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
Qjee_main_2026_21_jan_morningHeating Effect of Current
The heat generated in 1 minute between points A and B in the given circuit, when a battery of 9V with internal resistance of 1 Ω$1 \Omega$ is connected across these points is ____ J.
A bridge network with resistors 1 ohm, 2 ohms, 2 ohms, and 4 ohms forming the arms.
The circuit is a balanced Wheatstone bridge between A and B, because the ratio of adjacent arms is (1)/(2) = (2)/(4)$\frac{1}{2} = \frac{2}{4}$. Thus, the middle 1 Ω$1\,\Omega$ resistor (if there is one connecting the middle nodes) is ineffective.
A bridge network with resistors 1 ohm, 2 ohms, 2 ohms, and 4 ohms forming the arms.
The equivalent resistance across AB (RAB$R_{AB}$) involves the upper branch (1+2 = 3 Ω$1+2 = 3\,\Omega$) and the lower branch (2+4 = 6 Ω$2+4 = 6\,\Omega$) in parallel.
A bridge network with resistors 1 ohm, 2 ohms, 2 ohms, and 4 ohms forming the arms.
Total resistance in the circuit = RAB + rinternal = 2 + 1 = 3 Ω$R_{AB} + r_{\text{internal}} = 2 + 1 = 3\,\Omega$.
Total current from the battery i = 9V3 Ω = 3A$i = \frac{9\text{V}}{3\,\Omega} = 3\text{A}$.
The heat generated specifically between points A and B in 1 minute (60 seconds) is:
Check for balanced Wheatstone bridge first. Then use H = i² R t$H = i^2 R t$ strictly with the equivalent resistance of just the section AB$AB$ to find heat specifically generated across AB$AB$.
Chapter Mix
Class 12 Physics: Current Electricity
Q28jee_main_2026_21_jan_eveningPotentiometer
The total length of potentiometer wire AB is 50 cm$50 \text{ cm}$ in the arrangement as shown in figure. If P is the point where the galvanometer shows zero reading then the length AP is ________ cm.
Potentiometer circuit with a 50 cm wire AB, a galvanometer, and standard bridge resistors of 6 ohms and 4 ohms.
When the galvanometer shows zero reading, it implies that the bridge is balanced. The resistance of the wire segments is proportional to their lengths.
Let the length of segment AP be lAP$l_{AP}$ and PB be lPB$l_{PB}$.
Null deflection over a uniform wire directly translates resistance ratios into length ratios. Standard Wheatstone bridge application wrapped in potentiometer geometry.
Chapter Mix
Class 12 Physics: Current Electricity
Q35jee_main_2026_21_jan_eveningMaximum Power Transfer Theorem
P = I² R = ((E)/(R+r))² R$$P = I^2 R = \left(\frac{E}{R+r}\right)^2 R$$
Core Logic
To maximize power across the load resistance R$R$, we apply the Maximum Power Transfer Theorem. The power transferred to the load is maximum when the load resistance equals the internal resistance of the battery.
Power is maximum when the external resistance is strictly equal to the internal resistance (R=r$R=r$).
Pattern Recognition
Direct theoretical application of the Maximum Power Transfer Theorem.
Chapter Mix
Class 12 Physics: Current Electricity
Q40jee_main_2026_21_jan_eveningCombination of Resistors
Two known resistance of R Ω$R\,\Omega$ and 2R Ω$2R\,\Omega$ and one unknown resistance X Ω$X\,\Omega$ are connected in a circuit as shown in the figure. If the equivalent resistance between points A and B in the circuit is X Ω$X\,\Omega$, then the value of X is ________ Ω$\Omega$.
Bridge-like resistor circuit containing unknown resistance X and known resistances R and 2R.
A.(√(3)-1)R$(\sqrt{3}-1)R$
B.R$R$
C.2(√(3)-1)R$2(\sqrt{3}-1)R$
D.(√(3)+1)R$(\sqrt{3}+1)R$
Solution
Related Formula
For parallel combination: Req = (R₁ R₂)/(R₁ + R₂)$R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2}$
For series combination: Req = R₁ + R₂$R_{\text{eq}} = R_1 + R_2$
Core Logic
Based on the circuit topology, the 2R$2R$ and X$X$ resistors are in series. This combination is in parallel with the R$R$ resistor.
The total equivalent resistance across A and B is given as X$X$.
Self-referential equivalent resistance problems always collapse into a quadratic equation x² + ax + b = 0$x^2 + ax + b = 0$. Solve strictly for the positive root.
Chapter Mix
Class 12 Physics: Current Electricity
Q31jee_main_2026_22_january_morningMeter Bridge
A meter bridge with two resistances R₁$R_{1}$ and R₂$R_{2}$ as shown in figure was balanced (null point) at 40 cm from the point P. The null point changed to 50 cm from the point P, when 16~Ω$16~\Omega$ resistance is connected in parallel to R₂$R_{2}$. The values of resistances R₁$R_{1}$ and R₂$R_{2}$ are \_\_\_\_.
Meter bridge circuit with resistances R1 and R2 and balancing lengths.
Sees: Meter bridge balancing with parallel shunt resistance.
Shortcut: Set up length ratios for initial and modified states and solve linear simultaneous equations.
Check: Matches option (3). ✓
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.
