A meter bridge with two resistances R_1$R_{1}$ and R_2$R_{2}$ 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~Omega$16~\Omega$ resistance is connected in parallel to R_2$R_{2}$. The values of resistances R_1$R_{1}$ and R_2$R_{2}$ are \_\_\_\_.

### Related Formula R_texteq = fracR_1 R_2R_1 + R_2 text (parallel)$$R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2} \text{ (parallel)}$$ H = I^2 R t$H = I^2 R t$ I = fracVR_texteq + r$$I = \frac{V}{R_{\text{eq}} + r}$$ ### Core Logic The circuit is a balanced Wheatstone bridge between A and B, because the ratio of adjacent arms is frac12 = frac24$\frac{1}{2} = \frac{2}{4}$. Thus, the middle 1\,Omega$1\,\Omega$ resistor (if there is one connecting the middle nodes) is ineffective.

The equivalent resistance across AB (R_AB$R_{AB}$) involves the upper branch (1+2 = 3\,Omega$1+2 = 3\,\Omega$) and the lower branch (2+4 = 6\,Omega$2+4 = 6\,\Omega$) in parallel.

R_AB = frac3 times 63 + 6 = frac189 = 2\,Omega$$R_{AB} = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2\,\Omega$$ ### Step 1: Calculate Current and Heat Total resistance in the circuit = R_AB + r_textinternal = 2 + 1 = 3\,Omega$R_{AB} + r_{\text{internal}} = 2 + 1 = 3\,\Omega$. Total current from the battery i = frac9textV3\,Omega = 3textA$i = \frac{9\text{V}}{3\,\Omega} = 3\text{A}$. The heat generated specifically between points A and B in 1 minute (60 seconds) is: H_AB = i^2 R_AB t = (3)^2 times 2 times 60$$H_{AB} = i^2 R_{AB} t = (3)^2 \times 2 \times 60$$ H_AB = 9 times 120 = 1080text J$$H_{AB} = 9 \times 120 = 1080\text{ J}$$ ### Pattern Recognition Check for balanced Wheatstone bridge first. Then use H = i^2 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$AB$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity

### Related Formula For a balanced Wheatstone bridge configuration: fracR_1R_2 = fracl_1l_2$$\frac{R_1}{R_2} = \frac{l_1}{l_2}$$ ### 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 l_AP$l_{AP}$ and PB be l_PB$l_{PB}$. frac6\,OmegaR_AP = frac4\,OmegaR_PB$$\frac{6\,\Omega}{R_{AP}} = \frac{4\,\Omega}{R_{PB}}$$ Since R propto l$R \propto l$: fracR_APR_PB = fracl_APl_PB = frac64 = frac32$$\frac{R_{AP}}{R_{PB}} = \frac{l_{AP}}{l_{PB}} = \frac{6}{4} = \frac{3}{2}$$ ### Step 1: Applying Total Length Constraint We are given the total length: l_AP + l_PB = 50 text cm$$l_{AP} + l_{PB} = 50 \text{ cm}$$ Using the ratio l_AP = frac32 l_PB$l_{AP} = \frac{3}{2} l_{PB}$: frac32l_PB + l_PB = 50$$\frac{3}{2}l_{PB} + l_{PB} = 50$$ frac52l_PB = 50 implies l_PB = 20 text cm$$\frac{5}{2}l_{PB} = 50 \implies l_{PB} = 20 \text{ cm}$$ ### Step 2: Final Conclusion Solving for l_AP$l_{AP}$: l_AP = 50 - 20 = 30 text cm$$l_{AP} = 50 - 20 = 30 \text{ cm}$$ Alternatively, l_AP = frac33+2 times 50 = frac35 times 50 = 30 text 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity

### Related Formula P = I^2 R = left(fracER+rright)^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. Setting fracdPdR = 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity

### Related Formula For parallel combination: R_texteq = fracR_1 R_2R_1 + R_2$R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2}$ For series combination: R_texteq = R_1 + R_2$R_{\text{eq}} = R_1 + R_2$ ### 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$X$. R_texteq = frac(2R + X) cdot (R)(2R + X) + R$$R_{\text{eq}} = \frac{(2R + X) \cdot (R)}{(2R + X) + R}$$ ### Step 1: Setting up the Quadratic Set the equivalent resistance equal to X$X$: X = frac(2R + X)R3R + X$$X = \frac{(2R + X)R}{3R + X}$$ Cross-multiplying yields: X(3R + X) = 2R^2 + XR$$X(3R + X) = 2R^2 + XR$$ 3RX + X^2 = 2R^2 + XR$$3RX + X^2 = 2R^2 + XR$$ X^2 + 2RX - 2R^2 = 0$$X^2 + 2RX - 2R^2 = 0$$ ### Step 2: Solving for X Using the quadratic formula to solve for X$X$: X = frac-2R pm sqrt(2R)^2 - 4(1)(-2R^2)2$$X = \frac{-2R \pm \sqrt{(2R)^2 - 4(1)(-2R^2)}}{2}$$ X = frac-2R pm sqrt4R^2 + 8R^22 = frac-2R pm sqrt12R^22$$X = \frac{-2R \pm \sqrt{4R^2 + 8R^2}}{2} = \frac{-2R \pm \sqrt{12R^2}}{2}$$ X = frac-2R pm 2Rsqrt32 = -R pm Rsqrt3$$X = \frac{-2R \pm 2R\sqrt{3}}{2} = -R \pm R\sqrt{3}$$ Since resistance must be positive: X = (sqrt3 - 1)R$$X = (\sqrt{3} - 1)R$$ ### Pattern Recognition Self-referential equivalent resistance problems always collapse into a quadratic equation x^2 + ax + b = 0$x^2 + ax + b = 0$. Solve strictly for the positive root. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity