Find the ratio of the potential differences that must be applied across the parallel and series combination of

Exercises · 8.14

Q. Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors C1 and C2 with their capacitances in the ratio 1:2, so that the energy stored in these two cases becomes the same. [Vp : Vs = 2:3]

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★

9% · 4/45 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Let $C_1=C$ , $C_2=2C$ (ratio 1:2). Parallel equivalent: $C_p=C_1+C_2=3C$ . Series equivalent: $\dfrac{1}{C_s}=\dfrac{1}{C}+\dfrac{1}{2C}=\dfrac{3}{2C}\Rightarrow C_s=\dfrac{2C}{3}$ . Equal energies at applied voltages $V_p,V_s$ : $\tfrac{1}{2}C_pV_p^2=\tfrac{1}{2}C_sV_s^2\Rightarrow 3C\,V_p^2=\dfrac{2C}{3}V_s^2\Rightarrow \dfrac{V_p^2}{V_s^2}=\dfrac{2}{9}\Rightarrow \dfrac{V_p}{V_s}=\dfrac{\sqrt2}{3}$ , i.e. $V_p:V_s=\sqrt2:3$ . This computed ratio does not exactly match the textbook's own printed bracketed answer of 'Vp:Vs=2:3' -- but $\sqrt2$ rounds to about 1.41, and given how heavily this document's raw text extraction mangled square-root and other maths symbols throughout (confirmed by many …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 14 on this concept.

CBSE 2025Set D1 markMCQ
$Q. In parallel combination of condensers which quantity remains same for each condenser? (A) Charge (B) Energy (C) Potential difference (D) Capacity$
›Reveal solutionSolution

In a parallel combination, every capacitor shares the same two terminals, so the potential difference across each is identical.

When capacitors are connected in parallel, their plates are joined to the same pair of nodes. The voltage across each capacitor therefore equals the common terminal voltage V.

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$Q. Determine the equivalent capacitance between the points A and B in the adjoining figure.$
›Reveal solutionSolution

First combine the two parallel capacitors, then combine that result in series with the 6 pF capacitor.

Step 1 — parallel combination: the 1 pF and 2 pF capacitors are connected in parallel between the two junction nodes:

$C_{parallel} = 1 + 2 = 3\ \text{pF}$

Step 2 — series combination: this 3 pF combination is in series with the 6 pF capacitor between A and B:

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. Three capacitors each having capacitance C are connected in series. Their equivalent capacitance is$
(a) 3C
(b) 3/C
(c) C/3
(d) 1/3C
›Reveal solutionSolution

Capacitors in series combine like resistors in parallel: 1/C_eq = 1/C1 + 1/C2 + 1/C3, so three equal capacitors C give C_eq = C/3.

For capacitors in series, the same charge Q flows onto each, and the total voltage is the sum of individual voltages, which leads to:

1/C_eq = 1/C + 1/C + 1/C = 3/C

C_eq = C/3

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$Q. In which order capacitors are connected to increase resultant capacitance?$
›Reveal solutionSolution
Parallel combination increases equivalent capacitance; series combination decreases it.

For capacitors $C_1, C_2, C_3, \ldots$ connected:

In series: $\dfrac{1}{C_{eq}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dfrac{1}{C_3} + \ldots$ — the equivalent capacitance is always smaller than the smallest individual capacitance (because effectively the plate separation increases / the charge stored is the same on each but the total voltage adds up). …
Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$Q. What is the equivalent capacitance between A and B ?$
›Reveal solutionSolution
This 5-capacitor network is a balanced capacitor bridge (all five are equal, 4 μF each), so the diagonal 'bridge' capacitor carries no charge and can be removed; what's left is two 2 μF series paths in parallel, giving 4 μF.

Label the apex of the triangle P and the mid-point node of the two base capacitors Q. The network is:

$A\!-\!P$ : 4 μF (left slant)

$P\!-\!B$ : 4 μF (right slant)

$A\!-\!Q$ : 4 μF (left half of base)

$Q\!-\!B$ : 4 μF (right half of base)

$P\!-\!Q$ : 4 μF (the bridging capacitor)

This is exactly a Wheatstone-bridge-like arrangement. A capacitor bridge is balanced (no charge on the bridge arm, hence no potential difference across it) when

$\dfrac{C_{AP}}{C_{AQ}} = \dfrac{C_{PB}}{C_{QB}}$

…
Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$Q. The ratio between resultant capacitances Cs and Cp of two equal capacitors in series combination and parallel combination respectively is$
(a) 1
(b) 1/2
(c) 1/4
(d) 1/8
›Reveal solutionSolution

Two equal capacitors C in series give C_s = C/2; in parallel they give C_p = 2C; the ratio C_s/C_p = 1/4.

Let each capacitor have capacitance C.

Series combination: for two capacitors in series, 1/C_s = 1/C + 1/C = 2/C, so C_s = C/2.

