Question

In the circuit shown below, the current through the PMMC meter is assumed to be zero. The ideal switch toggles between position $1$ and position $2$. For each position, it is connected for time $T/2$. Assume $R_4{C}_x<<T/2$ and $R_2{C}_x>>T/2$

The capacitor $C_x$ can be charged up to the maximum voltage

• A. $\frac{V_{s}T}{2R_2{C}_x}$
• B. $\frac{V_{s}T}{2R_4{C}_x}$
• C. $\frac{2V_s{R}_2{C}_x}{T}$
• D. $\frac{2V_s{R}_4{C}_x}{T}$

Answer 1

The correct option is A $\frac{V_{s}T}{2R_2{C}_x}$

When switch is at position 1 and current through PMMC is zero then modified circuit will be as

time constant $\tau =R_{eq}C$

net resistance across C will be find as

$R_{eq}=R_2$

$therefore,\tau =R_2{C}_x$

$Now,V_{\infty }=V_s$

V(0) = 0

So, Charging equation for capacitor will be

$V_c\left(t\right)=V_{\infty }-[V_{\infty }-V(0\left)\right]e^{-t/\tau }$

= $V_s-[V_s-0]e^{-t/R_2{C}_x}$

= $V_s[1-e^{-t/R_2{C}_x}]$

as time increases, voltage across capacitor increases, so at t = T/2, we get maximum voltage during charging.

$so,V_c=V_s[1-e^{-T/2R_2{C}_x}]$

$\because T/2<<R_2{C}_x$

$so,\frac{T}{2R_2{C}_x}<<1$

so for $e^{-x}$, we can take linear approximation as = 1 - x

$where,x=\frac{T}{2R_2{C}_x}$

$Now,V_c=V_s[1-(1-\frac{T}{2R_2{C}_x})]$

$=\frac{V_{s}T}{2R_2{C}_x}$

During discharging, $R_4{C}_x<<T/2$, so it will discharge rapidly at position 2.