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Current Electricity appeared 50 times across 3 years — 5.8% of Physics. This question is from Combination of Resistors.

Year	2026	2025	2024	Total
Questions	18	13	19	50

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Find the equivalent resistance between two ends of the following circuit.
A complex looking resistive mesh whose node potentials reveal a simple symmetrical pattern.

A. $r$ $\mathrm{r}$
B. $r6$ $\frac{\mathrm{r}}{6}$
C. $r9$ $\frac{\mathrm{r}}{9}$
D. $r3$ $\frac{\mathrm{r}}{3}$

Solution & Explanation

Core Logic

Label the circuit nodes carefully to see how the elements are connected across terminal zones:

Node potential labelling steps for Q14 — A complex looking resistive mesh whose node potentials reveal a simple symmetrical pattern.

Node potential labelling steps for Q14 — A complex looking resistive mesh whose node potentials reveal a simple symmetrical pattern.

Tracing parallel loops reveals all inner subsets share identical path boundaries.

Req = r/33 = r9

Step 1: Final Expression

The overall simplified system network value equals $r9$ $\frac{\mathrm{r}}{9}$, matching option (3).

Pattern Recognition

Short-circuit tracking lines allow collapsing complex meshes into basic parallel branches. If all nodes connect symmetrically, apply simple division: $R/N$ $\mathrm{R}/\mathrm{N}$.

Chapter Mix

Class 12 Physics: Current Electricity

Reference Study Guides

Current Electricity

More Current Electricity Previous-Year Questions

Q jee_main_2026_21_jan_morning Heating 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

is connected across these points is ____ J.

Heating Effect of Current diagram for Q47 - JEE Main 2026 Morning — A bridge network with resistors 1 ohm, 2 ohms, 2 ohms, and 4 ohms forming the arms.

Numerical Answer. Answer: 1080 to 1080

Solution

Related Formula

Req = (R₁ R₂)/(R₁ + R₂) (parallel)

$H = I² R t$ $H = I^2 R t$

I = VReq + r

Core Logic

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.

Heating Effect of Current solution diagram for Q47 - JEE Main 2026 Morning — 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.

Heating Effect of Current solution diagram for Q47 - JEE Main 2026 Morning — A bridge network with resistors 1 ohm, 2 ohms, 2 ohms, and 4 ohms forming the arms.

RAB = (3 \times 6)/(3 + 6) = (18)/(9) = 2 Ω

Step 1: Calculate Current and Heat

Total resistance in the circuit = $RAB + rinternal = 2 + 1 = 3 Ω$ . Total current from the battery $i = 9V3 Ω = 3A$ .

The heat generated specifically between points A and B in 1 minute (60 seconds) is:

HAB = i² RAB t = (3)² \times 2 \times 60

HAB = 9 \times 120 = 1080 J

Pattern Recognition

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

Q28 jee_main_2026_21_jan_evening Potentiometer

The total length of potentiometer wire AB is

50 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 diagram for Q28 - JEE Main 2026 Evening — Potentiometer circuit with a 50 cm wire AB, a galvanometer, and standard bridge resistors of 6 ohms and 4 ohms.

A. $15$ $15$
B. $30$ $30$
C. $25$ $25$
D. $20$ $20$

Solution

Related Formula

For a balanced Wheatstone bridge configuration:

(R₁)/(R₂) = (l₁)/(l₂)

Core Logic

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 $lPB$ $l_{PB}$.

6 ΩRAP = 4 ΩRPB

Since $R \propto l$ $R \propto l$:

RAPRPB = lAPlPB = (6)/(4) = (3)/(2)

Step 1: Applying Total Length Constraint

We are given the total length:

lAP + lPB = 50 cm

Using the ratio $lAP = (3)/(2) lPB$ $l_{AP} = \frac{3}{2} l_{PB}$:

(3)/(2)lPB + lPB = 50

(5)/(2)lPB = 50 lPB = 20 cm

Step 2: Final Conclusion

Solving for $lAP$ $l_{AP}$:

lAP = 50 - 20 = 30 cm

Alternatively, $lAP = (3)/(3+2) \times 50 = (3)/(5) \times 50 = 30 cm$ .

