Question

The probes of a non-isolated, two channel oscilloscope are clipped to points A, B and C in the circuit of the adjacent figure. $V_{in}$ is a square wave of a suitable low frequency. The display on $C{h}_{1}andC{h}_2$ are as shown on the right. Then the 'Signal' and 'Ground' probes $S_1,G_{1}and{S}_2,G_{2}ofC{h}_{1}andC{h}_2$ respectively are connected to points:

• A. A, B, C, A
• B. A, B, C, B
• C. C, B, A, B
• D. B, A, B, C

Answer 1

The correct option is B A, B, C, B

Since, square wave is of low frequency. So, it can be assumed that time during which the wave forms are displayed on the screen, the voltage across R and L is $V_{in}$

$I\left(s\right)=\frac{V_{in}/s}{R+Ls}=\frac{V_{in}}{Ls(s+\frac{R}{L})}$

$=\frac{V_{in}}{L}(\frac{1}{s}-\frac{1}{s+\frac{R}{L}})\times \frac{L}{R}$

$I\left(s\right)=\frac{V_{in}}{R}(\frac{1}{s}-\frac{1}{s+\frac{R}{L}})$

$i\left(t\right)=\frac{V_{in}}{R}(1-e^{-Rt/L})$

$V_{AB}=Voltageacrossresistance$

$R\times i\left(t\right)=V_{in}[1-e^{-Rt/L}]$


$So,S_1$ is connected to point A and $G_1$ is connected to point B.
$Voltageacrossinductor{V}_{BC}=L\frac{di}{dt}$

$=L\frac{d}{dt}\left[\frac{V_{in}}{R}\right(1-e^{-Rt/L}\left)\right]$

$V_{BC}=V_{in}{e}^{-Rt/L}$


So, $S_2$ is connected to point C.
and $G_2$ is connected to point B.