Q.A network of four 10 μF capacitors is connected to a 500 V supply, as shown in Fig. 2.29.
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Capacitor Network Analysis
Imagine you have a bucket of water and a pipe. The bigger the bucket, the more water it can hold for a given water pressure. A capacitor does the same thing with electric charge — it stores charge when a voltage is applied. The "size" of the bucket is called capacitance (C), measured in farads (F).
Now, what happens when you connect several buckets together with pipes? That's a capacitor network. The analysis is about finding one equivalent bucket (a single capacitor) that behaves exactly like the whole network.
The Core Idea: Charge and Voltage Must Match
When you connect capacitors, two things are always true:
- Charge conservation: Total charge in a closed part of the circuit stays the same unless a battery pushes more in.
- Voltage is shared: The voltage across each capacitor depends on how they're wired.
There are only two basic ways to connect them. Everything else is a combination of these two.
Series Connection: One Path, Shared Charge
Connect capacitors end-to-end, like train cars. The same current flows through each, so each capacitor stores the same amount of charge Q.
But the total voltage across the combination is the sum of the individual voltages:
Vtotal=V1+V2+V3+…
Since V=Q/C for each capacitor, we get:
CeqQ=C1Q+C2Q+C3Q+…
Cancel Q (it's the same everywhere):
Ceq1=C11+C21+C31+…
Intuition: Adding capacitors in series makes the equivalent capacitance smaller than the smallest individual one. Why? Because you're effectively making the "bucket" longer and narrower — harder to fill.
A common mistake: students treat series capacitors like series resistors (adding reciprocals for resistors, but adding directly for capacitors). It's the opposite. For resistors in series: Req=R1+R2. For capacitors in series: 1/Ceq=1/C1+1/C2.
Parallel Connection: Multiple Paths, Same Voltage
Connect capacitors side by side, like buckets with their bottoms connected by a wide pipe. Each capacitor sees the same voltage V across it.
But the total charge stored is the sum of charges on each:
Qtotal=Q1+Q2+Q3+…
Since Q=CV for each:
CeqV=C1V+C2V+C3V+…
Cancel V:
Ceq=C1+C2+C3+…
Intuition: Adding capacitors in parallel makes the equivalent capacitance larger — you're just adding more bucket area. Easy to fill.
| Connection | Equivalent Formula | What happens to Ceq |
|------------|-------------------|----------------------------------|
| Series | 1/Ceq=∑1/Ci | Gets smaller than smallest |
| Parallel | Ceq=∑Ci | Gets larger than largest |
How to Analyze Any Network
- Spot the pattern: Look for capacitors that are clearly in series (only two terminals, no branching between them) or clearly in parallel (both ends connected together).
- Replace step by step: Replace each simple series or parallel group with its equivalent capacitor. Redraw the circuit after each step.
- Repeat until you have one capacitor.
This is exactly like simplifying resistor networks — but with the reciprocal formula for series. If you can do resistor networks, you can do capacitor networks. Just flip the series formula.
A Worked Example
Suppose you have three capacitors: C1=2 μF, C2=3 μF, C3=6 μF. C2 and C3 are in parallel, and that combination is in series with C1. …
The key idea is Capacitor Network Analysis — simplifying series and parallel combinations to find equivalent capacitance, then working backwards to find individual charges.
Step 1: Identify the network structure.
Capacitors C1, C2, and C3 (each 10 μF) form a series path between nodes A and D. Capacitor C4 (also 10 μF) is connected directly across A and D, in parallel with that series combination.
Step 2: Find equivalent capacitance of the series branch.
For three equal capacitors in series:
Cseries1=101+101+101=103⇒Cseries=310 μF
Step 3: Combine with the parallel capacitor.
C4 is in parallel with Cseries:
Ceq=Cseries+C4=310+10=340 μF≈13.33 μF
Step 4: Determine charges.
The supply voltage V=500 V appears across both C4 and the series branch. …
Equivalent capacitance Ceq=340 μF≈13.3 μF. Each of C1,C2,C3 carries 1.67×10−3 C; C4 carries 5×10−3 C.
Reading the network (Fig. 2.29). C1 (between A and B), C2 (between B and C) and C3 (between C and D) form a series chain running A to B to C to D. C4 is connected directly across A and D, so it is in parallel with that series chain. The 500 V supply is applied across A-D.
- Equivalent capacitance. Series combination of the three 10 μF capacitors:
C′ in parallel with C4:
C′1=101+101+101=103⟹C′=310 μF
Ceq=C′+C4=310+10=340 μF≈13.3 μF
- Charge on each capacitor. Series branch - C1,C2,C3 share the same charge, equal to the charge on their equivalent C′ held across 500 V: …
Method: Reducing a Capacitor Network to Find Equivalent Capacitance and Individual Charges
This method applies to any circuit diagram showing several capacitors wired together across a supply, where you must find the equivalent capacitance and then the charge on each individual capacitor.
Steps
Step 1: Trace the circuit to identify series and parallel groups
Follow the wiring node by node. Capacitors sharing a single unbroken path — one after another, with no branch point in between — are in series. Capacitors connected across the exact same pair of nodes are in parallel. Most networks reduce in stages: a sub-group of series capacitors first, then combined in parallel with another branch, and so on.
Step 2: Reduce each sub-group using the standard combination formulas
Series: Ceq1=∑iCi1Parallel: Ceq=∑iCi
Work from the innermost sub-group outward until a single equivalent capacitance for the whole network remains.
