Question

What current will flow through the 2 kilo ohm resistor in the circuit shown in the figure

Answer 1

Answer:

where is diagram ? ........

Answer 2

Answer:

E=(2I  

1

​  

+4I  

1

​  

+6I)×10  

3

 

⟹6(I  

1

​  

+I)×10  

3

=72V

⟹I  

1

​  

+I=12mA

Along loop BCDE−

2I  

1

​  

+4I  

1

​  

−3(I−I  

1

​  

)=0

⟹9I  

1

​  

=3I

⟹I=3I  

1

​  

 

And, I+I  

1

​  

=12mA

⟹3I  

1

​  

+I  

1

​  

=12mA

⟹I  

1

​  

=3mA=current in 2kΩ resistance

Explanation: