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CGPET 2018 Solved Paper

Section: Physics

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Question : 3 of 150

Marks: +1, -0

In the circuit shown in the figure below, the ammeter

A

, assumed to have negligible resistance, reads

0.1 A

R

$6\ \Omega$
$8\ \Omega$
$16\ \Omega$
$20\ \Omega$

Solution:

In the circuit shown in the figure below.

So, voltage across

50\ \Omega

and

30\ \Omega

resistor,

V_{1}=I_{1} \times 50 = 0.1 \times 50 = 5\ \mathrm{V}

V_{2}=I_{1} \times 30 = 0.1 \times 30 = 3\ \mathrm{V}

\because

Three branches are connected in parallel combination, therefore voltage across each branch will be equal.

\therefore

V_{1}+V_{2}=V_{3}

⇒

5+3=V_{3}

⇒

V_{3}=8\ \mathrm{V}

Applying KCL at point

B

,

I_{2}=I_{1}+I_{3}=0.1+\frac{2}{5}=0.5\ \mathrm{A}

Hence,

V_{1}+V_{2}=12-I_{2} R

\Rightarrow R=\frac{12-8}{0.5}=8\ \Omega

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