Capacitor Circuits — Every NEET Scenario

Part A — Series, Parallel, Mixed

Capacitors in series — equivalent C and charge sharing

NEET PYQ

Q. Three capacitors \(2\,\mu F\), \(3\,\mu F\) and \(6\,\mu F\) are connected in series across a 12 V battery. The equivalent capacitance and charge on each are:

(a) 1 μF, 12 μC(b) 11 μF, 132 μC(c) 6 μF, 72 μC(d) 1 μF, 24 μC

Given

\(C_1 = 2,\,C_2 = 3,\,C_3 = 6\,\mu F\) in series; \(V = 12\) V.

Asked

Equivalent C and charge on each capacitor.

Concept

Series → same charge Q flows through each cap; voltages add up to total V; reciprocals of C add.

Method

\[\dfrac{1}{C_{eq}} = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{3+2+1}{6} = 1\;\Rightarrow\;C_{eq} = 1\,\mu F\]

\[Q = C_{eq}\,V = 1 \times 12 = 12\,\mu C\;\text{(same on every cap)}\]

Easy trick

Series of capacitors behaves like resistor-parallel: \(1/C_{eq} = \sum 1/C_i\). The equivalent is smaller than the smallest cap. For three caps, just sum reciprocals over LCM.

Solution

\(C_{eq} = 1\,\mu F\); \(Q = 12\,\mu C\) on each. Answer: (a)

Capacitors in parallel — equivalent C and charge distribution

NEET PYQ

Q. Three capacitors \(2\,\mu F\), \(3\,\mu F\) and \(6\,\mu F\) are connected in parallel across a 12 V battery. The total charge stored is:

(a) 12 μC(b) 72 μC(c) 132 μC(d) 264 μC

Given

Three caps in parallel: 2, 3, 6 μF; V = 12 V.

Asked

Total charge \(Q_{\text{total}}\).

Concept

Parallel → same voltage across each; charges add; capacitances add directly.

Method

\[C_{eq} = 2+3+6 = 11\,\mu F,\quad Q_{tot} = C_{eq} V = 11 \times 12 = 132\,\mu C\]

Individual: \(Q_1 = 24,\;Q_2 = 36,\;Q_3 = 72\) μC (sum = 132 ✓).

Easy trick

Parallel of caps = "buckets side by side" → just add. Total charge is V × sum. Charge on each cap is in proportion to its own C (since V is shared).

Solution

\(Q_{tot} = 132\,\mu C\). Answer: (c)

Mixed series-parallel network

NEET style

Q. Two 6 μF capacitors are in parallel; this combination is in series with a 3 μF capacitor. The equivalent capacitance is:

(a) 2 μF(b) 3 μF(c) 4 μF(d) 15 μF

Given

Two 6 μF caps in parallel; this in series with a 3 μF cap.

Asked

Equivalent capacitance.

Concept

Collapse inner blocks first: parallel sum, then series of two using \(C_1 C_2/(C_1+C_2)\).

Method

Parallel block: \(6 + 6 = 12\,\mu F\).
Series with 3 μF: \(\dfrac{3 \times 12}{3 + 12} = \dfrac{36}{15} = 2.4\,\mu F\).

Easy trick

Two-cap series shortcut: product over sum. Always reduce parallel banks first — they sum cleanly.

Solution

\(C_{eq} = 2.4\,\mu F\) — closest to 2 μF. Answer: (a)

n identical capacitors — ratio of parallel to series equivalents

NEET classic

Q. n identical capacitors each of capacitance C are connected first all in parallel, then all in series. The ratio of the two equivalent capacitances \(C_p / C_s\) is:

(a) n(b) n²(c) 1/n(d) 1/n²

Given

n identical caps of capacitance C each.

Asked

Ratio \(C_p / C_s\).

Concept

Parallel of identical caps: just multiply by n. Series of identical caps: divide by n.

