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In the network shown

Option: 1

V_{AB}= 2V


Option: 2

I_1=1A


Option: 3

I_1=0.5A


Option: 4

I_2=0.5A


Answers (1)

best_answer

As we learnt

 

Current through any resistance -

 

i'=i(\frac{Resistance\, of \, opposite \, Branch}{total\, Resistance})

- wherein

 

 I_1=I.\frac{R_2}{R_1+R_2}\: \: and\: \: I_2=I.\frac{R_1}{R_1+R_2}

R_1=12\Omega

R_2=6\Omega

I_1=\left ( 1.5 \right )*\frac{6}{12+6}= 0.5A

I_2=\left ( 1.5 \right )*\frac{12}{12+6}= 1A

 

Posted by

Kuldeep Maurya

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