From the circuit given below, the capacitance between terminals A and B shown in the circuit is _______ $\mu\text{F}$.
(take $C_1 = C_2 = C_3 = 1 \mu\text{F}$ and $C_4 = 2 \mu\text{F}$.)
Step 1: Understanding the Concept:
By carefully analyzing the schematic nodes, we can redraw the circuit into a standard parallel/series representation. The key is tracing the continuous wires (nodes) to see which components are effectively bridged across the same potential differences.
Step 2: Key Formula or Approach:
Series capacitors: $\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \dots$
Parallel capacitors: $C_p = C_1 + C_2 + \dots$
Step 3: Detailed Explanation:
Let's analyze the circuit diagram connections carefully.
The main top branch consists of three capacitors in series: $C_1$, $C_2$, and $C_3$.
There are two vertical wires connecting a parallel bottom branch containing $C_4$.
- The first vertical wire drops down from the terminal A line, strictly *before* the plate of $C_1$. This means the left plate of $C_4$ is directly connected to Node A.
- The second vertical wire drops down from the terminal B line, strictly *after* the plate of $C_3$. This means the right plate of $C_4$ is directly connected to Node B.
Because $C_4$ spans the entire length from Node A to Node B, it is wired perfectly in parallel with the entire top series branch.
First, calculate the equivalent capacitance of the top series branch ($C_s$):
Since $C_1 = C_2 = C_3 = 1 \mu\text{F}$,
$\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{1} + \frac{1}{1} + \frac{1}{1} = 3 \mu\text{F}^{-1}$.
$C_s = \frac{1}{3} \mu\text{F}$.
Now, add the parallel capacitor $C_4 = 2 \mu\text{F}$:
The total equivalent capacitance $C_{eq} = C_s + C_4$
$C_{eq} = \frac{1}{3} + 2 = \frac{1 + 6}{3} = \frac{7}{3} \mu\text{F}$.
Step 4: Final Answer:
The equivalent capacitance is $7/3$ $\mu\text{F}$.
