Current measured by the ammeter (A) in the reported circuit when no current flows through $10 Ω$ resistance, will be A.

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Raj Pandey | Contributor-Level 9

4 months ago

Using whetstone

$\frac{R}{3} = \frac{4}{6} \Rightarrow R = 2 Ω$

$R_{e q u} = \frac{54}{15} Ω$

$i = \frac{36}{\frac{54}{15}} = 10 A$

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Sol. $\frac{1}{λ_{min}} = R [\frac{1}{1} - \frac{1}{\infty}] = R [n = \infty \to n = 1]$

$\frac{1}{λ_{m a x}} = R [\frac{1}{1} - \frac{1}{4}] = \frac{3 R}{4} [n = 2 \to n = 1]$

$\Rightarrow Δ λ \Rightarrow \frac{4}{3 R} - \frac{1}{R} \Rightarrow \frac{1}{3 R} = 340$

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Current measured by the ammeter (A) in the reported circuit when no current flows through 1 0 Ω resistance, will be A.