Step 3: Find the voltage across each branch …
- KEAM 2026Set eng-2026-04184 marksMCQQ.When a parallel combination of 2 capacitors of 100 pF each is connected across a series combination of 2 capacitors of 200 pF each, the effective capacitance is (A) 600 pF (B) 9400pF (C) 200 pF (D) 100 pF (E) 300 pF
›Reveal solutionSolution
The two 100 pF caps in parallel give 200 pF; the two 200 pF caps in series give 100 pF; connecting these two blocks across each other places them in parallel: 200+100=300pF.
Parallel combination of two 100pF:
C1=100+100=200pF
Series combination of two 200pF:
C2=200+200200×200=40040000=100pF …
- KEAM 2025Set eng-2025-04234 marksMCQQ.In a circuit, the capacitance C is connected. The effective capacitance of the circuit can be reduced by (A) introducing a metal plate between the plates of the capacitor (B) introducing a dielectric slab between the plates (C) reducing the potential difference between the plates (D) connecting another capacitor in series with it (E) connecting another capacitor in parallel with it
›Reveal solutionSolution
The effective capacitance is reduced by connecting another capacitor in series with it.
Concept and Intuition
Capacitors in series give a combined capacitance smaller than the smallest individual one, since Ceq1=C11+C21. All the other listed actions either increase capacitance or do not change it.
Step-by-Step Solution
- Series: Ceq1=C1+C′1⇒Ceq<C. …
- KEAM 2025Set eng-2025-04264 marksMCQQ.The equivalent capacitance of n capacitors of equal capacitance when connected in series and parallel are respectively 0.4 μF and 10 μF. The capacitance of each capacitor is (A) 2 μF (B) 4 μF (C) 5 μF (D) 6 μF (E) 1 μF
›Reveal solutionSolution
For n equal capacitors, series gives C/n and parallel gives nC. Multiplying the two results eliminates n: C2=0.4×10=4, so C=2 μF.
Series equivalent:
nC=0.4 μF
Parallel equivalent:
nC=10 μF
Multiply the two equations: …
- KEAM 2025Set pha-2025-0424F4 marksMCQQ.Three capacitors each of capacitance 12 μF, are connected in series. When this combination is connected to a battery of 12 V, the charge drawn from the battery is (A) 32 μC (B) 24 μC (C) 48 μC (D) 16 μC (E) 12 μC
›Reveal solutionSolution
Series equivalent =12/3=4μF; charge drawn Q=CeqV=4×12=48μC.
For n equal capacitors C in series, the equivalent capacitance is
Ceq=nC=312μF=4μF. …
- KEAM 2024Set eng-2024-06064 marksMCQQ.Three capacitances 1 µF, 4 µF and 5 µF are connected in parallel with a supply voltage. If the total charge flowing through the capacitors is 50 µC, then the supply voltage is (A) 2 V (B) 10 V (C) 6 V (D) 3 V (E) 5 V
›Reveal solutionSolution
Capacitances in parallel add; the supply voltage is total charge divided by total capacitance.
For capacitors in parallel the equivalent capacitance is the sum:
C=1+4+5=10 μF.
Using Q=CV with total charge Q=50 μC, …
- KEAM 2023Set eng-2023-P1-A14 marksMCQQ.N capacitors, each with 1μF capacitance, are connected in parallel to store a charge of 1 C. The potential across each capacitor is 100 V. If these N capacitors are now connected in series, the equivalent capacitance in the circuit will be: (A) 10−4 F (B) 10−6 F (C) 10−10 F (D) 5×10−8 F (E) 10−2 F
›Reveal solutionSolution
With N=104 capacitors, the series equivalent is 10−10 F.
Concept and Intuition
First find N from the parallel configuration: total parallel capacitance is Q/V, and it equals N times the individual 1 μF. For identical capacitors in series, the equivalent capacitance is the individual value divided by N.
Step-by-Step Solution
- Parallel: Cpar=VQ=1001=10−2 F.
- N×1 μF=10−2 F ⇒N=104. …
- KEAM 2022Set eng-2022-P1-A14 marksMCQQ.The effective capacitance between A and B in the given figure is [FIGURE] (A) 1.5μF (B) 1μF (C) 3μF (D) 2μF (E) 2.5μF
›Reveal solutionSolution
Solving the symmetric ladder (with the shorted rightmost branch removed) gives about 1 uF between A and B.
Concept and Intuition
The far-right wire ties the two rails together, so the 3 uF capacitor drawn across that shorted junction has both plates at the same node and carries no independent charge. What remains is a symmetric network: A and B feed 3 uF top and bottom rails whose internal junctions are bridged by 2 uF capacitors, terminating at a common right node.
Step-by-Step Solution
- Short the right ends: the rightmost 3 uF vertical branch is bypassed and drops out.
- Set A = V, B = 0; by top-bottom symmetry the right node sits at V/2. …
- KEAM 2021Set eng-2021-P1-A14 marksMCQQ.An air capacitor and identical capacitor filled with dielectric medium of dielectric constant 5 are connected in series to a voltage source of 12V. The fall of potential across C1 and C2 are respectively (A) 2 V and 10 V (B) 10 V and 2 V (C) 6 V and 6 V (D) 4 V and 8 V (E) 8 V and 4 V
›Reveal solutionSolution
The air capacitor drops 10 V and the dielectric one drops 2 V.
Concept and Intuition
Series capacitors carry the same charge Q, so V=Q/C: the smaller capacitor takes the larger voltage. The dielectric raises C2 five-fold, so it takes the smaller share.
Step-by-Step Solution
- C1=C (air), C2=5C (dielectric constant 5).
- Same Q: V1/V2=C2/C1=5. …
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