CBSE Class 12 Physics 2023 Outside Delhi Set 2 Paper - Question ...

CBSE Class 12 Physics 2023 Outside Delhi Set 2 Paper

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Question : 13 of 13

Marks: +1, -0

Two cells of emf

E_1

and

E_2

and internal resistances

r_1

and

r_2

are connected in parallel, with their terminals of the same polarity connected together. Obtain an expression for the equivalent emf of the combination.

Solution:

Terminal p.d of

1^{\;\text "st "\;}

cell,

V\;=E_1-I_1 r_1

∴ \;\; I_1\;=\;{E_1-V}/{r_1}

Terminal p.d of

2^{\;\text "nd "\;}

cell,

\;V\;=E_2-I_2 r_2

\; ∴ \;I_2\;=\;{E_2-V}/{r_2}

\;\;\text " Now, "\;\;I\;=I_1+I_2

\;\;\text " Or, "\;\;I\;=\;{E_1-V}/{r_1}+\;{E_2-V}/{r_2}

\;\;\text " Or, "\;\;I r_1 r_2\;=r_2 (E_1-V) +r_1 (E_2-V)

\;\;\text " Or, "\;\;V (r_1+r_2) \;=E_1 r_2+E_2 r_1-I r_1 r_2

\;\;\text " Or, "\;\;V\;=\;{E_1 r_2+E_2 r_1}/{r_1+r_2}- {\I} \;{r_2 r_1}/{r_1+r_2}

Comparing with

V=E-I r

Equivalent emf

=E={E_1 r_2+E_2 r_1}/{r_1+r_2}

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