Method

\[C_p = nC,\quad C_s = \dfrac{C}{n},\quad \dfrac{C_p}{C_s} = n^2\]

Easy trick

For identical caps: parallel multiplies, series divides, both by n. Ratio of ratios = n². Energy ratio (same V) = n² too.

Solution

\(C_p/C_s = n^2\). Answer: (b)

Part B — Battery Connected (V fixed) vs Disconnected (Q fixed)

Battery connected · dielectric inserted

NEET PYQ

Q. A parallel-plate capacitor connected to a battery has capacitance C, charge Q, voltage V, energy U. A dielectric slab of constant K is fully inserted while the battery stays connected. The new values are:

(a) KC, KQ, V, KU(b) C/K, Q/K, V, U/K(c) KC, Q, V/K, U(d) KC, KQ, V/K, U/K

Given

Battery connected ⇒ V is held fixed. Dielectric K fills the gap.

Asked

New C, Q, V, U.

Concept

Battery on ⇒ V doesn't change. Capacitance becomes KC (geometric). Battery pushes more charge in so \(Q = CV\) ⇒ \(Q \to KQ\). Energy \(U = \tfrac{1}{2}CV^2 \to KU\). The battery does extra work to provide the new charge.

Method

Quantity	New value	Change
C	KC	×K
Q = CV	KQ	×K
V	V	same
E = V/d	V/d	same
U = ½CV²	KU	×K

Easy trick

"Battery ON ⇒ V locked." Everything that involves V stays the same; everything that involves C scales by K. Battery is the boss.

Solution

C → KC, Q → KQ, V → V, U → KU. Answer: (a)

Battery disconnected · dielectric inserted

NEET PYQ

Q. A parallel-plate capacitor is charged by a battery to (C, Q, V, U). The battery is then disconnected and a dielectric K is fully inserted. The new values are:

(a) KC, Q, V/K, U/K(b) C/K, Q/K, V, U(c) KC, KQ, V, KU(d) KC, KQ, KV, KU

Given

Battery off ⇒ Q is held fixed. Dielectric K fills the gap.

Asked

New C, Q, V, U.

Concept

Q can't leave. C still becomes KC geometrically. So V = Q/C drops by K. Energy \(U = Q^2/(2C)\) also drops by K (the dielectric does negative work on the slab — it's pulled in for free).

Method

Quantity	New value	Change
C	KC	×K
Q	Q	same
V = Q/C	V/K	÷K
E = V/d	E/K	÷K
U = Q²/(2C)	U/K	÷K

Easy trick

"Battery OFF ⇒ Q locked." Everything that involves Q stays. C scales by K (geometric), so V and U fall by K. The "missing" energy went into sucking the slab in.

Solution

C → KC, Q → Q, V → V/K, U → U/K. Answer: (a)

Battery connected · dielectric removed

NEET style

Q. A fully-dielectric-filled parallel-plate capacitor (constant K) is connected to a battery V. The dielectric is now slowly pulled out while the battery remains connected. Which quantity DROPS?

(a) V(b) C and Q (and U)(c) E (field between plates)(d) Plate separation d

Given

Battery on ⇒ V fixed. K removed slowly.

Asked

Which quantities decrease?

Concept

Removing K drops C by K. Since V is fixed, Q = CV drops by K. Energy U = ½CV² drops by K. Field E = V/d is unchanged.

Method

Quantity	Change
C	÷K
Q	÷K
V	same
E = V/d	same
U	÷K

Excess charge flows back into the battery — the battery absorbs energy.

Easy trick

"Removing K (battery on) is the mirror image of inserting K (battery on)" — flip every arrow: C↓, Q↓, V same, U↓.

