Question

The circuit involves two ideal cells connected to a $1\mu F$ capacitor via a key $K$. Initially the key $K$ is in position $1$ and the capacitor is charged fully by $2V$ cell. The key is pushed to position $2$. Column I gives physical quantities involving the circuit after the key is pushed from position $1$ to position $2$. Column II gives corresponding results. Match the statements in Column I with the corresponding values in column II.



$$\begin{array}{cc}\text{Column I}& \text{Column II}\\ \text{(A) The net charge crossing the 4 volt cell in}\mu C\text{is}& \text{(p) 2}\\ \text{(B) The magnitude of work done by 4 volt cell in}\mu J\text{is}& \text{(q) 6}\\ \text{(C) The gain in potential energy of capacitor in}\mu J\text{is}& \text{(r) 8}\\ \text{(D) The net heat produced in}\mu J\text{is}& \text{(s) 16}\end{array}$$

• A. $A\to p,B\to q:C\to r:D\to s$
• B. $A\to p:B\to r:C\to q:D\to p$
• C. $A\to q:B\to p:C\to r:D\to s$
• D. $A\to s:B\to s:C\to p:D\to p$

Answer 1

The correct option is B $A\to p:B\to r:C\to q:D\to p$

When connected to $1$, charge on capacitor
$Q_i=C{V}_i=2\mu C(\because V_i=2V)\text{Energy stored in the capacitor}{U}_i=\frac{1}{2}C(V_i{)}^2=2\mu J$

When key is shifted to 2 (Polarity remains same)
$Q_f=C{V}_f=4\mu C(\because V_f=4V)U_f=\frac{1}{2}C{V}_f^2=8\mu J\text{Net charge crossing 4 volt cell}\Delta Q=Q_f-Q_i=4-2=2\mu C\text{Gain in energy stored in the capacitor}\Delta U=U_f-U_i\Delta U=8-2\Delta U=6\mu J\text{Work done by 4 Volt cell}{W}_{4V}=△Q{V}_f{W}_{4V}=(Q_f-Q_i)V_f{W}_{4V}=(4-2)4(\because △Q=2\mu C)W_{4V}=8\mu J$

Also from conservation of energy ,
$U_i+W_{4V}=U_f+H$,Here $H$ is heat dissipated.
$\Rightarrow 2+8=8+H\Rightarrow H=2\mu J$