Wheatstone Bridge with Capacitors — NEET Scenarios
Every NEET-relevant Wheatstone-bridge scenario built from capacitors (from Electrostatic Potential & Capacitance). Balance condition, charge on the bridging cap, equivalent C of a balanced / unbalanced bridge, capacitor cubes, dielectric in one arm, energy at balance. For each: Question · Circuit diagram · Given · Asked · Concept · Method · Easy trick · Solution.
Balance condition of a capacitor Wheatstone bridge
NEET concept
Q. Four capacitors C₁, C₂, C₃, C₄ are arranged as a Wheatstone bridge with a fifth capacitor C₅ between the mid-points B and D. A battery is connected across A and C. The bridge is "balanced" (no charge on C₅) when:
Four capacitor arms in a diamond, bridging capacitor C₅ between B and D.
Asked
Condition under which C₅ carries no charge (balance).
Concept
Steady state: no current flows. Bridging cap has no charge ⇔ \(V_B = V_D\). Going A→B→C through C₁,C₃ and A→D→C through C₂,C₄, same Q flows on each pair in series.
Method
Series chains have same charge: \(Q_1 = Q_3 = Q_{ABC}\) and \(Q_2 = Q_4 = Q_{ADC}\). Equate voltages:
Capacitor bridge balance has the SAME shape as resistor bridge: \(C_1/C_2 = C_3/C_4\). (Not the cross-product PS = QR form — that's the resistor mnemonic; for capacitors it's the direct-pair-ratio form.)
Solution
\(C_1/C_2 = C_3/C_4\). Answer: (b)
10-yr-old
Imagine four water bottles meeting in a diamond shape, with a fifth bottle bridging the middle. If the upper pair of bottles has the same size-ratio as the lower pair, then both middle points end up at the same water level — so water has no reason to flow through the bridge bottle. That perfect-matching rule is what "balanced" means.
2
Find the unknown capacitor in a balanced bridge
NEET style
Q. In a capacitor Wheatstone bridge, three arms have C₁ = 2 μF, C₂ = 4 μF, C₃ = 3 μF and the bridge is balanced (no charge on C₅). The fourth capacitor C₄ is:
Same recipe as resistor bridge: unknown = (product of its two neighbours) ÷ (the arm diagonally opposite). Here opposite to C₄ is C₁ (2 μF).
Solution
\(C_4 = 6\,\mu F\). Answer: (b)
10-yr-old
It's like a "fill-in-the-pattern" puzzle. The first two numbers go 2 then 4 (small to twice as big). The next two have to follow the same doubling pattern. So after 3 comes 6. Easy!
3
Charge on the bridging capacitor at balance
NEET PYQ
Q. A capacitor Wheatstone bridge is balanced. A capacitor C₅ is connected between the mid-points B and D. The charge stored on C₅ is:
(a) C₅ V(b) zero(c) ½ C₅ V(d) C₅ V/2
Given
Bridge balanced (\(C_1/C_2 = C_3/C_4\)). Battery V across A-C. Capacitor C₅ between B-D.
Asked
Charge on C₅.
Concept
Balance ⇒ \(V_B = V_D\) ⇒ PD across C₅ is zero ⇒ Q₅ = C₅ × 0 = 0.
Method
Check balance: \(C_1/C_2 = C_3/C_4\) is given → bridge is balanced.
Balance ⇒ both midpoints B and D are at the SAME potential.
Voltage across C₅ = V_B − V_D = 0.
Apply Q = CV: Q₅ = C₅ × 0 = 0.
\[Q_5 = C_5\,(V_B - V_D) = C_5 \times 0 = 0\]
Easy trick
"Balanced bridge ⇒ bridging element does nothing." For a resistor it carries zero current; for a capacitor it carries zero charge. You can remove or replace it without changing the rest of the circuit.
Solution
Q₅ = 0. Answer: (b)
10-yr-old
If both ends of a slide are at the same height, a ball placed on the slide just sits there — nothing rolls. Same with the bridge cap: if both its ends are at the same voltage, no charge crosses, so it stores zero charge. The bridge cap might as well not exist.
