A cell of emf 6 V and internal resistance 1 Ω…

A cell of emf 6 V and internal resistance 1 Ω is joined in parallel (like terminals together) with a cell of emf 3 V and internal resistance 2 Ω.…

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A cell of emf

6\ \mathrm{V}

and internal resistance

1\ \Omega

is joined in parallel (like terminals together) with a cell of emf

3\ \mathrm{V}

and internal resistance

2\ \Omega

. This combination drives current through an external resistor of

\dfrac{4}{3}\ \Omega

. The current through the

3\ \mathrm{V}

cell is

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Answer: (C) $\dfrac16\ \mathrm{A}$ , charging the cell

Let

V

be the common terminal voltage. Kirchhoff's current law at the junction:

\dfrac{6-V}{1}+\dfrac{3-V}{2}=\dfrac{V}{4/3}

, i.e.

6-V+1.5-0.5V=0.75V

, so

2.25V=7.5

and

V=\dfrac{10}{3}\ \mathrm{V}

. Current through the

3\ \mathrm{V}

cell

=\dfrac{3-10/3}{2}=-\dfrac16\ \mathrm{A}

: the negative sign means current enters its positive terminal, so it is being charged by the stronger cell. (Check:

\tfrac83\ \mathrm{A}-\tfrac16\ \mathrm{A}=2.5\ \mathrm{A}=V/R

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