1.

For the circuit shown in the figure, potential difference between points A and B is 16 V. Find the current passing through 2Omega.

Answer»

SOLUTION :`V_(A)-V_(B)=16V`
`therefore 4i_(1)+2(i_(1)+i_(2))-3+4i_(1)=16V"…(1)"`
USING Kirchhoff.s second LAW in the closed loop, we have `9-i_(2)-2(i_(1)+i_(2))=0"….(3)"`
SOLVING eqs (1) and (2), we GET `i_(1)=1.5A and i_(2)=2A`
`therefore " Current through "2Omega" resistor "=2+1.5=3:5A`


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