Find the equivalent resistance betwen P and Q in the given circuit.

$4 Ω$

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$3 Ω$

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$2.4 Ω$

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$2 Ω$

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Solution

The correct option is C
$2.4 Ω$

In the given circuit,
$S Q_{e q} = \frac{(R_{1} \times R_{2})}{(R_{1} + R_{2})} = \frac{4 \times 4}{4 + 4} = 2 o h m$ (parallel connection)
$P {, Q}_{e q} = (R_{1} + R_{2}) = (4 + 2) = 6 o h m$ (series connection)
so that, the circuit reduces to
$P Q_{e q} = \frac{(R_{1} \times R_{2})}{(R_{1} + R_{2})} = \frac{(4 \times 6)}{(4 + 6)} = 2.4 o h m$ (parallel connection)

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Find the equivalent resistance betwen P and Q in the given circuit.

The correct option is C 2.4Ω In the given circuit, SQeq=(R1×R2)(R1+R2)=4×44+4=2ohm (parallel connection) P,Qeq=(R1+R2)=(4+2)=6ohm (series connection) so that, the circuit reduces to PQeq=(R1×R2)(R1+R2)=(4×6)(4+6)=2.4ohm (parallel connection)