A 16 Ω$16 \, \Omega$ wire is bend to form a square loop. A 9V battery with internal resistance 1 Ω$1 \, \Omega$ is connected across one of its sides. If a 4 μ F$4 \, \mu \mathrm{F}$ capacitor is connected across one of its diagonals, the energy stored by the capacitor will be (x)/(2) μ J$\frac{x}{2} \, \mu \mathrm{J}$. where x =$x = $ _________.
Numerical Answer Type:
Enter a numerical valueAnswer: 81 to 81+4 marks
Solution & Explanation
Related Formula
Energy stored (U$U$) by a capacitor of capacitance C$C$ under steady state voltage VC$V_C$:
U = (1)/(2) C VC²$$U = \frac{1}{2} C V_C^2$$
Core Logic
A wire of resistance 16 Ω$16 \, \Omega$ is bent into a square, so each of the 4 sides has a resistance of:
The battery is connected across one side. Let this side have resistance 4 Ω$4 \, \Omega$. The remaining three sides are connected in series, creating a parallel branch with a combined resistance of:
Step 3: Find Potential Difference across the Diagonal
In a steady state, the capacitor acts as an open circuit. Let the diagonal link nodes be A$A$ and B$B$. The path contains two sides of the 12 Ω$12 \, \Omega$ branch (total resistance = 8 Ω$8 \, \Omega$):
Comparing this directly with the expression (x)/(2) μ J$\frac{x}{2} \, \mu \mathrm{J}$:
x = 81$x = 81$
Therefore, the value of x$x$ is 81$81$.
Pattern Recognition
Remember that capacitors act as standard open circuits when reaching steady-state DC conditions. Simply calculate the node potentials across the bridge connection points using standard current distribution laws first, then execute the energy equation.
Chapter Mix
Class 12 Physics: Current Electricity
Class 12 Physics: Electrostatic Potential and Capacitance
Circuit network layout showing the square loop resistance divisions and diagonal capacitor connection for Q57
More Current Electricity Previous-Year Questions — Page 10
Q39jee_main_2024_31_jan_eveningMeter Bridge
The resistance per centimeter of a meter bridge wire is r$r$, with X Ω$X \, \Omega$ resistance in left gap. Balancing length from left end is at 40 cm$40 \text{ cm}$ with 25 Ω$25 \, \Omega$ resistance in right gap. Now the wire is replaced by another wire of 2r$2r$ resistance per centimeter. The new balancing length for same settings will be at
For a meter bridge, the balancing condition is independent of the absolute resistance of the bridge wire as long as it is uniform. The ratio of the resistances in the gaps balances with the ratio of lengths.
Meter Bridge diagram for Q39 - JEE Main 2024 Evening
Since the ratio X/25$X/25$ remains identical, the balancing length ratio / (100- )$\ell / (100-\ell)$ also remains identical.
Therefore, ₂ = ₁ = 40 cm$\ell_2 = \ell_1 = 40 \text{ cm}$.
Pattern Recognition
Meter bridge balance point strictly depends on length ratio, NOT the specific resistivity or thickness of the wire (provided it is uniform). If external resistors don't change, the balance point never changes.
Chapter Mix
Class 12 Physics: Current Electricity
Q43jee_main_2024_31_jan_eveningPower Dissipation
By what percentage will the illumination of the lamp decrease if the current drops by 20%$20\%$?
A.46%$46\%$
B.26%$26\%$
C.36%$36\%$
D.56%$56\%$
Solution
Related Formula
Power (Illumination) is proportional to the square of the current for a constant resistance:
P = I² R$P = I^2 R$
P₂ = (0.8 I₁)² R = 0.64 I₁² R = 0.64 P₁$$P_2 = (0.8 I_1)^2 R = 0.64 I_1^2 R = 0.64 P_1$$
Step 1: Calculate Percentage Change
Δ P % = (P₂ - P₁)/(P₁) × 100%$$\Delta P \% = \frac{P_2 - P_1}{P_1} \times 100\%$$Δ P % = (0.64 P₁ - P₁)/(P₁) × 100%$$\Delta P \% = \frac{0.64 P_1 - P_1}{P_1} \times 100\%$$Δ P % = (0.64 - 1) × 100% = -36%$$\Delta P \% = (0.64 - 1) \times 100\% = -36\%$$
Step 2: Final Statement
The negative sign indicates a decrease. The illumination drops by 36%$36\%$.
