Figure below shows a portion of an electric circuit with the currents in amperes and their directions. The magnitude and direction of the current in the portion PQ is
A
0A
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B
3A from P to Q
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C
4A from Q to P
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D
6A from Q to P
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Solution
The correct option is D6A from Q to P At junction the net current arriving should be equal to the net current leaving
The net current 8+2=4+2+1+x, x=3A
Now at junction Q the net current arriving is 3+3=6A. This will flow toward P.