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Physics-

General

Easy

Question

Two cells having emf 4V, 2V and internal resistances 1 $blank capital omega comma$ 1 $blank capital omega$ are connected as shown in figure below. Current through 6 $blank capital omega$ resistance is

$\frac{1}{3} A$
$\frac{2}{3} A$
1A
$\frac{2}{9} A$

The correct answer is: $fraction numerator 1 over denominator 3 end fraction A$

The emf of the circuit is
$E equals E subscript 1 end subscript plus E subscript 2 end subscript$
$equals 4 V plus 2 V equals 6 V$
In the given circuit, 3 $capital omega$ and 6 $capital omega$ are connected in parallel, hence equivalent resistance is
$fraction numerator 1 over denominator R to the power of ´ end exponent end fraction equals fraction numerator 1 over denominator 3 end fraction plus fraction numerator 1 over denominator 6 end fraction equals fraction numerator 3 over denominator 6 end fraction equals fraction numerator 1 over denominator 2 end fraction$
$⟹ R to the power of ´ end exponent equals 2 capital omega$
Total resistance of circuit is
$R equals 1 capital omega plus 1 capital omega plus 2 capital omega equals 6 capital omega$
From Ohm’s law $V equals i R$
$⟹ i equals fraction numerator V over denominator R end fraction equals fraction numerator 6 over denominator 6 end fraction equals 1 A$
The 3 $capital omega blank a n d blank 6 capital omega$ resistors are in parallel, hence
$i subscript 1 end subscript R subscript 1 end subscript equals i subscript 2 end subscript R subscript 2 end subscript equals V$
$therefore i subscript 1 end subscript cross times 3 equals i subscript 2 end subscript cross times 6$
$⟹ i subscript 1 end subscript equals 2 i subscript 2 end subscript a n d i subscript 1 end subscript plus i subscript 2 end subscript equals 1$
$2 i subscript 2 end subscript plus i subscript 2 end subscript equals 1$
$3 i subscript 2 end subscript equals 1 ⟹ i subscript 2 end subscript equals fraction numerator 1 over denominator 3 end fraction A$

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