Pattern Recognition

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

Q35 jee_main_2026_21_jan_evening Maximum Power Transfer Theorem

A battery with EMF E and internal resistance r is connected across a resistance R. The power consumption in R will be maximum when :

A. $R=2r$ $R=2r$
B. $R=(r)/(2)$ $R=\frac{r}{2}$
C. $R=\sqrt(2)r$ $R=\sqrt{2}r$
D. $R=r$ $R=r$

Solution

Related Formula

P = I² R = ((E)/(R+r))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.

Setting $(dP)/(dR) = 0$ $\frac{dP}{dR} = 0$ yields: $R = r$ $R = r$

Step 1: Final Conclusion

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

Q40 jee_main_2026_21_jan_evening Combination of Resistors

Two known resistance of

R Ω

and

2R Ω

and one unknown resistance

X Ω

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

, then the value of X is ________

Ω

.

Circuit diagram of resistors for Q40 - JEE Main 2026 Evening — Bridge-like resistor circuit containing unknown resistance X and known resistances R and 2R.

A. $(\sqrt(3)-1)R$ $(\sqrt{3}-1)R$
B. $R$ $R$
C. $2(\sqrt(3)-1)R$ $2(\sqrt{3}-1)R$
D. $(\sqrt(3)+1)R$ $(\sqrt{3}+1)R$

Solution

Related Formula

For parallel combination: $Req = (R₁ R₂)/(R₁ + R₂)$ 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 $X$ $X$.

Req = ((2R + X) \cdot (R))/((2R + X) + R)

Step 1: Setting up the Quadratic

Set the equivalent resistance equal to $X$ $X$:

X = ((2R + X)R)/(3R + X)

Cross-multiplying yields:

X(3R + X) = 2R² + XR

3RX + X² = 2R² + XR

X² + 2RX - 2R² = 0

Step 2: Solving for X

Using the quadratic formula to solve for $X$ $X$:

X = -2R \pm \sqrt((2R)² - 4(1)(-2R²))2

X = -2R \pm \sqrt(4R² + 8R²)2 = -2R \pm \sqrt(12R²)2

X = -2R \pm 2R\sqrt(3)2 = -R \pm R\sqrt(3)

Since resistance must be positive:

X = (\sqrt(3) - 1)R

Pattern Recognition

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

Q31 jee_main_2026_22_january_morning Meter Bridge

A meter bridge with two resistances

R₁

and

R₂

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

resistance is connected in parallel to

R₂

. The values of resistances

R₁

and

R₂

are \_\_\_\_.

Current Electricity diagram for Q31 - JEE Main 2026 January Morning — Meter bridge circuit with resistances R1 and R2 and balancing lengths.

A. $R₂ = 16Ω, R₁ = (16)/(3)Ω$
B. $R₂ = 4Ω, R₁ = (4)/(3)Ω$
C. $R₂ = 8Ω, R₁ = (16)/(3)Ω$
D. $R₂ = 12Ω, R₁ = (12)/(3)Ω$

Solution

Related Formula

(R₁)/(R₂) = (l₁)/(100 - l₁)

Core Logic

Solution diagram for Q31 - JEE Main 2026 Morning — Meter bridge circuit with resistances R1 and R2 and balancing lengths.

Initially:

$(R₁)/(R₂) = (40)/(60) = (2)/(3) R₁ = (2)/(3)R₂$ ---- (1)

After connecting $16~Ω$ $16~\Omega$ in parallel with $R₂$ $R_2$:

$(R₁)/(((R₂ \times 16)/(R₂ + 16))) = (50)/(50) R₁ = (16 R₂)/(16 + R₂)$ ---- (2)

Solving equations (1) and (2) yields:

R₂ = 8~Ω, R₁ = (16)/(3)~Ω

Pattern Recognition

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

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