Pattern Recognition
For squared relations y = x²$y = x^2$, if x$x$ changes by a factor k$k$ (0.8$0.8$), y$y$ changes by a factor k²$k^2$ (0.64$0.64$). 1 - 0.64 = 36%$1 - 0.64 = 36\%$. This bypasses algebraic limits usually done for small changes (like 2Δ x / x$2\Delta x / x$) since 20% is too large for approximation.
Chapter Mix
Class 12 Physics: Current Electricity
Q51jee_main_2024_31_jan_eveningPower in DC Circuits
In the following circuit, the battery has an emf of 2 V$2 \text{ V}$ and an internal resistance of (2)/(3) Ω$\frac{2}{3} \, \Omega$. The power consumption in the entire circuit is ______ W.
The image shows a DC circuit with multiple resistors in parallel/series connected to a 2V battery with 2/3 ohm internal resistance.
Numerical Answer.Answer: 3 to 3
Solution
Related Formula
P = V²Req$$P = \frac{V^2}{R_{eq}}$$
Core Logic
To find total power, collapse the entire external circuit and battery internal resistance into a single equivalent resistance Req$R_{eq}$ across the battery's ideal terminals.
Step 1: Equivalent Resistance Calculation
Analyzing the diagram:
The circuit simplifies to an equivalent resistance Req$R_{eq}$ combining the parallel/series elements along with the internal resistance r = 2/3 Ω$r = 2/3 \, \Omega$.
The final simplified equivalent resistance of the entire system calculates to:
Whenever "entire circuit" power is asked, include the battery's internal resistance inside Req$R_{eq}$ so you can use P = E² / Rtotal$P = E^2 / R_{total}$ directly.
Chapter Mix
Class 12 Physics: Current Electricity
Qjee_main_2024_31_jan_morningResistor Circuits
Equivalent resistance of the following network is ________ Ω$\Omega$
A complex resistor bridge network connecting nodes A and B with multiple branches.
A complex resistor bridge network connecting nodes A and B with multiple branches.
By carefully identifying the nodes, we can see that a 6 Ω$6\,\Omega$ resistor in the middle branch is short-circuited by a direct zero-resistance wire path across it.
A complex resistor bridge network connecting nodes A and B with multiple branches.
Removing the short-circuited 6 Ω$6\,\Omega$ resistor simplifies the circuit into three identical branches connected directly between the terminals A$A$ and B$B$.
A complex resistor bridge network connecting nodes A and B with multiple branches.
Step 2: Equivalent Calculation
The simplified circuit consists of three identical 3 Ω$3\,\Omega$ resistors in parallel.
Always trace nodes directly connected by straight wires (zero resistance). Any resistor with both ends connecting to the exact same electrical node is shorted out and can be erased.
Chapter Mix
Class 12 Physics: Current Electricity
Q48jee_main_2024_31_jan_morningTemperature Dependence Of Resistance
Two conductors have the same resistances at 0°C$0^{\circ}\mathrm{C}$ but their temperature coefficients of resistance are α₁$\alpha_1$ and α₂$\alpha_2$. The respective temperature coefficients for their series and parallel combinations are :
1 + αeq,p Δ T ≈ [1 + (α₁ + α₂)Δ T] [ 1 - (α₁ + α₂)/(2)Δ T ]$$1 + \alpha_{\text{eq,p}} \Delta T \approx [1 + (\alpha_1 + \alpha_2)\Delta T] \left[ 1 - \frac{\alpha_1 + \alpha_2}{2}\Delta T \right]$$1 + αeq,p Δ T ≈ 1 + (α₁ + α₂)Δ T - (α₁ + α₂)/(2)Δ T$$1 + \alpha_{\text{eq,p}} \Delta T \approx 1 + (\alpha_1 + \alpha_2)\Delta T - \frac{\alpha_1 + \alpha_2}{2}\Delta T$$αeq,p Δ T = (α₁ + α₂)/(2)Δ T$$\alpha_{\text{eq,p}} \Delta T = \frac{\alpha_1 + \alpha_2}{2}\Delta T$$αeq,p = (α₁ + α₂)/(2)$$\alpha_{\text{eq,p}} = \frac{\alpha_1 + \alpha_2}{2}$$
Pattern Recognition
For two identical base resistances, the effective temperature coefficient is simply the arithmetic mean of their individual coefficients, regardless of whether they are in series or parallel.
Chapter Mix
Class 12 Physics: Current Electricity
More Current Electricity Questions — jee_main_2024_29_jan_morning
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
