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 The circuit shown here has two batteries of 8.0 V and 16.0 V and three resistors 3 \Omega, 9 \Omega and 9 \Omega and a capacitor 5.0 \muF.

How much is the current I in the circuit in steady state ?

  • Option 1)

     1.6 A
     

     

  • Option 2)

     0.67 A

  • Option 3)

     2.5 A

     

  • Option 4)

     0.25 A

 

Answers (2)

best_answer

As we have learned

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

 

 

I'_1 = I_2 = I  since in steady state current through capacitor is  0 

USing kirchoff's rule 

16-9I -3I -8V = 0   or I = 8/12 A = 2/3 A 

 = 0.67 A 

 

 


Option 1)

 1.6 A
 

 

Option 2)

 0.67 A

Option 3)

 2.5 A

 

Option 4)

 0.25 A

Posted by

Avinash

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