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Physics-

General

Easy

Question

A 3 V battery with negligible internal resistance is connected in a circuit as shown in the figure. The current I, in the circuit will be

1A
1.5A
2A
$\frac{1}{3} A$

The correct answer is: 1.5A

Resistance in the arms AC and BC are in series,

∴ R’=3+3=6 $capital omega$
Now, R’ and 3 $capital omega$ are in parallel,
$therefore blank R subscript e q end subscript equals fraction numerator 6 cross times 3 over denominator 6 plus 3 end fraction equals 2 capital omega$
Now, V=IR
$⟹ I equals fraction numerator 3 over denominator 2 end fraction equals 1.5 A$

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