A wire of length 25~m$25\mathrm{~m}$ and cross-sectional area 5~mm²$5\mathrm{~mm}^2$ having resistivity of 2× 10⁻⁶Ω~m$2\times 10^{-6}\Omega\mathrm{~m}$ is bent into a complete circle. The resistance between diametrically opposite points will be:

Related Formula

Core Logic

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.

Step 1: First Condition

Given ₁ = 40 cm$\ell_1 = 40 \text{ cm}$:

Step 2: Second Condition

When replaced by a wire of 2r$2r$ per cm, the new lengths ₂$\ell_2$ will satisfy:

Notice that the 2r$2r$ terms cancel out entirely from both sides, leaving:

Step 3: Conclusion

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

Related Formula

Power (Illumination) is proportional to the square of the current for a constant resistance: P = I² R$P = I^2 R$

Core Logic

Initial power: P₁ = I₁² R$P_1 = I_1^2 R$

If current drops by 20%, the new current is:

New power:

Step 1: Calculate Percentage Change

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

Related Formula

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:

Step 2: Calculate Power

Pattern Recognition

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

Related Formula

Core Logic

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.

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$B$.

Step 2: Equivalent Calculation

The simplified circuit consists of three identical 3 Ω$3\,\Omega$ resistors in parallel.

Pattern Recognition

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

Related Formula

Step 1: Series Combination

Let base resistance at 0^° C$0^\circ \text{C}$ be R$R$. For series:

Step 2: Parallel Combination

For parallel at 0^° C$0^\circ \text{C}$, Req,0 = R/2$R_{\text{eq,0}} = R/2$.

Using binomial expansion for small Δ T$\Delta T$:

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