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CBSE Class 12 Physics 2016 Delhi Set 1 Paper

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Question : 9 of 26
Marks: +1, -0
Two cells of emfs 1.5 V1.5\ \mathrm{V} and 2.0 V2.0\ \mathrm{V} having internal resistances 0.2 Ω0.2\ \Omega and 0.3 Ω0.3\ \Omega respectively are connected in parallel. Calculate the emf and internal resistance of the equivalent cell.
Solution:  
Calculation of emf
Calculation of internal resistance
   E. M. F   \;\text{ E. M. F }\;   =  E12r2+E21r1+r2\;=\;\frac{E_{12} r_2+E_{21}}{r_1+r_2}
  =  1.5×0.3+2×0.20.2+0.3 V\;=\;\frac{1.5 \times 0.3+2 \times 0.2}{0.2+0.3}\ \mathrm{V}
  =  0.45+0.400.5 V=1.7 V\;=\;\frac{0.45+0.40}{0.5}\ \mathrm{V}=1.7\ \mathrm{V}
r  =  r1r2r1+r2r\;=\;\frac{r_1 r_2}{r_1+r_2}
  =  0.2×0.30.2+0.3 Ω\;=\;\frac{0.2 \times 0.3}{0.2+0.3}\ \Omega
  =  0.060.5 Ω\;=\;\frac{0.06}{0.5}\ \Omega
  =  0.12 Ω\;=\;0.12\ \Omega
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