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Standard XII

Physics

Kirchhoff's Junction Law

Three wires h...

Question

Three wires having resistances $1 Ω, 2 Ω$ and $6 Ω$ are joined in parallel across a battery. If a current
0.1A flows through $6 Ω$ resistor, the total current drawn by the combination is:

0.6 A

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0.3 A

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0.1 A

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1 A

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Solution

The correct option is D $1 A$
$0.1 \times 6 = E$
$E = 0.6$
$\Rightarrow I_{2} = \frac{0.6}{2}$
$= 0.3$
$I_{1} = \frac{0.6}{1}$
$= 0.6$
$\Rightarrow T o t a l = 0.3 + 0.6 + 0.1$
$= 1 A$

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Three wires having resistances 1Ω,2Ω and 6Ω are joined in parallel across a battery. If a current0.1A flows through 6Ω resistor, the total current drawn by the combination is:

The correct option is D 1A0.1×6=E E=0.6⇒I2=0.62 =0.3I1=0.61 =0.6⇒Total=0.3+0.6+0.1 =1A

Three wires having resistances $1 Ω, 2 Ω$ and $6 Ω$ are joined in parallel across a battery. If a current
0.1A flows through $6 Ω$ resistor, the total current drawn by the combination is:

The correct option is D $1 A$
$0.1 \times 6 = E$
$E = 0.6$
$\Rightarrow I_{2} = \frac{0.6}{2}$
$= 0.3$
$I_{1} = \frac{0.6}{1}$
$= 0.6$
$\Rightarrow T o t a l = 0.3 + 0.6 + 0.1$
$= 1 A$