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ICSE Class 10 Physics 2014 Paper

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Two resistors of 4Ω4 \Omega and 6Ω6 \Omega are connected in parallel to a cell to draw 0.5A0.5 \mathrm{A} current from the cell.
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Question : 50 of 66
Marks: +1, -0
Calculate the current in each resistor.
Solution:  
Let current through 4 Ω4\ \Omega resistance is ii then current through 6 Ω6\ \Omega resistance is (0.5−i)(0.5-i)
∴    i×4  =(0.5−i)×6\therefore\;\; i \times 4\;=(0.5-i) \times 6
4i  =3−6i4 i\;=3-6 i
10i  =310 i\;=3
i  =0.3 Ai\;=0.3\ \mathrm{A}
∴\therefore Current through 4 Ω4\ \Omega resistance
=0.3 A=0.3\ \mathrm{A}
andcurrent through 6 Ω6\ \Omega resistance
=0.5−0.3=0.2 A=0.5-0.3=0.2\ \mathrm{A}
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