4
Equivalent capacitance of a balanced bridge
NEET PYQ
Q. In a balanced bridge with C₁ = 2, C₂ = 4, C₃ = 3, C₄ = 6 μF and a bridging cap C₅ = 5 μF, the equivalent capacitance across A and C is:
Balance ⇒ drop C₅. Upper arm C₁ series C₃; lower arm C₂ series C₄; two combinations in parallel.
Method
Upper series: \(C_{13} = \dfrac{C_1 C_3}{C_1 + C_3} = \dfrac{2 \times 3}{5} = 1.2\,\mu F\). Wait — A→B is C₁, B→C is C₃: actually C₁ and C₃ are series only if the bridge geometry is A-C₁-B-C₃-C. Check the geometry: in our diamond, C₁ is A-B (top-left), C₃ is D-B (bottom-left), so series chain A→D→B→C goes through C₂ then C₃ — be careful. Below we use the standard "upper path A-B-C through C₁ then C₂" convention.
Using standard convention (top arms = C₁, C₂ on path A-B-C; bottom arms = C₃, C₄ on path A-D-C):
Because the diamond is balanced, the bridge cap C₅ is doing nothing — pretend it isn't there. Now you just have two simple chains running between A and C. Combine each chain (two caps in series = product-over-sum), then add the two chains together (parallel = simple sum). That's the whole answer.
5
All four arms equal + bridging C₅ — equivalent capacitance
NEET PYQ
Q. Four equal capacitors C are arranged as a Wheatstone bridge with a fifth capacitor C₅ between B and D. Find the equivalent capacitance between A and C.
(a) C(b) C/2(c) 2C(d) C/2 + C₅ (depends on C₅)
Given
All four arms = C. Bridging cap C₅ between B and D.
Asked
Equivalent C between A and C.
Concept
All arms equal ⇒ \(C/C = C/C\) ⇒ balanced. C₅ carries no charge regardless of its value — drop it.
Upper arm (A→B→C): two caps C in series ⇒ \(C_{\text{top}} = \dfrac{C \cdot C}{C+C} = \dfrac{C}{2}\).
Lower arm (A→D→C): same ⇒ \(C_{\text{bot}} = \dfrac{C}{2}\).
The two arms are in parallel between A and C ⇒ add: \(\dfrac{C}{2} + \dfrac{C}{2} = C\).
\[C_{AC} = \dfrac{C}{2} + \dfrac{C}{2} = C\]
Easy trick
"4 equal caps in a Wheatstone diamond → equivalent = C (regardless of C₅)." Standard NEET fact — memorise it.
Solution
\(C_{AC} = C\). Answer: (a)
10-yr-old
All four caps are identical — the diamond is perfectly symmetric, like a properly-built ice-cream sandwich. The middle bridge cap is bored: no charge passes through it. Each arm has two identical cups in a row (= half a cup), and the two arms stand side by side (= back to one full cup). Answer = C. The value of C₅ doesn't matter at all.
6
Unbalanced bridge — find PD across the bridging cap
NEET style
Q. A Wheatstone bridge has C₁ = 2, C₂ = 4, C₃ = 4, C₄ = 2 μF (so it is NOT balanced) and a bridging cap C₅ between B and D. Battery 12 V across A-C. Before connecting C₅, the PD between B and D is:
(a) 0 V(b) 2 V(c) 4 V(d) 6 V
Given
C₁ = 2, C₂ = 4, C₃ = 4, C₄ = 2 μF. V = 12 V. C₅ not yet connected.
Asked
\(V_B - V_D\).
Concept
Without C₅ the bridge is two independent series chains A→B→C and A→D→C, each carrying its own charge. Compute V at B and at D, subtract.
Method
Layout: A (top, battery +), B (right), C (bottom, battery −), D (left). Right arm A→B→C carries C₂ then C₄. Left arm A→D→C carries C₁ then C₃. Take \(V_A = 12\) V, \(V_C = 0\) V.
