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Physics-

General

Easy

Question

In the circuit shown the value of I in ampere is

1
060
0.4
1.5

The correct answer is: 0.4

We can simplify the network as shown

So, net resistance,
R=2.4+1.6=4.0 $capital omega$
Therefore, current from the battery.
$i equals fraction numerator V over denominator R end fraction equals fraction numerator 4 over denominator 4 end fraction equals 1 A$
Now, from the circuit (b),
4I’ =6I
$⟹ I to the power of ´ end exponent equals fraction numerator 3 over denominator 2 end fraction I$
But $i$ =I+I’
$equals I plus fraction numerator 3 over denominator 2 end fraction I equals fraction numerator 5 over denominator 2 end fraction I$
$therefore blank 1 equals fraction numerator 5 over denominator 2 end fraction I$
$⟹ I equals fraction numerator 2 over denominator 5 end fraction equals 0.4 A$

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