Test Index

CBSE Class 10 Science Exam 2014 Term 1 Paper

© examsnet.com
Question : 16 of 36
Marks: +1, -0
Show four different ways in which four resistors of rr ohm each may be connected in a circuit. In which case is the equivalent resistance of the combination.
(i) maximum;
(ii) minimum ?
Solution:  
(a)
Resultant resistance, R=r+r+r+rR = r + r + r + r
R=4rR = 4 r
(b)
  1R  =  1r  +  1r  +  1r  +  1r  =  4r\;\frac{1}{R}\;=\;\frac{1}{r}\;+\;\frac{1}{r}\;+\;\frac{1}{r}\;+\;\frac{1}{r}\;=\;\frac{4}{r}
  =  4r\;=\;\frac{4}{r}
R  =  r4R\;=\;\frac{r}{4}
(c)
Resistance (AB)=R1=r+r=2r(AB) = R_1 = r + r = 2 r
Resistance (PQ)=R2=r+r=2r(PQ) = R_2 = r + r = 2 r
Resultant resistance, R=R = ?
  1R  =  1R1  +  1R2  =  12r  +  12r  =  1+12r  =  22r  =  1r\;\frac{1}{R}\;=\;\frac{1}{R_1}\;+\;\frac{1}{R_2}\;=\;\frac{1}{2r}\;+\;\frac{1}{2r}\;=\;\frac{1+1}{2r}\;=\;\frac{2}{2r}\;=\;\frac{1}{r}
R=rR = r
(d)
Resistance (PQ)=R1=r+r+r=3r(PQ) = R_1 = r + r + r = 3 r
Resistance (AB)=R2=r(AB) = R_2 = r
Resultant Resistance, R=R = ?
  1R  =  1R1  +  1R2  =  13r  +  1r  =  1+33r  =  43r\;\frac{1}{R}\;=\;\frac{1}{R_1}\;+\;\frac{1}{R_2}\;=\;\frac{1}{3r}\;+\;\frac{1}{r}\;=\;\frac{1+3}{3r}\;=\;\frac{4}{3r}
R  =  3r4R\;=\;\frac{3r}{4}
(i) Maximum resistance is in case (1) where all the resistors are combined in series.
(ii) Minimum resistance is in case (2) where all the resistors are combined in parallel combination.
© examsnet.com
Go to Question:
Prev Question
Next Question