the answer 5.52 x 10^-3A. This is the correct answer, so I assume my thinking of adding the 0.33 volts the ammeter "took" to the battery voltage was correct? Thanks for your help!

You obtained the correct solution but your reasoning is incorrect?
When you multiplied 5.25mA times the sum of the two resistors you tell us "you get a battery voltage of 6.45 volts"
That's not true. That is the voltage drop across those two resistors with the ammeter connected.
To determine the voltage on the battery, you may apply KVL (Kirchoff's Voltage Law), to this circuit (with the ammeter in line).
From KVL: V_batt = i_A(R_int+750+480), where V_batt is battery's voltage
i_A is given = 5.52mA and R_int is given = 63 ohm
So you may then solve for V_batt.
Once you have V_batt, you remove the ammeter and solve for current (i) in this circuit
with only the two resistors in series.
Using KVL: V_batt = i (750+480) so i = V_batt/(750+480)

 

[saw176]

Oh okay I see, that makes sense, thanks gneill!

Edit: I see Ouabache's reply now too, you guys are right in pointing out my mistake of assuming the the voltage drop was the voltage of the battery, I knew it seemed kind of fishy but didn't really know what I was doing. Thanks guys!