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Numericals · Q31

Q.A set of three coils having resistances 10 \Omega, 12 \Omega and 15 \Omega are connected in parallel. This combination is connected in series with a series combination of three coils of the same resistances. Calculate the total resistance and the current through the circuit, if a battery of emf 4.1 Volt is used for drawing current.

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Three coils of 10 \Omega, 12 \Omega and 15 \Omega are first combined in PARALLEL:

1Rp=110+112+115=6+5+460=1560=14  ⇒  Rp=4 Ω\frac{1}{R_p} = \frac{1}{10}+\frac{1}{12}+\frac{1}{15} = \frac{6+5+4}{60} = \frac{15}{60} = \frac{1}{4} \;\Rightarrow\; R_p = 4\ \Omega

and this parallel combination is connected in SERIES with a separate series combination of the SAME three resistances:

Rs=10+12+15=37 ΩR_s = 10+12+15 = 37\ \Omega

so the total resistance of the whole circuit is

Rtotal=Rp+Rs=4+37=41 ΩR_{total} = R_p+R_s = 4+37 = 41\ \Omega …

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