Solution

C, Q, U all drop by factor K. V and E stay constant. Answer: (b)

Battery disconnected · dielectric removed (work done by external agent)

NEET PYQ

Q. A capacitor is charged with battery to energy \(U_0\), filled with dielectric K. Battery is disconnected; then the dielectric is fully pulled out. Work done by external agent in pulling out the slab is:

(a) (K−1) U₀(b) U₀(K−1)/K(c) K U₀(d) 0

Given

Initial state: \(C_0 = KC\), \(Q\) fixed, energy \(U_0 = Q^2/(2KC)\). Pull dielectric out.

Asked

Work done by external agent.

Concept

Q is locked. After removal, capacitance is C (was KC), so energy is \(Q^2/(2C) = K U_0\). The slab is sucked in by the field, so pulling it out costs work.

Method

\[U_{\text{final}} = K U_0\;\Rightarrow\;W_{\text{ext}} = U_{\text{final}} - U_0 = (K-1)U_0\]

Easy trick

"Battery OFF + slab removed ⇒ energy goes UP by factor K." Work done by you = increase in energy = (K−1) × original.

Solution

\(W_{ext} = (K-1)U_0\). Answer: (a)

Battery connected · plate separation doubled

NEET style

Q. A parallel-plate capacitor of capacitance C connected to a battery V has its plate separation slowly doubled while the battery stays connected. The change in stored energy is:

(a) halved(b) doubled(c) unchanged(d) quartered

Given

V fixed. \(C \propto 1/d\), so doubling d halves C.

Asked

Change in stored energy.

Concept

\(C \to C/2\). V fixed. \(U = \tfrac{1}{2}CV^2 \to U/2\).

Method

Qty	Change
C	÷2
Q	÷2
V	same
E = V/d	÷2
U	÷2

Easy trick

"Battery ON, pull plates apart" → C and everything proportional to C drops. V stays. Energy halves when d doubles.

Solution

U halves. Answer: (a)

Battery disconnected · plate separation doubled

NEET PYQ

Q. A capacitor is charged to (C, Q, V, U), disconnected from battery, and then plate separation is doubled. The new values are:

(a) C/2, Q, 2V, 2U(b) 2C, Q, V/2, U/2(c) C/2, Q/2, V, U(d) C/2, 2Q, 2V, 4U

Given

Battery off ⇒ Q fixed. d → 2d ⇒ C → C/2.

Asked

New C, Q, V, U.

Concept

Q fixed; C halves; V = Q/C doubles; U = Q²/(2C) doubles. Field E = σ/ε₀ unchanged because surface charge density is unchanged.

Method

Qty	Change
C	÷2
Q	same
V	×2
E	same
U	×2

Extra U comes from work done against the plate-attraction force while pulling them apart.

Easy trick

"Battery OFF + plates apart" → Q same, V up, U up. You do positive work against the attraction; that energy enters the capacitor.

Solution

C → C/2, Q → Q, V → 2V, U → 2U. Answer: (a)

Part C — Dielectric at Constant Speed

Dielectric slab pushed in at constant speed (battery on) — constant current

NEET PYQ

Q. A parallel-plate capacitor (plates of length L, breadth b, gap d) is connected to a battery V. A dielectric slab of constant K, same area, is inserted from one side at constant speed v while the battery remains connected. The current flowing in the circuit is:

(a) zero(b) varies with time(c) \(I = \dfrac{\varepsilon_0 b V (K-1) v}{d}\) (constant)(d) \(I = \dfrac{\varepsilon_0 b V K v}{d}\)

Given

Slab inserted at constant speed v, battery V connected.

Asked

Current in circuit.

Concept

As slab enters by length x = vt, the capacitor splits into two parallel strips (with K and air). Capacitance grows linearly with time. V fixed, so charge \(Q = CV\) grows linearly with time, and current \(I = dQ/dt = V \cdot dC/dt\) is constant.

Method

Strips in parallel: \(C(x) = \dfrac{\varepsilon_0 b}{d}\bigl[x K + (L - x)\bigr] = \dfrac{\varepsilon_0 b}{d}\bigl[L + (K-1)x\bigr]\).
Differentiate w.r.t. t: \(\dfrac{dC}{dt} = \dfrac{\varepsilon_0 b (K-1)}{d} \cdot v\).
Current: \(I = V \cdot \dfrac{dC}{dt} = \dfrac{\varepsilon_0 b V (K-1) v}{d}\).

Easy trick

Constant speed of insertion ⇒ constant rate of C change ⇒ constant current. The current = V × (rate of capacitance change). For K = 1 (no dielectric) current is zero — sanity check.

Solution

\(I = \dfrac{\varepsilon_0 b V (K-1) v}{d}\) — constant. Answer: (c)

Part D — Two-Capacitor Sharing & Switching

Two charged capacitors connected in parallel — common potential & energy loss

NEET PYQ

Q. A 3 μF capacitor charged to 100 V is connected in parallel with an uncharged 6 μF capacitor. Common voltage and energy lost are:

(a) 33.3 V, 0.01 J(b) 50 V, 0(c) 33.3 V, 0.005 J(d) 66.7 V, 0.01 J

Given

\(C_1 = 3\,\mu F\), \(V_1 = 100\) V; \(C_2 = 6\,\mu F\), \(V_2 = 0\). Joined same polarity.

Asked

Common voltage and energy lost.

Concept

Charge is conserved; the two caps end up at the same voltage. Energy is NOT conserved — some is lost as heat/EM during the transient redistribution.

Method

\[V_c = \dfrac{C_1 V_1 + C_2 V_2}{C_1 + C_2} = \dfrac{3(100) + 6(0)}{9} = 33.3\,\text{V}\]

\[\Delta U = \tfrac{1}{2}\,\dfrac{C_1 C_2}{C_1 + C_2}\,(V_1 - V_2)^2 = \tfrac{1}{2}\cdot\dfrac{18}{9}\cdot 100^2 = 10000\,\mu J = 0.01\,J\]

Easy trick

Common V = (weighted average by C). Loss = \(\tfrac{1}{2}\,\dfrac{C_1 C_2}{C_1+C_2}(V_1-V_2)^2\). If voltages equal, no loss.

Solution

\(V_c = 33.3\) V, \(\Delta U = 0.01\) J. Answer: (a)

Switch transfer — one cap charges another (battery is then removed)

NEET PYQ

Q. A 2 μF capacitor is fully charged to 10 V from a battery (switch S at position P). The switch is then thrown to Q, connecting it across an uncharged 8 μF capacitor. The final voltage across the 8 μF cap is:

(a) 10 V(b) 2 V(c) 8 V(d) 5 V

Given

First step: 2 μF cap charged to 10 V → \(Q_0 = 20\,\mu C\). Switch flipped to Q: now 2 μF and 8 μF in parallel, no battery.

Asked

Final voltage across the 8 μF cap.

Concept

After flip, battery is out. Total charge \(Q_0 = 20\,\mu C\) redistributes over total capacitance \(2 + 8 = 10\,\mu F\). Common voltage = total Q / total C.

Method

\[V_{\text{final}} = \dfrac{Q_0}{C_1 + C_2} = \dfrac{20}{10} = 2\,\text{V}\]

Easy trick

After-switch problems: charge conservation across the now-parallel banks. The final V is the same on both. Energy lost: \(\tfrac{1}{2}\dfrac{C_1 C_2}{C_1+C_2}V_1^2\) (since second cap starts at 0).

Solution

\(V_{\text{final}} = 2\) V. Answer: (b)

Part E — Dielectrics & Slabs of Various Shapes

Partial dielectric (slab of thickness t inside the gap)

NEET PYQ

Q. A parallel-plate capacitor has plate area A and gap d. A dielectric slab of constant K and thickness \(t < d\) is inserted parallel to the plates. The new capacitance is:

(a) \(\dfrac{\varepsilon_0 A}{d}\)(b) \(\dfrac{\varepsilon_0 A}{d - t + t/K}\)(c) \(\dfrac{K \varepsilon_0 A}{d}\)(d) \(\dfrac{\varepsilon_0 A}{d - t}\)

Given

Plate area A, gap d, slab of K with thickness t (\(t

Asked

Equivalent capacitance.

