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In the circuit shown, the current in the 1\Omega resistor is :

 

  • Option 1)

    1.3 A ,from P to Q

  • Option 2)

    0 A

  • Option 3)

    0.13 A ,from Q to P

  • Option 4)

    0.13 A ,from P to Q

 

Answers (4)

best_answer

As we discussed in

In closed loop -

-i_{1}}R_{1} + i_{2}}R_{2} -E_{1}-i_{3}}R_{3}+E_{2}+E_{3}-i_{4}}R_{4}=0

- wherein

 

 Applying KVL in loop PQCDP

-1I2-3I2+9-2I2+-1I1=0

6I2-I1=9-(i)

Applying KVL in loop PQBAP

 

4I1-I2=6-(ii)

\therefore from equation (i) and (ii) we get 

I_{1}=1.83A    I_{2}=1.95A

\therefore the current in the 1\Omega resistor is 0.13 A from Q to P

 


Option 1)

1.3 A ,from P to Q

Option 2)

0 A

Option 3)

0.13 A ,from Q to P

Option 4)

0.13 A ,from P to Q

Posted by

solutionqc

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answer will b option 4 

 

Posted by

Shruti Aggarwal

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answer is option 3)0.13A from Q  to P

  •  Let, Current i1 from 6v battery passes through 3 ohm resistance . At Q current breaks and current i2 goes through 3 ohm and ( i1-i2 ) goes through 1 ohm ,.
  • Applying, kirchoffs law in 1st and 2nd loop,we get two eqns as follows
  • 4i1-i2=6
  • 6i2-i1=9
  • solving ,,  i1-i2 =0.13 A
Posted by

vishal dange

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