Question

Find the equivalent resistance across the two ends A and B of this circuit.​

Answer 1

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Equivalent resistance across the ends A and B of the circuit (R) = $\bold{1\Omega}$

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Resistors $\bold{R_1}$ and $\bold{R_2}$ are connected in parallel.

Hence, the equivalent resistance of the parallel combination of $\bold{R_{12}}$ :-

$\bold{\frac{1}{R_{12}}=\frac{1}{R_1}+\frac{1}{R_2}}$

=$\bold{\frac{1}{2}+\frac{1}{2}=1\Omega}$

$\implies\bold{R_{12}=1\Omega}$

Similarly, $\bold{R_{34}=1\Omega}$

$\bold{R_{56}=1\Omega}$

$\bold{R_{78}=1\Omega}$

Now, $\bold{R_{12}}$ and $\bold{R_{34}}$ are connected in series

$\therefore\bold{R_{1234}=R_{12}+R_{34}=1+1=2\Omega}$

Again, $\bold{R_{56}}$ and $\bold{R_{78}}$ are connected in series

$\therefore\bold{R_{5678}=R_{56}+R_{78}=1+1=2\Omega}$

Now equivalent resistance,

$\bold{\frac{1}{R}=\frac{1}{R_{1234}}+\frac{1}{R_{5678}}=}$

$\bold{\frac{1}{2}+\frac{1}{2}=1\Omega}$

i.e., $\bold{R=1\Omega}$