Two capacitors C1 and C2 are connected in parallel to a battery. Charge-time graph is shown below for the two capacitors. The energy stored with them are U1 and U2 , respectively. Which of the given statements is true?

Since both capacitors are connected in parallel across the same battery, their potential difference V is identical. From the given charge-time graph, at any specific instant of time, the charge accumulated on the second capacitor is greater than that on the first: Using the relation q = CV, since V is identical, we get: Now, the electrostatic energy stored in a capacitor is given by: Since C2 > C1 and V is constant, the stored energy satisfies: Thus, the correct relationships are C2 > C1 and U2 > U1. Sees parallel connection → Immediately lock potential difference V as constant. This simplifies q propto C and U propto C, creating a direct linear bridge from graph height to energy capacity. Class 12 Physics: Electrostatic Potential and Capacitance

Two capacitors C1 and C2 are connected in parallel to a battery. Charge-time graph is shown below for the two capacitors. The energy stored with them are U1 and U2 , respectively. Which of the given s

Since both capacitors are connected in parallel across the same battery, their potential difference V is identical. From the given charge-time graph, at any specific instant of time, the charge accumulated on the second capacitor is greater than that on the first: Using the relation q = CV, since V is identical, we get: Now, the electrostatic energy stored in a capacitor is given by: Since C2 > C1 and V is constant, the stored energy satisfies: Thus, the correct relationships are C2 > C1 and U2

Two capacitors C1 and C2 are connected in parallel to a battery. Charge-time graph is shown below for the two capacitors. The energy stored with them are U1 and U2 , respectively. Which of the given statements is true?

Since both capacitors are connected in parallel across the same battery, their potential difference V is identical. From the given charge-time graph, at any specific instant of time, the charge accumulated on the second capacitor is greater than that on the first: Using the relation q = CV, since V is identical, we get: Now, the electrostatic energy stored in a capacitor is given by: Since C2 > C1 and V is constant, the stored energy satisfies: Thus, the correct relationships are C2 > C1 and U2 > U1. Sees parallel connection → Immediately lock potential difference V as constant. This simplifies q propto C and U propto C, creating a direct linear bridge from graph height to energy capacity. Class 12 Physics: Electrostatic Potential and Capacitance

Two capacitors C1 and C2 are connected in parallel to a battery. Charge-time graph is shown below for the two capacitors. The energy stored with them are U1 and U2 , respectively. Which of the given s

Since both capacitors are connected in parallel across the same battery, their potential difference V is identical. From the given charge-time graph, at any specific instant of time, the charge accumulated on the second capacitor is greater than that on the first: Using the relation q = CV, since V is identical, we get: Now, the electrostatic energy stored in a capacitor is given by: Since C2 > C1 and V is constant, the stored energy satisfies: Thus, the correct relationships are C2 > C1 and U2

Two capacitors C1 and C2 are connected in parallel to a battery. Charge-time graph is shown below for the two…

Year	2026	2025	2024	Total
Questions	3	7	3	13

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

V = same

U = (1)/(2) CV²

q = CV

Core Logic

Since both capacitors are connected in parallel across the same battery, their potential difference $V$ $\mathrm{V}$ is identical.

From the given charge-time graph, at any specific instant of time, the charge accumulated on the second capacitor is greater than that on the first:

q₂ > q₁

Using the relation $q = CV$ $\mathrm{q} = \mathrm{C}\mathrm{V}$, since $V$ $\mathrm{V}$ is identical, we get:

C₂ > C₁

Now, the electrostatic energy stored in a capacitor is given by:

U = (1)/(2) CV²

Since $C₂ > C₁$ $\mathrm{C}_2 > \mathrm{C}_1$ and $V$ $\mathrm{V}$ is constant, the stored energy satisfies:

U₂ > U₁

Step 1: Final Conclusion

Thus, the correct relationships are $C₂ > C₁$ $\mathrm{C}_2 > \mathrm{C}_1$ and $U₂ > U₁$ $\mathrm{U}_2 > \mathrm{U}_1$.

Pattern Recognition

Sees parallel connection → Immediately lock potential difference $V$ $\mathrm{V}$ as constant. This simplifies $q \propto C$ $\mathrm{q} \propto \mathrm{C}$ and $U \propto C$ $\mathrm{U} \propto \mathrm{C}$, creating a direct linear bridge from graph height to energy capacity.

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance

More Electrostatic Potential and Capacitance Previous-Year Questions — Page 3

Q35 jee_main_2024_01_february_morning Capacitors and Capacitance

Two identical capacitors have same capacitance

C

. One of them is charged to the potential

V

and other to the potential

2V

. The negative ends of both are connected together. When the positive ends are also joined together, the decrease in energy of the combined system is:

A. $(1)/(4) CV²$ $\frac{1}{4} CV^2$
B. $2 CV²$ $2 CV^2$
C. $(1)/(2) CV²$ $\frac{1}{2} CV^2$
D. $(3)/(4) CV²$ $\frac{3}{4} CV^2$

Solution

Related Formula

Energy stored in a capacitor:

U = (1)/(2)CV²

Common potential after sharing charges:

VC = (q₁ + q₂)/(C₁ + C₂)

Loss of energy:

Δ U = (1)/(2) (C₁ C₂)/(C₁ + C₂) (V₁ - V₂)²

Core Logic

Initial energy of the individual systems:

Uᵢ = (1)/(2)CV² + (1)/(2)C(2V)² = (1)/(2)CV² + 2CV² = (5)/(2)CV²

Common potential $VC$ $V_C$ when joined in parallel:

