Question

An AC ammeter is used to measure current in a circuit. When a given direct constant current passes through the circuit, the AC ammeter reads 3 Ampere. When an alternating current passes through the circuit, the AC ammeter reads 4 Ampere. Then the reading of this ammeter if DC and AC flow through the circuit simultaneously is:
A) 3A
B) 4A
C) 7A
D) 5A

Answer 1

Hint: An AC ammeter measures the RMS value of the AC current. The RMS value of the AC current is the same value of direct current which when passed through a resistor for the same time produces the same amount of heat as the AC current.
Therefore, we can calculate the total heat current due to the AC and DC currents and then calculate a virtual DC current that will produce the same amount of heat per second of time.
Formula used:
RMS value of AC = DC that produces the same amount of heat as the AC per second.
$H\text{ }=\text{ }{{I}^{2}}R$
where H is the heat energy produced per second when a current (I) passes through a resistor of resistance (R).

Complete step by step answer:
Given, RMS value of AC current (measured by ammeter) = $3A$
Given value of DC current = $4A$
Using (1)
Heat produced by the AC current in a resistor R per second = ${{3}^{2}}R=9R$
Heat produced by the DC current in a resistor R per second = ${{4}^{2}}R=16R$
Therefore, total heat produced = $9R+16R=25R$ ------(2)
Now, we have to find the value of current shown by the ammeter when the above two currents will flow simultaneously will produce the same heat as the two currents.
This will be the value shown by the ammeter. Let this value be I.
Therefore using (1) and (2),
$\therefore {{I}^{2}}R=25R$
$\therefore {{I}^{2}}=25$
$\therefore I=\sqrt{25}=5A$
Therefore, the required current and the value shown by the ammeter is 5A.
Therefore, the correct option is D) 5A.

Note: This problem could also be solved by the traditional definition of RMS that is root of the mean of square of a quantity, by integrating a function in terms of $\sin \theta $ or $\cos \theta $ for the AC current, however that would have become a long and confusing process requiring some confusing calculations. However, that is the most general way of approaching a problem involving RMS of a quantity.
Since, in electricity RMS can be defined in terms of the heat produced, it is always better to apply the definition and process stated above. This saves a lot of time and enables better understanding and an easier thought process for the student.