A DC ammeter and a hot wire ammeter are connected to a circuit in series. When a direct current is passed through a circuit, the DC ammeter shows 6 A. When AC current flows through a circuit, the AC ammeter shows 8 A. What will be reading of each ammeter if DC and AC currents flow simultaneously through the circuit?
$\begin{align}
  & \text{A}\text{. }DC=6A,AC=10A \\
 & \text{B}\text{. }DC=3A,AC=5A \\
 & \text{C}\text{. }DC=5A,AC=8A \\
 & \text{D}\text{. }DC=2A,AC=3A \\
\end{align}$

Hint: DC ammeter gives peak value of current while AC ammeter gives RMS value of current. The total current must be the sum of AC and DC current. Calculate the average value of DC current when it will flow through the circuit and calculate root means square value of AC current over a period of 0 to T. This way, you will get the current in the circuit.

Complete step by step solution:
In this question, we need to find out the reading of ammeter if DC and AC currents flow simultaneously through the circuit.
Here, DC current shows current $6A$and we know that DC always shows constant peak value.
$\therefore {{I}_{0}}={{I}_{1}}=6A$
Now talking about hot wire ammeter which is nothing but AC ammeter showed current $8A$. AC ammeter shows the root mean square value of current because AC ammeter reads root means square value only, not peak value.
$\therefore {{I}_{rms}}=6A$
Now calculate peak value of AC ammeter
${{I}_{0}}={{I}_{2}}=\sqrt{2}{{I}_{rms}}=8\sqrt{2}A$
If both the currents (AC and DC) flow through circuit then take superposition of this two current
$I={{I}_{1}}+{{I}_{2}}$
Put the value of current in the above equation
$\begin{align}
  & I={{I}_{1}}+{{I}_{2}} \\
 & I=6+8\sqrt{2}\sin \omega t \\
\end{align}$
Now, if $I$current flows through the circuit, then DC ammeter shows the average value of current and AC ammeter shows root means the square value of ammeter.
Now find the average value of AC ammeter
\[{{I}_{avg}}=\dfrac{1}{T}\int\limits_{0}^{T}{Idt}\]
Put the value of $I$in the above equation, we get
\[\begin{align}
  & {{I}_{avg}}=\dfrac{1}{T}\int\limits_{0}^{T}{(6+8\sqrt{2}\sin \omega t)dt} \\
 & {{I}_{avg}}=\dfrac{1}{T}\int\limits_{0}^{T}{(6)+\dfrac{1}{T}\int\limits_{0}^{8}{8}\sqrt{2}\sin \omega t)dt} \\
 & \because \omega =\dfrac{2\pi }{T} \\
 & {{I}_{avg}}=\dfrac{1}{T}\int\limits_{0}^{T}{(6)+\dfrac{1}{T}\int\limits_{0}^{8}{8}\sqrt{2}\sin \dfrac{2\pi }{T}tdt} \\
 & {{I}_{avg}}=\dfrac{1}{T}\left[ 6T \right]_{0}^{T}+\dfrac{8\sqrt{2}}{T}\left[ -\dfrac{\cos 2\pi t}{T} \right]_{0}^{T} \\
 & {{I}_{avg}}=\dfrac{6T}{T}-0 \\
 & {{I}_{avg}}=6A \\
\end{align}\]
As we have already discussed that hot wire ammeter shows the root means the square value of current
\[\begin{align}
  & {{I}_{rms}}=\sqrt{\dfrac{1}{T}{{\left( 6+8\sqrt{2}\sin \omega t \right)}^{2}}}dt \\
 & I_{rms}^{2}T=\int\limits_{0}^{T}{{{\left( 6+8\sqrt{2}\sin \omega t \right)}^{2}}dt} \\
 & I_{rms}^{2}T=\int\limits_{0}^{T}{\left( 36+48\sqrt{2}\sin \omega t+{{\left( 8\sqrt{2}\sin \omega t \right)}^{2}} \right)dt} \\
\end{align}\]
Solve integration, we get,
\[\begin{align}
 & I_{rms}^{2}T=36T+0+64(T-0) \\
 & I_{rms}^{2}=100 \\
 & {{I}_{rms}}=10 \\
\end{align}\]
Therefore,
$DC=6A,AC=10A$
Answer- (A)

Note: The direct current can increase, can be stable, or can reduce also, but it is not oscillatory. Alternating current is oscillatory. In this question, the time period is taken from \[0\text{ }to\text{ }T\]. AC current increases to a maximum, then reduces to zero towards the positive direction of the y-axis and again increases to a maximum in the other direction and again reduces to zero towards the negative direction of the y-axis. Direct current flows in one direction, but the alternating current flows in a periodic manner, in one cycle, in forward and reverse directions.