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