Right arm (C₂ series C₄): \(C_R = \dfrac{C_2 C_4}{C_2 + C_4} = \dfrac{4 \times 2}{6} = \dfrac{4}{3}\,\mu F\). Charge in this arm: \(Q_R = C_R \cdot V = \dfrac{4}{3}\times 12 = 16\,\mu C\).
Voltage drop across C₂ (A → B): \(V_{AB} = \dfrac{Q_R}{C_2} = \dfrac{16}{4} = 4\) V. So \(V_B = V_A - 4 = 12 - 4 = 8\) V.
Left arm (C₁ series C₃): \(C_L = \dfrac{C_1 C_3}{C_1 + C_3} = \dfrac{2 \times 4}{6} = \dfrac{4}{3}\,\mu F\). Charge \(Q_L = 16\,\mu C\).
Voltage drop across C₁ (A → D): \(V_{AD} = \dfrac{Q_L}{C_1} = \dfrac{16}{2} = 8\) V. So \(V_D = V_A - 8 = 12 - 8 = 4\) V.
\(V_B - V_D = 8 - 4 = +4\) V ⇒ magnitude 4 V.
Easy trick
In a non-balanced cap bridge (no bridging element), use the series-chain voltage divider for capacitors: voltage on cap k = \(V \times \dfrac{C_{\text{other}}}{C_1 + C_2}\). Apply to each arm, subtract midpoint voltages.
Solution
|V_B − V_D| = 4 V. Answer: (c)
10-yr-old
When the diamond is NOT balanced, the two middle points sit at different heights. To find the difference, work out the height at B and the height at D separately — like sliding down two different slides from the same starting roof. Bigger cap at the top of a chain = less of the voltage falls on it, more falls on the bottom one. Compare the two midpoint heights and subtract.
7
Dielectric inserted into one arm — does balance survive?
NEET concept
Q. A balanced capacitor bridge has all arms = C. A dielectric slab of constant K is fully inserted into the arm C₁ (top-left). What happens?
(a) Bridge stays balanced(b) C₁ becomes KC and bridge becomes unbalanced(c) Bridge becomes balanced only for K = 1(d) Q on C₅ becomes infinite
Given
All arms originally = C. C₁ becomes KC.
Asked
Does the bridge stay balanced?
Concept
Balance: \(C_1/C_2 = C_3/C_4\) ⇒ before: \(C/C = C/C\) ✓. After: \(KC/C = C/C\) ⇒ \(K = 1\). Only K = 1 (no dielectric) keeps the balance.
Method
Before inserting: \(C_1/C_2 = C/C = 1\) and \(C_3/C_4 = C/C = 1\). Equal → balanced ✓
After inserting K into C₁: new \(C_1 = KC\).
New ratio \(C_1/C_2 = KC/C = K\). But \(C_3/C_4\) is still 1.
Balance demands K = 1. For any K > 1 (real dielectric), bridge is unbalanced.
Now C₅ has a non-zero PD across it ⇒ Q₅ = C₅(V_B − V_D) ≠ 0.
Easy trick
To stay balanced after inserting dielectric in one arm, you'd need to insert the same K in the diagonally-opposite arm too. Single-arm changes always break balance.
Solution
Bridge becomes unbalanced. Answer: (b)
10-yr-old
Stick a sponge inside one of the four cups. That cup now holds more water for the same height. The other three cups haven't changed. The perfect 1-1-1-1 match is broken, so the diamond is no longer balanced. Trick to put it back: stick the same sponge in the cup diagonally opposite as well.
8
Cube of 12 identical capacitors — across body diagonal
NEET PYQ
Q. Each edge of a cube has a capacitor C. The equivalent capacitance between two body-diagonally-opposite corners is:
(a) 6C/5(b) 5C/6(c) 12C/5(d) 7C/12
Given
12 identical capacitors C on cube edges; measure between A and B (body-diagonal).
Asked
\(C_{AB}\).
Concept
By symmetry of capacitor cube (dual of resistor cube): the 3 neighbours of A are at the same potential, and the 3 neighbours of B are at the same potential. Merge each triplet. Capacitor "series" and "parallel" are swapped versus resistors.
Method
From A → 3 parallel caps C ⇒ 3C.
Between the two merged middle nodes → 6 parallel edges ⇒ 6C.
From middle to B → 3 parallel ⇒ 3C.
Three in series (3C, 6C, 3C): \(\dfrac{1}{C_{AB}} = \dfrac{1}{3C} + \dfrac{1}{6C} + \dfrac{1}{3C} = \dfrac{5}{6C}\;\Rightarrow\;C_{AB} = \dfrac{6C}{5}\).
Easy trick
Memorise: capacitor cube across body diagonal = 6C/5. (Resistor cube body diagonal is 5R/6 — exact reciprocal-style swap.)
Solution
\(C_{AB} = 6C/5\). Answer: (a)
10-yr-old
Picture a cube of 12 identical caps — 12 little springs along the edges. Stand at one corner, want to push charge to the corner on the opposite side. From your corner, three edges shoot out — that's a parallel group worth 3C. They meet at three "middle" corners (which all sit at the same voltage by symmetry, so we merge them). From there, six edges bridge across to three more "middle" corners (another merge) → 6C in parallel. Finally three edges arrive at the opposite corner → 3C. Stack the three groups in series: 1/(3C) + 1/(6C) + 1/(3C) = 5/(6C), so the whole cube acts like 6C/5.
9
Capacitor cube — across an edge
NEET style
Q. Each edge of a cube has capacitance C. The equivalent capacitance across one EDGE (two adjacent corners) is:
(a) 12C/7(b) 7C/12(c) 3C/4(d) 4C/3
Given
Cube of 12 identical caps C; A-B is one edge.
Asked
\(C_{AB}\) across the edge.
Concept
Symmetry argument identical to the resistor case but with the parallel/series rules swapped. Standard result for capacitor cube across an edge: 12C/7.
Method
Identify the two endpoints: A and B are joined by ONE direct edge (C), plus many longer paths through the rest of the cube.
Standard result (from symmetry + Kirchhoff): the 11 other edges combine into an effective \(5C/7\) in parallel with the direct edge C.
Total: \(C_{AB} = C + 5C/7 = 12C/7\).
(Quicker: use the duality with the resistor cube — resistor cube across edge = 7R/12, swap to get 12C/7 for the capacitor cube.)
Easy trick
Memorise three answers — capacitor cube: edge = 12C/7, face-diagonal = 4C/3, body-diagonal = 6C/5. They are the reciprocals (in the form-swap sense) of the resistor cube answers 7R/12, 3R/4, 5R/6.
Solution
\(C_{AB} = 12C/7\). Answer: (a)
10-yr-old
Same cube, but now you're going from one corner to the corner right next door. There's a direct edge between them and lots of longer paths through the cube. The longer paths all together act like a "bonus" cap of 5C/7. Add it to the direct edge C → answer 12C/7. Just memorise it next to the body-diagonal answer.
10
Battery on a balanced bridge — total charge drawn
NEET PYQ
Q. In Scenario 4 (\(C_{AC} = 10/3\,\mu F\)), a battery of 6 V is connected across A and C. Total charge drawn from the battery is:
(a) 5 μC(b) 10 μC(c) 20 μC(d) 60 μC
Given
Balanced bridge; \(C_{eq} = 10/3\,\mu F\); battery V = 6 V.
Asked
Total Q drawn from battery.
Concept
Charge from battery = \(C_{eq} \times V\).
Method
From Scenario 4 we already know the balanced bridge collapses to \(C_{eq} = 10/3\,\mu F\).
Battery EMF = 6 V is fully applied across this equivalent capacitor.
Battery only "sees" the equivalent capacitance. Once you've reduced the bridge, the charge formula is just \(Q = C_{eq} V\).
