Question

An ac ammeter is used to measure current in a circuit. When a given direct current passes through the circuit, the ac ammeter reads 3A. When another alternating current passes through the circuit, the ac ammeter reads 4A. Then find the reading of this ammeter (in A), if dc and ac flow through the circuit simultaneously.

Answer 1

Hint: Use the formula for root mean square value i.e. ${{i}_{rms}}=\sqrt{\dfrac{\int\limits_{0}^{T}{{{i}^{2}}}.dt}{T}}$ for an alternating current and for a dc current. When the both currents flow simultaneously, then the net current becomes $i={{i}_{0}}+{{i}_{max}}\sin \omega t$. Then find the rms value of this current.

Formula used:
${{i}_{rms}}=\sqrt{\dfrac{\int\limits_{0}^{T}{{{i}^{2}}}.dt}{T}}$
${{P}_{dc}}=i_{0}^{2}R$
${{P}_{ac}}=i_{rms}^{2}R$

Complete step-by-step answer:
Let us first understand how and what value does an ac ammeter measure.
When an alternating current passes through an ac ammeter, it measures the rms value of the alternating current for one complete cycle.
The full form of rms is root mean square. Rms value is equal to the square root of the means of the squares of the given values.
Suppose the alternating current flowing through the ammeter is given as $i={{i}_{max}}\sin \omega t$.
Then ${{i}_{rms,ac}}=\sqrt{\dfrac{\int\limits_{0}^{T}{{{i}^{2}}}.dt}{T}}=\sqrt{\dfrac{\int\limits_{0}^{T}{i_{\max }^{2}{{\sin }^{2}}(\omega t).dt}}{T}}$ …. (i).
T is the time taken for one complete cycle i.e time period.
When a direct current flows through the ac ammeter, the current is a constant value. Let $i={{i}_{0}}$. Therefore, in this case,
${{i}_{rms,dc}}=\sqrt{\dfrac{\int\limits_{0}^{T}{{{i}^{2}}.dt}}{T}}=\sqrt{\dfrac{i_{0}^{2}\int\limits_{0}^{T}{dt}}{T}}=\sqrt{\dfrac{i_{0}^{2}T}{T}}={{i}_{0}}$. …. (ii)
Now, when both ac and dc current flow through flow through the ammeter, the two currents will superimpose giving a new current i.e. $i={{i}_{0}}+{{i}_{max}}\sin \omega t$.
Let find the rms value of current i.
 ${{i}_{rms}}=\sqrt{\dfrac{\int\limits_{0}^{T}{{{i}^{2}}}.dt}{T}}$
$\Rightarrow {{i}_{rms}}=\sqrt{\dfrac{\int\limits_{0}^{T}{{{\left( {{i}_{0}}+{{i}_{max}}\sin \omega t \right)}^{2}}}.dt}{T}}$
$\Rightarrow i_{rms}^{2}=\dfrac{\int\limits_{0}^{T}{{{\left( {{i}_{0}}+{{i}_{max}}\sin \omega t \right)}^{2}}}.dt}{T}$
$\Rightarrow i_{rms}^{2}=\dfrac{\int\limits_{0}^{T}{\left( i_{0}^{2}+2{{i}_{0}}{{i}_{max}}\sin \omega t+i_{\max }^{2}{{\sin }^{2}}(\omega t) \right)}.dt}{T}$.
$\Rightarrow i_{rms}^{2}=\dfrac{\int\limits_{0}^{T}{i_{0}^{2}dt}+\int\limits_{0}^{T}{2{{i}_{0}}{{i}_{max}}\sin \omega tdt}+\int\limits_{0}^{T}{i_{\max }^{2}{{\sin }^{2}}(\omega t)dt}}{T}$.
$\Rightarrow i_{rms}^{2}=\dfrac{\int\limits_{0}^{T}{i_{0}^{2}dt}}{T}+\dfrac{\int\limits_{0}^{T}{2{{i}_{0}}{{i}_{max}}\sin \omega tdt}}{T}+\dfrac{\int\limits_{0}^{T}{i_{\max }^{2}{{\sin }^{2}}(\omega t)dt}}{T}$
From equations (i) and (ii) we get,
$\Rightarrow i_{rms}^{2}=i_{0}^{2}+\dfrac{2{{i}_{0}}{{i}_{max}}\int\limits_{0}^{T}{\sin \omega tdt}}{T}+i_{rms,ac}^{2}$.
Note that integration of sine function for one complete cycle is zero. Therefore, $\int\limits_{0}^{T}{\sin \omega tdt}=0$,
 $\Rightarrow i_{rms}^{2}=i_{0}^{2}+i_{rms,ac}^{2}$.
It is given that when only the alternating current flows through the ammeter, it measures 4A. Therefore, ${{i}_{rms,ac}}=4A$.
It is given that when only the direct current flows through the ammeter, it measures 3A. Therefore, ${{i}_{0}}=3A$.
This means that $i_{rms}^{2}={{3}^{2}}+{{4}^{2}}=9+16=25$
${{i}_{rms}}=5A$.
Therefore, when both the ac and dc flow through the ammeter, the ammeter will read 5A.

Note: We can also use the power of heat produced by both the currents.
The power of heat dissipated by a direct current is given as ${{P}_{dc}}=i_{0}^{2}R$.
The power of heat dissipated by an alternating current is given as ${{P}_{ac}}=i_{rms}^{2}R$.
When both ac and dc are flowing through the circuit, the total power will be ${{P}_{t}}={{i}^{2}}R$. This power will be the sum of the power due the ac and dc.
This means ${{P}_{t}}={{P}_{dc}}+{{P}_{ac}}$.
$\Rightarrow {{i}^{2}}R=i_{0}^{2}R+i_{rms}^{2}R$
$\Rightarrow {{i}^{2}}=i_{0}^{2}+i_{rms}^{2}$
$\Rightarrow {{i}^{2}}={{3}^{2}}+{{4}^{2}}=25$
$\Rightarrow i=5A$.
Hence, the ammeter will read 5A.