In a typical Wheatstone network, resistances in cyclic order are A = 10 ,Ω , B = 5 ,Ω , C = 4 ,Ω

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3.Current Electricity

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In a typical Wheatstone network, the resistances in cyclic order are $A = 10 \,\Omega $, $B = 5 \,\Omega $, $C = 4 \,\Omega $ and $D = 4 \,\Omega $ for the bridge to be balanced

A

$10 \,\Omega $ should be connected in parallel with $A$

B

$10 \,\Omega $ should be connected in series with $A$

C

$5 \,\Omega $ should be connected in series with $B$

D

$5 \,\Omega $ should be connected in parallel with $B$

Solution

For a balance Wheatstone bridge.

$\frac{A}{B} = \frac{D}{C} \Rightarrow \frac{{10}}{5} \ne \frac{4}{4}$ (Unbalanced)

$\frac{{A'}}{B} = \frac{D}{C} \Rightarrow \frac{{A'}}{5} = \frac{4}{4}$$ \Rightarrow $ $A' = 5\,\Omega $

$A'$ $(5\,\Omega )$ is obtained by connecting a $10\,\Omega $ resistance in parallel with $A$.

Standard 12

Physics

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