Resistors P and Q are connected in the gaps of the meter bridge. The balancing point is obtained 1 3 m from the zero end. If a 6 Ω resistance is connected in series with P ; the balance point shifts to 2 3 m from the same end. P and Q are

⇒ P : Q = 1 : 2 P = k , Q = 2 k P + 6 Q = 2 3 1 - 2 3 ⇒ P + 6 Q = 2 1 k + 6 2 k = 2 1 ⇒ k + 6 = 4 k ⇒ k = 2 ∴ P = 2 Ω , Q = 4 Ω

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PhysicsCurrent ElectricityCircuit AnalysisEasy2 minPYQ_2020

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Resistors $P$ and $Q$ are connected in the gaps of the meter bridge. The balancing point is obtained $\frac{1}{3} m$ from the zero end. If a $6 Ω$ resistance is connected in series with $P$ ; the balance point shifts to $\frac{2}{3} m$ from the same end. $P$ and $Q$ are

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Solution:

$\Rightarrow P : Q = 1 : 2$

$P = k, Q = 2 k$

$\frac{P + 6}{Q} = \frac{\frac{2}{3}}{1 - \frac{2}{3}}$

$\Rightarrow \frac{P + 6}{Q} = \frac{2}{1}$

$\frac{k + 6}{2 k} = \frac{2}{1}$

$\Rightarrow k + 6 = 4 k$

$\Rightarrow k = 2$

$∴ P = 2 Ω, Q = 4 Ω$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Current ElectricitySubtopic:Circuit Analysis

⏱ 2mℹ️ Source: PYQ_2020

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