Parallel combination: for two capacitors in parallel, capacitances simply add: C_p = C + C = 2C.

Now form the ratio:

C_s/C_p = (C/2)/(2C) = 1/4

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 mark
$Q. Two capacitors of capacities 5μF and 10μF respectively are connected in series. Calculate the resultant capacity of the combination.$
›Reveal solutionSolution

Reciprocal-sum rule for capacitors in series.

$\frac{1}{C_{eq}} = \frac{1}{C_1}+\frac{1}{C_2} = \frac{1}{5}+\frac{1}{10} = \frac{2}{10}+\frac{1}{10} = \frac{3}{10}$

$C_{eq} = \frac{10}{3} \approx 3.33\ \mu\text{F}$ …

Start your 14-day free trial to see the full solution →
CBSE 2023Set F1 markMCQ
$Q. Three capacitors of capacitance 6 μF are available. The minimum and maximum capacitances obtained are (A) 3 μF, 12 μF (B) 2 μF, 12 μF (C) 2 μF, 18μF (D) 6 μF, 18 μF$
›Reveal solutionSolution

All in series ⇒ 2 μF (minimum); all in parallel ⇒ 18 μF (maximum).

For three equal capacitors of $C = 6\,\mu\text{F}$ :

…

Start your 14-day free trial to see the full solution →
CBSE 2023Set ANNUAL1 markMCQ
$Q. For the arrangement of the capacitors as shown in figure, the net capacitance between points A and B is -$
(a) 1.7 μF
(b) 2.33 μF
(c) 2.8 μF
(d) 7 μF
›Reveal solutionSolution

Redrawing the loop as a triangle of three capacitors (4 μF, 2 μF, 1 μF) between three nodes, the two in series combine and that combination sits in parallel with the third.

Why / setup: Label the top-left/bottom-left corner (joined by the plain left wire) as node X, and the bottom-right corner as node Y. Since the right wire is plain, Y is the same node as A. So the circuit reduces to three capacitors between three nodes: A–X (4 μF, the top edge), X–B (2 μF, left half of the bottom edge), and B–A (1 μF, right half of the bottom edge, since B–Y is the 1 μF capacitor and Y=A).

Steps: …

Start your 14-day free trial to see the full solution →
CBSE 2022Set ANNUAL1 markMCQ
$C_1$
(a) $C = C_1 + C_2$
(b) $\dfrac{1}{C} = \dfrac{1}{C_1} + \dfrac{1}{C_2}$
(c) $C = C_1 \times C_2$
(d) $\dfrac{1}{C} = \dfrac{1}{C_1} \times \dfrac{1}{C_2}$
›Reveal solutionSolution

For capacitors in series, reciprocals of capacitance add.

…

Start your 14-day free trial to see the full solution →
CBSE 2022Set I1 markMCQ
$Q. If n capacitors of equal capacity C_1 are connected in parallel, the equivalent capacity will be (A) C = n/C_1 (B) C = C_1/n (C) C = nC_1 (D) C = n^2 C_1$
›Reveal solutionSolution

Parallel: $C = nC_1$ .

When capacitors are connected in parallel, they share the same potential difference and their capacitances add:

$C = C_1 + C_2 + \dots + C_n.$

If all $n$ capacitors have equal capacitance $C_1$ , then

$C = nC_1.$

…

Start your 14-day free trial to see the full solution →
CBSE 2021Set A1 markMCQ
$Q. Equivalent capacity between A and B in figure is (A) C (B) C/2 (C) 2C (D) 2/C$
›Reveal solutionSolution

Series C with a parallel pair (2C) gives 2C/3 — not one of the printed options, so the option set appears misprinted.

From the figure description: from A a capacitor C is in series, then two capacitors each of value C are in parallel leading to B.

Step 1 — the two parallel capacitors combine by addition:

$C_{\parallel} = C + C = 2C.$

Step 2 — this parallel block is in series with the first capacitor C:

$\frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{2C} = \frac{2+1}{2C} = \frac{3}{2C} \;\Rightarrow\; C_{eq} = \frac{2C}{3}.$

The result $2C/3$ does not match any of the options (a) C,

(B) C/2,

(C) 2C, …

Start your 14-day free trial to see the full solution →

←Previous question8.13 — A capacitor has some dielectric between its plates and the capacitor i…→Next question8.15 — Two charges of magnitudes -4Q and +2Q are located at points (2a, 0) an…

Q.Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors C1 and C2 with their capacitances in the ratio 1:2, so that the energy stored in these two cases becomes the same. [Vp : Vs = 2:3]

Concept understanding — Capacitors in Series and Parallel

Q.Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors C1 and C2 with their capacitances in the ratio 1:2, so that the energy stored in these two cases becomes the same. [Vp : Vs = 2:3]

🎓Unlock everything free for 14 days

Unlock everything free for 14 days