The effective capacitance between points \(X\) and \(Y\) in the figure shown below is. Assume all the capacitors are \(4\,\mu\text{F}\) each.
Show Hint
For capacitor networks that are not simple series-parallel combinations, assign potentials to internal junctions and apply charge balance at floating nodes.
\[
\sum C(V_{\text{node}}-V_{\text{connected node}})=0
\]
This method quickly gives the equivalent capacitance.
Step 1: Assign potentials to the terminals.
Let the potential at \(X\) be
\[
V_X=V
\]
and the potential at \(Y\) be
\[
V_Y=0.
\]
Let the potential at the junction of \(C_1\) and \(C_2\) be \(V_A\), and the potential at the junction of \(C_2\) and \(C_3\) be \(V_B\).
All capacitors have same capacitance,
\[
C=4\,\mu\text{F}.
\]
Step 2: Apply charge balance at internal junctions.
At junction \(A\),
\[
C(V_A-V)+C(V_A-V_B)+C(V_A-0)=0
\]
Dividing by \(C\),
\[
(V_A-V)+(V_A-V_B)+V_A=0
\]
\[
3V_A-V_B=V
\]
At junction \(B\),
\[
C(V_B-V_A)+C(V_B-0)+C(V_B-V)=0
\]
Dividing by \(C\),
\[
(V_B-V_A)+V_B+(V_B-V)=0
\]
\[
3V_B-V_A=V
\]
Step 3: Solve for \(V_A\) and \(V_B\).
From symmetry of the two equations,
\[
V_A=V_B.
\]
Substituting in
\[
3V_A-V_B=V,
\]
we get
\[
3V_A-V_A=V
\]
\[
2V_A=V
\]
\[
V_A=V_B=\frac{V}{2}.
\]
Step 4: Find total charge drawn from terminal \(X\).
Capacitors connected directly to terminal \(X\) are \(C_1\) and \(C_4\).
So the total charge drawn from \(X\) is
\[
Q=C(V-V_A)+C(V-V_B)
\]
Since
\[
V_A=V_B=\frac{V}{2},
\]
\[
Q=C\left(V-\frac{V}{2}\right)+C\left(V-\frac{V}{2}\right)
\]
\[
Q=\frac{CV}{2}+\frac{CV}{2}
\]
\[
Q=CV.
\]
Step 5: Calculate equivalent capacitance.
Equivalent capacitance is
\[
C_{\text{eq}}=\frac{Q}{V}
\]
\[
C_{\text{eq}}=\frac{CV}{V}
\]
\[
C_{\text{eq}}=C
\]
Since
\[
C=4\,\mu\text{F},
\]
\[
C_{\text{eq}}=4\,\mu\text{F}.
\]
Step 6: Final conclusion.
Therefore, the effective capacitance between \(X\) and \(Y\) is
\[
\boxed{4\,\mu\text{F}}
\]
