Question

In column I, variation of current $i$ with time $t$ is given in the figure. In column II, root mean square current $i_{\text {rms }}$ and average current are given. Match column I with corresponding quantities given in column II

Column I		Column II
i		a	$i_{ rms }=\frac{i_{0}}{\sqrt{3}}$
ii		b	Average current for positive half cycle is $i_{0}$
iii		c	Average current for positive half cycle is $\frac{i_{0}}{2}$
iv		d	Full cycle average current is zero

• A. i. $\rightarrow$ a.; ii. $\rightarrow$ b., d.; iii. $\rightarrow$ b., c.; iv. $\rightarrow$ b.
• B. i. $\rightarrow$ d.; ii. $\rightarrow$ a., d.; iii. $\rightarrow$ b., d.; iv. $\rightarrow$ b.
• C. i. $\rightarrow$ b.; ii. $\rightarrow$ a., c.; iii. $\rightarrow$ c., d.; iv. $\rightarrow$ d.
• D. i. $\rightarrow$ c.; ii. $\rightarrow$ c., d.; iii. $\rightarrow$ b., d.; iv. $\rightarrow$ b.

Answer 1

Answer: (B)

i. For sinusoidal curve $i_{ rms }=i_{0} / \sqrt{2}$ and average current for positive half cycle is $2 i_{0} / \pi$.
ii. Let us find the average current (or rms current) for $t=0$ to $T / 4$. Then it will also be average current (or rms current) for half cycle, i.e., from $t=0$ to $T / 2$.
For $t=0$ to $T / 4: i=\frac{4 i_{0}}{T} t$
then $i_{ av }=\frac{\int\limits_{0}^{T / 4} i d t}{T / 4}=\frac{\int\limits_{0}^{T / 4} \frac{4 i_{0}}{T} t d t}{T / 4}=\frac{i_{0}}{2}$
and $i_{ ms }=\sqrt{\frac{\int\limits_{0}^{T / 4} i^{2} d t}{T / 4}}=\sqrt{\frac{64 i_{0}^{2}}{T^{3}} \int\limits_{0}^{T / 4} t^{2} d t}=\frac{i_{0}}{\sqrt{3}}$
Full cycle average current is zero.
iii. For positive half cycle, average current
$=\frac{\int i d t}{\int d t}=\frac{i_{0}(T / 2)}{T / 2}=i_{0}$
Full cycle average current is zero.
iv. Average current for positive half cycle is $i_{0}$.
For full cycle, average current
$=\frac{\int i d t}{\int d t}=\frac{i_{0}(T / 2)+0}{T}=\frac{i_{0}}{2}$
and $i_{ rms }=\sqrt{\frac{\int\limits_{0}^{T / 2} i_{0}^{2} d t}{T}}=\frac{i_{0}}{\sqrt{2}}$