Jump in RC current response to a square wave

Question

Periodic square wave feeding an RC circuit:

I am currently studying RC filter response to a square wave input with a period of \$ T_1 + T_2 \$.

I have understood about voltage expressions and current expressions, but I don’t understand why:

\$ \textbf{$\Delta$ = $ \frac{V_0}{R}\ $} \$

The right most graph in green is of \$i_c\$ current through the capacitor and \$\Delta\$ is the current jump. I don’t understand the jump.

Answer 1

Δ is that current jump and that is what I don’t understand

To understand change in Current during switching instant -

just before t=T1

Voltage drop across resistor =$$V_R=V_s-V_c$$ $$\implies$$ $$I_1=\frac{V_s-V_c}{R}$$

Just after t=T1

Voltage drop across resistor =$$V_R=-V_c$$ $$\implies$$ $$I_2=\frac{-V_c}{R}$$

Hence ,$$∆=I_1-I_2 =\frac{V_s}{R}$$