SATHEE: current-electricity Question 36

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Question: Q. 14. A student connects a cell, of emf $E_{2}$ and internal resistance $r_{2}$ with a cell of emf $E_{1}$ and internal resistance $r_{1}$ , such that their combination has a net internal resistance less than $r_{1}$ . This combination is then connected across a resistance $R$ .

Draw a diagram of the ‘set-up’ and obtain an expression for the current flowing through the resistance.

R [Foreign Comptt., 2016]

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Solution:

Ans. Since $r_{e q} < r_{1}$ , the two cells must have been connected in parallel.

The diagram of the setup is as shown

Note :

[Award $(1 / 2 + 1 / 2 + = 1)$ mark even if the student directly draws the diagram of the setup]

Equivalent internal resistance, $r = \frac{r_{1} r_{2}}{(r_{1} + r_{2})}$

$1 / 2$

Equivalent emf, $E$ is given by

$E = (\frac{E_{1}}{r_{1}} + \frac{E_{2}}{r_{2}}) r$

The current, flowing through $R$ , is

$I = \frac{E}{(R + r)}$

where, $E$ and $r$ are given by the above expressions. $1 / 2$ [Note : Award 2 marks if the student obtains / writes the expression, $I = \frac{E_{1} r_{2} + E_{2} r_{1}}{[R (r_{1} + r_{2}) + r_{1} r_{2}]}$ for the current flowing through $R]$ .

[CBSE Marking Scheme, 2016]

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