Calculate the Equivalent Capacitance of the Given Capacitor Combination

Question: The equivalent capacitance of the following assembly of capacitors is ______ μF. (2018)

Correct Answer: 1 μF

Understanding the Given Capacitor Network

To calculate the equivalent capacitance of this circuit, the first step is to identify which capacitors are connected in parallel and which equivalent combinations are connected in series. The circuit may initially appear complicated because it contains several capacitors arranged in rectangular branches, but it becomes very simple when each parallel group is reduced separately.

The left part of the circuit contains two capacitors of 2 μF each connected between the same two points. Therefore, these two capacitors are connected in parallel. The central-right part of the circuit contains three capacitors of 1 μF, 2 μF, and 1 μF connected between the same two points, so these three capacitors are also connected in parallel.

After simplifying these parallel groups, the complete network reduces to four identical 4 μF capacitances connected in series.

Basic Rules for Combining Capacitors

Capacitors Connected in Parallel

When capacitors are connected in parallel, the potential difference across every capacitor is the same. Their equivalent capacitance is obtained by directly adding the individual capacitances:

C_eq = C₁ + C₂ + C₃ + …

Therefore, a parallel combination always has an equivalent capacitance greater than any individual capacitor in that group.

Capacitors Connected in Series

When capacitors are connected in series, the same magnitude of charge appears on each capacitor. The reciprocal of the equivalent capacitance is equal to the sum of the reciprocals of the individual capacitances:

1/C_eq = 1/C₁ + 1/C₂ + 1/C₃ + …

The equivalent capacitance of a series combination is always smaller than the smallest individual capacitance in the series network.

Step-by-Step Calculation of the Equivalent Capacitance

Step 1: Simplify the Left Parallel Combination

On the left side of the circuit, the two 2 μF capacitors are connected in parallel because both capacitors are connected across the same pair of nodes.

For capacitors connected in parallel:

C_L = 2 + 2

Therefore:

C_L = 4 μF

Thus, the entire left rectangular combination can be replaced by a single capacitor of 4 μF.

Step 2: Identify the First Individual 4 μF Capacitor

After the left parallel combination, the circuit contains an individual capacitor of 4 μF. This capacitor is connected in series with the equivalent 4 μF capacitance obtained from the left section.

At this stage, the first two capacitances along the main path are therefore:

4 μF and 4 μF

Step 3: Simplify the 1 μF, 2 μF, and 1 μF Parallel Combination

The next rectangular section contains three branches. The upper branch has a 1 μF capacitor, the middle branch has a 2 μF capacitor, and the lower branch has another 1 μF capacitor.

All three capacitors are connected between the same two nodes. Therefore, they are connected in parallel.

The equivalent capacitance of this group is:

C_M = 1 + 2 + 1

Therefore:

C_M = 4 μF

Thus, the entire three-capacitor parallel combination can also be replaced by a single 4 μF capacitor.

Step 4: Identify the Final 4 μF Capacitor

After the second parallel combination, there is one more individual 4 μF capacitor connected along the main line. Therefore, after all parallel groups have been simplified, the entire original circuit becomes:

4 μF — 4 μF — 4 μF — 4 μF

These four 4 μF capacitors are connected in series.

Step 5: Calculate the Equivalent Capacitance of Four 4 μF Capacitors in Series

For capacitors connected in series:

1/C_eq = 1/4 + 1/4 + 1/4 + 1/4

Adding the terms:

1/C_eq = 4/4

Therefore:

1/C_eq = 1

Hence:

C_eq = 1 μF

Alternative Quick Calculation

Once the two parallel groups are simplified, the circuit contains four identical capacitors of 4 μF connected in series. For n identical capacitors, each having capacitance C, connected in series, the equivalent capacitance is:

C_eq = C/n

Here:

C = 4 μF

and:

n = 4

Therefore:

C_eq = 4/4 = 1 μF

This gives the same result immediately.

Why Are the Capacitors Inside the Rectangular Sections in Parallel?

The physical appearance of a circuit diagram can sometimes make a capacitor network look more complicated than it really is. The correct way to identify a parallel connection is to examine the nodes rather than the visual shape of the circuit.

Two or more capacitors are in parallel when both terminals of every capacitor are connected to the same two nodes. In the left section, both 2 μF capacitors share the same two nodes, so they are parallel. Similarly, in the other rectangular section, the 1 μF, 2 μF, and 1 μF capacitors all share the same two nodes, so they are also parallel.

After replacing each parallel group by its equivalent capacitance, the remaining four 4 μF capacitances lie one after another along a single path. Therefore, they form a series combination.

Complete Circuit Reduction

The entire simplification can be summarized as follows:

Left parallel group = 2 μF + 2 μF = 4 μF

Middle parallel group = 1 μF + 2 μF + 1 μF = 4 μF

Therefore, the original network reduces to:

4 μF, 4 μF, 4 μF, and 4 μF in series

Hence:

C_eq = 1 μF

Final Answer

The equivalent capacitance of the given capacitor assembly is 1 μF.

The two 2 μF capacitors on the left combine in parallel to give 4 μF, while the 1 μF, 2 μF, and 1 μF capacitors combine in parallel to give another 4 μF. The circuit then reduces to four 4 μF capacitors connected in series, giving a final equivalent capacitance of 1 μF.