Question

$FIGUREP31.38$shows the voltage across a $0.10mF$capacitor. Draw a graph showing the displacement current through the capacitor as a function of time.

Short answer

Required graph is given below.

Step-by-step solution

Given information

We are given that$Q=CV$

So, $I=\frac{dQ}{dt}=C\frac{dV}{dt}$

Step2:Simplify

So, it will roughly follow the slope of the voltage curve

So, from $0$to $1$:$\frac{dV}{dt}=\frac{100}{1}uS$

So, $I=0.1\times 100=10A$

from $1$to $2$:$\frac{dV}{dt}=0$

so $I=0$

from $2$to $4$: $\frac{dV}{dt}=\frac{100}{2uS}$

So, $I=0.1\times 50=-5A$