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Physics - Current Electricity Question with Solution | TestHub

PhysicsCurrent ElectricityCircuit AnalysisMedium2 minPYQ_2020

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The unknown resistances are connected in two gaps of a metre bridge. The null point is at $20 cm$ from zero end. A resistance of $15Ω$ is connected in series with the smaller of the two. The null point shifts to $40 cm$ . The smaller resistance is

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Solution:

$\frac{20}{80} = \frac{r _{1}}{r _{2}} = \frac{1}{4} \Rightarrow 4 r_{1} = r_{2}$ $\frac{40}{60} = \frac{r _{1} + 15}{r _{2}} = \frac{2}{3} \Rightarrow (r_{1} + 15) 3 = 2 r_{2} = 2 \times 4 r_{1} = 8 r_{1}$ $3 r_{1} + 45 = 8 r_{1}$ $5 r_{1} = 45 ∴ r_{1} = 9Ω$

Stream:MHT_CETSubject:PhysicsTopic:Current ElectricitySubtopic:Circuit Analysis

⏱ 2mℹ️ Source: PYQ_2020

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