Concept

Treat as TWO capacitors in SERIES: an air gap of thickness (d − t) and a dielectric slab of thickness t. Same plate area A; same Q on both.

Method

\[\dfrac{1}{C} = \dfrac{d - t}{\varepsilon_0 A} + \dfrac{t}{K \varepsilon_0 A}\;\Rightarrow\;C = \dfrac{\varepsilon_0 A}{d - t + t/K}\]

Limits: t=0 → bare cap; t=d → fully filled (KC₀); K→∞ → \(\dfrac{\varepsilon_0 A}{d-t}\) (slab acts as a metal layer).

Easy trick

"Effective gap" = \(d - t + t/K\). The slab replaces thickness t of air with thickness t/K of effective air. Memorise the formula in this form.

Solution

\(C = \dfrac{\varepsilon_0 A}{d - t + t/K}\). Answer: (b)

Two dielectrics — side-by-side (parallel) vs stacked (series)

NEET PYQ

Q. Two dielectrics of constants K₁ and K₂ fill a parallel-plate gap of area A and separation d.
(i) They lie side by side (each occupies half the plate area). Equivalent C is:
(ii) They are stacked, each layer of thickness d/2. Equivalent C is:

Given

Two dielectrics K₁ and K₂, plate area A, gap d. Two geometries.

Asked

Equivalent C in each case.

Concept

Side-by-side: same voltage across each sub-cap → parallel. Stacked: same charge flows through each layer → series.

Method

(i) Side-by-side: two parallel halves each with area A/2.

\[C = \dfrac{\varepsilon_0(A/2)K_1}{d} + \dfrac{\varepsilon_0(A/2)K_2}{d} = \dfrac{\varepsilon_0 A (K_1 + K_2)}{2 d}\]

(ii) Stacked: two series layers each of thickness d/2.

\[\dfrac{1}{C} = \dfrac{d/2}{\varepsilon_0 A K_1} + \dfrac{d/2}{\varepsilon_0 A K_2}\;\Rightarrow\;C = \dfrac{2\varepsilon_0 A K_1 K_2}{d(K_1 + K_2)}\]

Easy trick

"Same V ⇒ parallel ⇒ arithmetic mean". "Same Q ⇒ series ⇒ harmonic mean". Side-by-side = arithmetic; stacked = harmonic. Memorise this pair.

Solution

Side-by-side: \(C = \dfrac{\varepsilon_0 A(K_1+K_2)}{2d}\). Stacked: \(C = \dfrac{2\varepsilon_0 A K_1 K_2}{d(K_1+K_2)}\).

Metal (conductor) slab inserted into the gap

NEET PYQ

Q. A parallel-plate capacitor has gap d. A conducting (metal) slab of thickness t is inserted parallel to the plates. The new capacitance is:

(a) \(\dfrac{\varepsilon_0 A}{d}\)(b) \(\dfrac{\varepsilon_0 A}{d - t}\)(c) \(\dfrac{\varepsilon_0 A}{d + t}\)(d) 0

Given

Metal slab of thickness t in a gap of d. (Equivalent to K → ∞.)

Asked

New capacitance.

Concept

Inside a conductor E = 0 — it short-circuits its own thickness. Effective gap becomes (d − t).

Method

\[C = \dfrac{\varepsilon_0 A}{d - t}\]

Key fact: C does NOT depend on the position of the slab — only on its thickness.

Easy trick

Metal slab = "the air gap is just shorter by t". Plug into \(C = \varepsilon_0 A/(d - t)\) and ignore where the slab sits in the gap.

Solution

\(C = \dfrac{\varepsilon_0 A}{d - t}\). Answer: (b)