CBSE Class 10 Science 2015 Term I Set 1

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Question : 22
Total: 36
For the series combination of three resistors establish the relation :
R=R1+R2+R3
where the symbols have their usual meanings.
Calculate the equivalent resistance of the combination of three resistors of 6Ω,9Ω and 18Ω joined in parallel.
Solution:  
Same current (I) flows through different resistances, when these are joined in series, as shown in the figure.
Let R be the combined resistance, then
V=IR
Also, ‌‌V1=IR1,V2=IR2,V3=IR3
∵‌‌V=V1+V2+V3
∴‌‌IR=IR1+IR2+IR3
⇒‌‌IR=I(R1+R2+R3)
∴‌‌R=R1+R2+R3
Now, R1=6Ω,R2=9Ω,R3=18Ω
In parallel combination,
‌
1
R
‌
=‌
1
R1
+‌
1
R2
+‌
1
R3

⇒‌‌‌
1
R
‌
=‌
1
6
+‌
1
9
+‌
1
18
=‌
3+2+1
18

‌=‌
6
18
=‌
1
3

⇒‌‌‌
1
R
‌
=‌
1
3

⇒‌‌R‌=3Ω
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