# The current  I   drawn from the 5  volt source will be : Option 1) $0.17 A$ Option 2) $0.33 A$ Option 3) $0.5 A$ Option 4) $0.67 A$

As we learnt in

Wheat stone Bridge -

It is an arrangement of four resistances which can be used to measure one of them in terms of rest

- wherein

The equivalent circuit is a balanced Wheatstone's bridge. Hence no current flows through arm BD

AB and BC are in series

$\therefore R_{ABC}= 5+10=15\Omega$

$AD \; and \; AC \: are \; in \; series$

$\therefore R_{ADC}= 10+20=30\Omega$

$ABC \; and\; ADC \; are\; in\; paralle$

$\therefore R_{eq}= \frac{\left ( R_{ABC} \right )\left ( R_{ADC} \right )}{\left (R_{ABC}+R_{ADC} \right )}$

$or\: \: \: \: R_{eq}=\frac{15\times 30}{15+30}= \frac{15\times 30}{45}= 10\Omega$

$\therefore Current\; \; I= \frac{E}{R_{eq}}= \frac{5}{10}= 0.5A$

Option 1)

$0.17 A$

This option is incorrect.

Option 2)

$0.33 A$

This option is incorrect.

Option 3)

$0.5 A$

This option is correct.

Option 4)

$0.67 A$

This option is incorrect.

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