Solution
\(Q_{\text{total}} = 20\,\mu C\). Answer: (c)
10-yr-old
The battery doesn't see all five cups individually — it just sees one giant cup equal to the bridge's "total capacity" of 10/3 μF. Total charge it pushes out = capacity × voltage = 10/3 × 6 = 20 μC. Done.
11
Energy stored at balance
NEET style
Q. Same bridge as Scenario 10 (Q drawn from battery = 20 μC, V = 6 V). Total energy stored in the bridge is:
(a) 30 μJ(b) 60 μJ(c) 120 μJ(d) 10 μJ
Given
\(C_{eq} = 10/3\,\mu F\), V = 6 V.
Asked
Total energy U.
Concept
Treat bridge as a single \(C_{eq}\) once you've reduced it. Then \(U = \tfrac{1}{2} C_{eq} V^2\).
Method
Use the energy formula \(U = \tfrac{1}{2} C V^2\) on the equivalent capacitance.
"Energy of a balanced bridge" = energy of its equivalent single capacitor. No need to add up energies on each arm — they sum to the same thing.
Solution
U = 60 μJ. Answer: (b)
10-yr-old
Energy stored in a capacitor is like the energy stored in a stretched rubber band — the more you stretch (higher voltage), the much more energy is locked in (because of the SQUARE). Use the formula half × capacity × voltage squared on the single big cup we found, and we get 60 μJ.
12
Symmetry shortcut — balanced bridges from inspection
NEET technique
Q. Which of the following capacitor bridges is balanced? (i) C₁=2, C₂=3, C₃=4, C₄=6 μF (ii) C₁=1, C₂=2, C₃=3, C₄=5 μF (iii) C₁=5, C₂=10, C₃=15, C₄=30 μF
(a) Only (i)(b) Only (iii)(c) (i) and (iii)(d) All three
Given
Three sets of capacitor values.
Asked
Which set(s) give a balanced bridge?
Concept
Check \(C_1/C_2 = C_3/C_4\) for each set.
Method
(i) 2/3 vs 4/6 = 2/3 ✓
(ii) 1/2 vs 3/5 ✗
(iii) 5/10 = 1/2 vs 15/30 = 1/2 ✓
Easy trick
Reduce both ratios to lowest terms. If they match, balanced. Multiplying all four arms by the same factor never changes balance — that's why (iii) (×5 of an obvious 1,2,3,6 set) works.
Solution
(i) and (iii) are balanced. Answer: (c)
10-yr-old
Want to check if a bridge is balanced without doing real work? Just turn both pairs into simplest fractions: 2:4 → 1:2, and 3:6 → 1:2. Same fraction ⇒ balanced. For (ii) 1:2 ≠ 3:5 ⇒ not balanced. For (iii) 5:10 → 1:2 and 15:30 → 1:2 ⇒ balanced. Trick: if you multiply all four caps by the same number, balance never changes — that's why (iii) (×5 of an obvious 1,2,3,6 pattern) still works.
🧠 Everything to Memorise — Wheatstone Bridge with Capacitors
One-page revision sheet. If you can recite this table cold, you've got every NEET PYQ from this topic covered.
🔷 Core Bridge Rules
Topic
What to memorise
Why / When
Balance condition (capacitors)
\(\dfrac{C_1}{C_2} = \dfrac{C_3}{C_4}\)
Same ratio form as resistor bridge. At balance, \(V_B = V_D\) so the bridging cap C₅ sees zero PD.
Balance condition (resistors)
\(\dfrac{P}{Q} = \dfrac{R}{S}\) (or PS = QR)
Same physical idea — null deflection on galvanometer between B and D.
Charge on C₅ at balance
\(Q_5 = 0\)
No PD across it. So you can remove or change C₅ without affecting the rest.
Current through galvanometer at balance
\(I_g = 0\)
Resistor-bridge equivalent of the above. Null deflection.
Diagonal swap (cell ↔ galvanometer)
Balance unchanged
The two diagonals are interchangeable in a balanced bridge.
Multiply all arms by k
Balance unchanged
Ratios survive any common scaling factor — useful for spotting balance fast.
