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CBSE Class 10 Science 2015 Term I Set 1

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Question : 22 of 36
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For the series combination of three resistors establish the relation :
R=R1+R2+R3R=R_1+R_2+R_3
where the symbols have their usual meanings.
Calculate the equivalent resistance of the combination of three resistors of 6 Ω, 9 Ω6\,\Omega,\ 9\,\Omega and 18 Ω18\,\Omega joined in parallel.
Solution:  
Same current (I) flows through different resistances, when these are joined in series, as shown in the figure.
Let RR be the combined resistance, then
V=IRV=IR
Also, V1=IR1, V2=IR2, V3=IR3V_1=I R_1,\ V_2=I R_2,\ V_3=I R_3
∵V=V1+V2+V3\because V=V_1+V_2+V_3
∴IR=IR1+IR2+IR3\therefore IR=IR_1+IR_2+IR_3
⇒IR=I(R1+R2+R3)\Rightarrow IR=I(R_1+R_2+R_3)
∴R=R1+R2+R3\therefore R=R_1+R_2+R_3
Now, R1=6 Ω, R2=9 Ω, R3=18 ΩR_1=6\,\Omega,\ R_2=9\,\Omega,\ R_3=18\,\Omega
In parallel combination,
1R=1R1+1R2+1R3\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}
⇒1R=16+19+118=3+2+118\Rightarrow \frac{1}{R}=\frac{1}{6}+\frac{1}{9}+\frac{1}{18}=\frac{3+2+1}{18}
=618=13=\frac{6}{18}=\frac{1}{3}
⇒1R=13\Rightarrow \frac{1}{R}=\frac{1}{3}
⇒R=3 Ω\Rightarrow R=3\,\Omega
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