A wire of resistance R$R$ is bent into an equilateral triangle and an identical wire is bent into a square. The ratio of resistance between the two end points of an edge of the triangle to that of the square is

Related Formula

H = I² R t$H = I^2 R t$

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.

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.

Step 1: Calculate Current and Heat

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:

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

Related Formula

For a balanced Wheatstone bridge configuration:

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$l_{PB}$.

Since R ∝ l$R \propto l$:

Step 1: Applying Total Length Constraint

We are given the total length:

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

Step 2: Final Conclusion

Solving for lAP$l_{AP}$:

Alternatively, lAP = (3)/(3+2) × 50 = (3)/(5) × 50 = 30 cm$l_{AP} = \frac{3}{3+2} \times 50 = \frac{3}{5} \times 50 = 30 \text{ cm}$.

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

Related Formula

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

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$X$.

Step 1: Setting up the Quadratic

Set the equivalent resistance equal to X$X$:

Cross-multiplying yields:

Step 2: Solving for X

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

Since resistance must be positive:

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

Related Formula

Core Logic

Initially:

(R₁)/(R₂) = (40)/(60) = (2)/(3) R₁ = (2)/(3)R₂$$\frac{R_1}{R_2} = \frac{40}{60} = \frac{2}{3} \implies R_1 = \frac{2}{3}R_2$$ ---- (1)

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

(R₁)/(((R₂ × 16)/(R₂ + 16))) = (50)/(50) R₁ = (16 R₂)/(16 + R₂)$$\frac{R_1}{\left(\frac{R_2 \times 16}{R_2 + 16}\right)} = \frac{50}{50} \implies R_1 = \frac{16 R_2}{16 + R_2}$$ ---- (2)

Solving equations (1) and (2) yields:

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