VC = (CV + C(2V))/(C + C) = (3CV)/(2C) = (3)/(2)V

Step 1: Calculate Final Energy and Loss

Final energy of the combined system:

Uf = (1)/(2)(2C)VC² = C ((3)/(2)V)² = (9)/(4)CV²

Decrease in energy (loss):

Δ U = Uᵢ - Uf = (5)/(2)CV² - (9)/(4)CV² = (1)/(4)CV²

Pattern Recognition

Direct Formula Shortcut:

Δ U = (1)/(2) (C \cdot C)/(2C) (2V - V)² = (1)/(4)C(V)² = (1)/(4)CV²

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance

Q56 jee_main_2024_29_january_evening Capacitance and Resistor Networks at Steady State

In the given figure, the charge stored in

6

capacitor, when points A and B are joined by a connecting wire is ________

Capacitor-resistor bridge circuit for Q56 - JEE Main 2024 29 January Shift 2 — The diagram displays a bridge-like network containing a 6 Ohm resistor, 3uF capacitor, 6uF capacitor, and 3 Ohm resistor powered by a 9V supply.

Numerical Answer. Answer: 36 to 36

Solution

Related Formula

At steady state, a capacitor acts as an open circuit to DC current. The charge on a capacitor is:

$Q = C Δ V$ $Q = C \Delta V$

Core Logic

When node $A$ $A$ and node $B$ $B$ are connected by a wire, they reach the same electrical potential ( $VA = VB = VM$ $V_A = V_B = V_M$).

At DC steady state, capacitors block current, so current only flows through the resistors from the $9 V$ $9\text{ V}$ source to ground:

The $6 Ω$ $6\ \Omega$ resistor is connected between $9 V$ $9\text{ V}$ and node $M$ $M$.
The $3 Ω$ $3\ \Omega$ resistor is connected between node $M$ $M$ and ground.

Hence, the equivalent series resistance for the DC current path is:

Req = 6 + 3 = 9 Ω

The steady state current is:

I = VReq = 9 V9 Ω = 1 A

Simplified equivalent steady-state circuit for Q56 — The diagram displays a bridge-like network containing a 6 Ohm resistor, 3uF capacitor, 6uF capacitor, and 3 Ohm resistor powered by a 9V supply.

Step 1: Calculate the Potential Difference and Charge

The potential at node $M$ $M$ ( $VM$ $V_M$) is:

VM = I \times 3 Ω = 1 A \times 3 Ω = 3 V

The $6$ $6\mu\text{F}$ capacitor is connected between the $9 V$ $9\text{ V}$ source and node $B$ $B$ (which is at potential $VM = 3 V$ $V_M = 3\text{ V}$).

Thus, the potential difference across the $6$ $6\mu\text{F}$ capacitor is:

Δ V₆ = 9 V - 3 V = 6 V

The charge stored in the $6$ $6\mu\text{F}$ capacitor is:

Q = C Δ V = 6 \times 6 V = 36

Thus, the charge stored is $36$ $36\ \mu\text{C}$.

Pattern Recognition

Shorting $A$ $A$ and $B$ $B$ makes the network a simple voltage divider for resistors at steady state. Once potential of the middle node is found ( $3 V$ $3\text{ V}$), the capacitor charge is calculated instantly using $Q = C Δ V$ $Q = C \Delta V$.

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance

Q57 jee_main_2024_30_jan_morning Energy Loss in Capacitors

A capacitor of capacitance C and potential V has energy E. It is connected to another capacitor of capacitance 2C and potential 2V. Then the loss of energy is

(x)/(3)E

, where x is ________.

Numerical Answer. Answer: 2 to 2

Solution

Related Formula

E = (1)/(2) C V²

Δ E = (1)/(2) (C₁ C₂)/(C₁ + C₂) (V₁ - V₂)²

Core Logic

When two charged capacitors are connected in parallel, charge redistributes until they reach a common potential. During this redistribution, energy is dissipated as heat, strictly governed by the standard loss formula.

Step 1: Assign Values

Initial energy of first capacitor: $E = (1)/(2) C V²$ $E = \frac{1}{2} C V^2$ Capacitor 1: $C₁ = C$ $C_1 = C$, $V₁ = V$ $V_1 = V$ Capacitor 2: $C₂ = 2C$ $C_2 = 2C$, $V₂ = 2V$ $V_2 = 2V$

Step 2: Calculate Energy Loss

Δ E = (1)/(2) ((C)(2C))/(C + 2C) (V - 2V)²

Δ E = (1)/(2) (2C²)/(3C) (-V)²

Δ E = (1)/(2) ((2C)/(3)) V²

Δ E = (2)/(3) ((1)/(2) C V²)

Δ E = (2)/(3) E

Step 3: Match the Pattern

Given loss is $(x)/(3)E$ $\frac{x}{3}E$. Comparing, we get $x = 2$ $x = 2$.

Pattern Recognition

The loss formula $Δ E = (1)/(2) Ceq(Δ V)²$ elegantly bypasses recalculating common potential $Vc$ $V_c$ and summing final state energies.

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More Electrostatic Potential and Capacitance Previous-Year Questions — Page 3

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Related Formula

Core Logic

Step 1: Calculate Final Energy and Loss

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate the Potential Difference and Charge

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Assign Values

Step 2: Calculate Energy Loss

Step 3: Match the Pattern

Pattern Recognition

Chapter Mix

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Related Formula

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