Calculate I for the given circuit diagram.

10 A

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5 A

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2.5 A

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20 A

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Solution

The correct option is C 10 A
For the wheatstone bridge part, the bridge resistance can be ignored. For this part, we have two branches each having two 5 ohm resistors.
$\frac{1}{R_{W}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$

where, $R_{1} = R_{2} = 5 + 5 = 10 Ω$ because they are in series.

Hence, $R_{W} = 5 Ω$

The wheatstone bridge is parallel to a 5 ohm resistor. Hence total equivalent resistance,
$\frac{1}{R_{e q}} = \frac{1}{R_{W}} + \frac{1}{5}$

$\Rightarrow R_{e q} = 2.5 Ω$

$I = \frac{V}{R} = \frac{25}{2.5} = 10 A$

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Similar questions

a_{1}, a_{2}, a_{3}, . . . . . . . . i s a n A . P . s u c h t h a t a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225

a_{1} + a_{2} + a_{3} + . . . . . . . + a_{23} + a_{24} i s e q u a l t o

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a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24}

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a_{1}, a_{2}, a_{3}

a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} =

a_{1} + a_{2} + a_{3} + . . . + a_{23} + a_{24} =

If $a_{1}, a_{2}, a_{3} . . . . .$ are in $A . P .$ such that $a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225$ .

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a_{1}, a_{2}, a_{3}, . . . .

a_{1}, a_{5}, a_{10} + a_{15} + a_{20} + a_{24} = 225

a_{1} + a_{2} + a_{3} + . . . + a_{24}

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