Question

The current in a certain circuit varies with time, as shown in the figure.

• A. The average current from $t=0$ to $t=2\text{s}$ is $2.5\text{A}$
  
• B. The average current for a full cycle is zero.
• C. The rms current from $t=0$ to $t=4\text{s}$ is $\frac{5}{\sqrt{2}}\text{A}$
  
• D. The rms current from $t=0$ to $t=4\text{s}$ is $\frac{5}{\sqrt{6}}\text{A}$
  
  

Answer 1

Answer: (A), (B), (D)

The rms current from $t=0$ to $t=4\text{s}$ is $\frac{5}{\sqrt{6}}\text{A}$



$\text{The average current from t=0 to t=2 s}:$


$i_{avg}=\frac{\text{Area under}i-t\text{curve}}{\text{Total time}}$

$\Rightarrow i_{avg}=\frac{\frac{1}{2}\times 2\times 5}{2}=2.5\text{A}$

$\text{Average current for full cycle:}$


From the figure, we can see that, the area under the shaded region for the first half is equal to that of the second half of the full cycle but of opposite sign.

Hence, for the full cycle, net area is zero $i.e.$ average current is zero because average current is equal to total area under $i-t$ curve divided by total time.

$\text{rms current:}$ The variation of current with respect to time $t$ for the given time range is,

$I=\{\begin{array}{ccc}& \frac{5t}{2}& 0\le t\le 2\text{s}\\ & -\frac{5(4-t)}{2}& 2\le t\le 4\text{s}\end{array}$

So, rms value of current,

$I_{rms}^2=\frac{1}{T}{\int }_0^T{I}^{2}dt$

$=\frac{1}{4}[{\int }_0^2{\left(\frac{5t}{2}\right)}^{2}dt+{\int }_2^4{(-\frac{5(4-t)}{2})}^{2}dt]$

$=\frac{25}{16}[{\left[\frac{t^3}{3}\right]}_0^2-{\left[\frac{(4-t{)}^3}{3}\right]}_2^4]$

$=\frac{25}{48}[(2^3-0)-(0-2^3)]$

$=\frac{25}{6}$

$\therefore I_{rms}=\frac{5}{\sqrt{6}}\text{A}$

Hence, options $\left(A\right),\left(B\right)$ and $\left(D\right)$ are correct.