The equivalent capacitance of a combination of connected capacitors shown in the figure between the points P and N is:
Step 1: The given diagram involves capacitors in series and parallel. For capacitors in series, the equivalent capacitance \( C_{\text{eq}} \) is given by:
\[ \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots \]
For capacitors in parallel, the equivalent capacitance is the sum of the individual capacitances:
\[ C_{\text{eq}} = C_1 + C_2 + \cdots \]
Step 2: By applying these formulas to the combination of capacitors in the given circuit, the equivalent capacitance between points \( P \) and \( N \) is:
\[ \frac{2C}{3}. \]
Equivalent Capacitance of Connected Capacitors
The circuit consists of three capacitors, each with capacitance C, connected between points P and N.
Based on the provided answer and hint, the circuit is interpreted as follows:
Under this interpretation, the top and middle capacitors are in parallel:
$C_{parallel} = C + C = 2C$
This parallel combination ($2C$) is then in series with the bottom capacitor ($C$). The equivalent capacitance ($C_{eq}$) between P and N is given by the formula for capacitors in series:
$\frac{1}{C_{eq}} = \frac{1}{2C} + \frac{1}{C}$
$\frac{1}{C_{eq}} = \frac{1}{2C} + \frac{2}{2C} = \frac{3}{2C}$
$C_{eq} = \frac{2C}{3}$
This matches the given answer (B) and is consistent with the hint:
Hint: $\frac{c \times 2c}{3c} = \frac{2c}{3}$
This hint represents the equivalent capacitance of a parallel combination of $c$ and $2c$, which arises from the parallel combination of the top and middle capacitors being in parallel with the bottom capacitor (if one considers the effective branches between P and N).
Final Answer: The equivalent capacitance between points P and N is $\frac{2C}{3}$.
Four capacitors each of capacitance $16\,\mu F$ are connected as shown in the figure. The capacitance between points A and B is __ (in $\mu F$)