In wheatstone's bridge $P = 9$ ohm, $R = 4$ ohm and $S = 6$ ohm. How much resistance must be put in parallel to the resistance S to balance the bridge?

24

ohm

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13.5

ohm

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7.2

ohm

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18.7

ohm

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Solution

The correct option is B $13.5$ ohm
ACC, to what stone's bridge
At balance condition
$P S = Q R$
$\frac{P S}{R} = Q$
$\frac{9 \times 6}{4} = Q = \frac{54}{4}$
$Q = 13.5 Ω$

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In wheatstone's bridge P=9 ohm, R=4 ohm and S=6 ohm. How much resistance must be put in parallel to the resistance S to balance the bridge?

The correct option is B 13.5 ohmACC, to what stone's bridgeAt balance conditionPS=QRPSR=Q9×64=Q=544Q=13.5Ω

In wheatstone's bridge $P = 9$ ohm, $R = 4$ ohm and $S = 6$ ohm. How much resistance must be put in parallel to the resistance S to balance the bridge?

The correct option is B $13.5$ ohm
ACC, to what stone's bridge
At balance condition
$P S = Q R$
$\frac{P S}{R} = Q$
$\frac{9 \times 6}{4} = Q = \frac{54}{4}$
$Q = 13.5 Ω$