Q.47 The equivalent capacitance of following assembly of capacitors is ________ μF.
Understanding how to systematically reduce such networks is essential for quickly solving numerical questions and gaining confidence in circuit analysis.
Step 1: Identify series and parallel groups
From left to right, the circuit can be read as:
- Two 2 μF capacitors connected in parallel on the left.
- A single 4 μF capacitor in series with the rest of the circuit.
- A central block containing one 2 μF capacitor in the middle and two 1 μF capacitors, one on the top branch and one on the bottom branch, all in parallel.
- Another single 4 μF capacitor in series at the far right.
Symbolically, the network becomes:
(2 μF ∥ 2 μF) – 4 μF – (1 μF ∥ 2 μF ∥ 1 μF) – 4 μF.
Step 2: Simplify the left parallel combination
For capacitors in parallel, the equivalent capacitance is the sum of the individual capacitances:
Cleft = 2 μF + 2 μF = 4 μF.
Thus, the left block behaves like a single 4 μF capacitor.
Step 3: Simplify the central parallel combination
The middle 2 μF and the two 1 μF capacitors are also connected in parallel:
Cmiddle = 1 μF + 2 μF + 1 μF = 4 μF.
So, the central block is equivalent to a single 4 μF capacitor, and the entire circuit reduces to four 4 μF capacitors in series:
4 μF – 4 μF – 4 μF – 4 μF.
Step 4: Combine the series capacitors
For capacitors in series, the reciprocals of the capacitances add:
1 / Ceq = 1/4 + 1/4 + 1/4 + 1/4 = 4/4 = 1.
Therefore, Ceq = 1 μF, which is the equivalent capacitance of the entire assembly.
Explanation of common options
Many exam questions based on this diagram typically offer options such as:
- (A) 0.5 μF
- (B) 1 μF
- (C) 2 μF
- (D) 4 μF
Why 0.5 μF is incorrect
The value 0.5 μF often appears when the effective number of series capacitors is miscounted or when unequal equivalent capacitances are assumed during simplification.
In this problem, all four series capacitors after reduction are equal (4 μF each), leading to 1 μF, not 0.5 μF.
Why 2 μF is incorrect
A result of 2 μF would occur if only two 4 μF capacitors were taken in series, because 1/C = 1/4 + 1/4 = 1/2 ⇒ C = 2 μF.
However, the reduced circuit clearly contains four such capacitors in series.
Why 4 μF is incorrect
Each simplified block has a capacitance of 4 μF, but when capacitors are placed in series, the overall capacitance must be smaller than the smallest individual capacitor.
Hence, the final answer cannot be 4 μF.
Why 1 μF is correct
With four identical 4 μF capacitors in series, the total inverse capacitance is the sum of their inverses, so 1/Ceq = 4 × (1/4) = 1, giving Ceq = 1 μF.
Therefore, 1 μF is the correct equivalent capacitance of the given assembly of capacitors.
