Question

Say switches $S_1,S_2$ and so on upto $S_6$ are closed, one after other in order (first $S_1$, then $S_2$) at regular intervals of $1\text{minute}$ starting from $t=0$. The switch $S_1$ is closed at $t=1\text{min}$. The graph of current versus time is best represented as

• A.
• B.
• C.
• D. None of these

Answer 1

The correct option is A

Let $V$ be the potential difference across the battery + its adjacent resistance $R$ combination.
At $t=0$, all the switches are open $\therefore i_0=\frac{V}{15R}+\frac{V}{6R}$
At $t=1\text{minute}$, $S_1$ is closed. The current flows through minimum resistant path, therefore
$i_1=\frac{V}{15R}+\frac{V}{5R}$
Similarly, at $t=2\text{minute}$, $S_2$ is closed, then
$i_2=\frac{V}{11R}+\frac{V}{5R}$
and so on.
The current increases as the equivalent resistance decreases with every time interval of $1\text{minute}$.
After $5$ mins, the effective resistance is $R$ and the current will remain constant after that.
Here $V$ is not constant though, since current changes after each minute.
Alternatively, you can find $R_{eq}$ for each case and comment on current.
$\therefore$ Option (a) is correct.