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`        in the figure a part of electric circuit is given. The value of each resistance is R and the current in some parts is shown in the diagram then current in AB branch is?`
2 years ago

Dibyakanti Kumar
59 Points
```							Take the voltage of centre be 0 . Then voltage of the point just above it(let me call that D) is -R since current of 1A is flowing across a resistance of R therefore voltage difference should be R . Similarly for the point to its right (let’s call that E)is +R since current is flowing towards the centre and current travels from high to low voltage .Similarly for the point to it’s left(let’s call that F) is +R.
Current between ED is {R-(-R)}/R = 2A . Simlarly from F to D .
Apply Kirchoff’s circuital law for the point E and F .You will get the Voltage of all points and hence at the end all you have to do is to apply KCL at point A.

The answer at the end comes out to be 29 A .
```
2 years ago
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### Course Features

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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions