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In the given circuit, what will be the equivalent resistance between the points $A$ and $B$ ?
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Verified Answer
The correct answer is:
$10 / 3 \Omega$
$\because \quad \frac{2}{3}=\frac{4}{6}$
$\therefore$ Given, circuit is a balanced Wheatstone's Bridge.
So, $10 \Omega$ resistance will be ineffective. Equivalent resistance of upper arms
$R_1=2+3=5 \Omega$
$\therefore$ Equivalent resistance of lower arms
$\begin{aligned} R_2 & =4+6=10 \Omega \\ R_{A B} & =\frac{R_1 R_2}{R_1+R_2} \\ & =\frac{5 \times 10}{5+10} \\ & =\frac{50}{10}=\frac{10}{3} \Omega\end{aligned}$
$\therefore$ Given, circuit is a balanced Wheatstone's Bridge.
So, $10 \Omega$ resistance will be ineffective. Equivalent resistance of upper arms
$R_1=2+3=5 \Omega$
$\therefore$ Equivalent resistance of lower arms
$\begin{aligned} R_2 & =4+6=10 \Omega \\ R_{A B} & =\frac{R_1 R_2}{R_1+R_2} \\ & =\frac{5 \times 10}{5+10} \\ & =\frac{50}{10}=\frac{10}{3} \Omega\end{aligned}$
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