In a typical Wheatstone network,the resistanc | vedclass

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(A) For a balanced Wheatstone bridge,the condition is $\frac{A}{B} = \frac{D}{C}$.Given values: $A = 10 \,\Omega $,$B = 5 \,\Omega $,$C = 4 \,\Omega $

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

Vedclass · Answer

(A) For a balanced Wheatstone bridge,the condition is $\frac{A}{B} = \frac{D}{C}$.Given values: $A = 10 \,\Omega $,$B = 5 \,\Omega $,$C = 4 \,\Omega $,$D = 4 \,\Omega $.Checking the balance condition: $\frac{A}{B} = \frac{10}{5} = 2$ and $\frac{D}{C} = \frac{4}{4} = 1$.Since $\frac{A}{B} 
eq \frac{D}{C}$,the bridge is currently unbalanced.To balance the bridge,we need to change $A$ to a new value $A'$ such that $\frac{A'}{B} = \frac{D}{C}$.$\frac{A'}{5} = \frac{4}{4} \Rightarrow A' = 5 \,\Omega $.To change $A$ from $10 \,\Omega $ to $5 \,\Omega $,we connect a resistance $R$ in parallel with $A$ such that $\frac{1}{A'} = \frac{1}{A} + \frac{1}{R}$.$\frac{1}{5} = \frac{1}{10} + \frac{1}{R} \Rightarrow \frac{1}{R} = \frac{1}{5} - \frac{1}{10} = \frac{1}{10}$.Thus,$R = 10 \,\Omega $ must be connected in parallel with $A$.

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

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