Drop C₅. Series in each arm, then add the two arms in parallel.
All 4 arms equal to C (+ any C₅)
\(C_{eq} = C\)
Bridge is auto-balanced. Each arm = C/2 series, two arms = C parallel. C₅ irrelevant.
Series of two caps
\(\dfrac{C_1 C_2}{C_1 + C_2}\) (product over sum)
Same charge flows. Result is smaller than the smallest cap.
Parallel of two caps
\(C_1 + C_2\) (just add)
Same voltage across each. Result is larger than the largest.
🔷 Capacitor Cube — Three Standard Answers
Measured across
Capacitor cube (each edge = C)
Resistor cube (each edge = R)
Edge (adjacent corners)
\(\dfrac{12\,C}{7}\)
\(\dfrac{7\,R}{12}\)
Face diagonal
\(\dfrac{4\,C}{3}\)
\(\dfrac{3\,R}{4}\)
Body diagonal (longest)
\(\dfrac{6\,C}{5}\)
\(\dfrac{5\,R}{6}\)
⚡ Pattern to lock in: the capacitor-cube and resistor-cube answers are reciprocals of each other (with C and R swapped). So if you remember the resistor numbers 7/12, 3/4, 5/6 you can flip them for capacitors: 12/7, 4/3, 6/5.
Capacitor voltage divider — the SMALLER cap drops the MORE voltage.
PD across the bridging cap (before connecting C₅)
\(V_B - V_D\) = (V at B from top chain) − (V at D from bottom chain)
Use the voltage-divider above for each chain.
Charge on bridging cap (unbalanced)
\(Q_5 = C_5\,(V_B - V_D)\)
Once C₅ is in place steady-state, it equilibrates to whatever PD remains.
Battery charge drawn (any bridge)
\(Q_{\text{total}} = C_{eq}\,V\)
Battery only "sees" the equivalent capacitance.
🔷 Energy at Balance
Energy form
Use when…
Note
\(U = \tfrac{1}{2} C_{eq} V^2\)
You know battery V and the bridge's \(C_{eq}\)
Easiest when battery is connected.
\(U = \dfrac{Q^2}{2 C_{eq}}\)
You know Q drawn from battery and \(C_{eq}\)
Easiest when battery is disconnected and Q is fixed.
\(U = \tfrac{1}{2} Q V\)
You know both Q and V
Bridge-form of \(W = qV\).
🔷 Dielectric in the Bridge
Action
Effect on balance
How to restore balance
Insert K into ONE arm
Bridge becomes UN-balanced
Insert the same K into the diagonally-opposite arm.
Insert K into ALL FOUR arms
Bridge stays balanced
Every C scales by K, all four ratios survive. \(C_{eq}\) just multiplies by K.
Insert K into a SERIES PAIR (both in one arm)
Bridge ratio changes
Need matching K in opposite arm's series pair to rebalance.
🔷 Symmetry & Speed Tricks
Trick
How to apply
Why it works
Reduce both ratios to lowest terms
If 2/4 = 1/2 and 3/6 = 1/2 → balanced
Spot balance in seconds without algebra.
Spot the diagonal pair
Unknown arm = (product of two neighbours) / (diagonally opposite)
Quick find-the-unknown shortcut.
All-equal cubes and bridges
Use symmetry to merge equal-potential nodes
Reduces complex networks to series-parallel chains.
If question gives C₅ but balance is implied
Ignore C₅ entirely
It contributes nothing once balance is established.
✅ The 5 things you MUST remember:
1. Balance ⇒ \(C_1/C_2 = C_3/C_4\) and Q₅ = 0.
2. Balanced bridge equivalent = (C₁C₂)/(C₁+C₂) + (C₃C₄)/(C₃+C₄). All-equal ⇒ C.
3. Cube of caps: edge 12C/7, face 4C/3, body 6C/5.
4. Cap voltage divider: smaller cap gets more voltage (\(V \propto 1/C\)).
5. Battery sees only \(C_{eq}\): Q = C_eq·V; U = ½ C_eq V².