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ICSE Class 10 Physics 2018 Paper

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Question : 59 of 78
Marks: +1, -0
Find the equivalent resistance between AA and BB .
Solution:  
From the figure, 6 Ω6\,\Omega and 3 Ω3\,\Omega are in parallel, so,
1Rp=16+13=1+26=36=12\frac{1}{R_{p}} = \frac{1}{6} + \frac{1}{3} = \frac{1+2}{6} = \frac{3}{6} = \frac{1}{2}
∴  RP=2 Ω\therefore\; R_{P} = 2\,\Omega
Also, 4 Ω4\,\Omega and 12 Ω12\,\Omega are in parallel, so
1RP′=14+112=3+112=412=13\frac{1}{R_{P}'} = \frac{1}{4} + \frac{1}{12} = \frac{3+1}{12} = \frac{4}{12} = \frac{1}{3}
∴  RP′=3 Ω\therefore\; R_{P'} = 3\,\Omega
Now, the given parallel combinations are in series. So we get,
Req=2+3=5 ΩR_{eq} = 2+3 = 5\,\Omega
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