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  • A cell of internal resistances r is connected to a load of resistance R. Energy is dissipated in the load, but some thermal energy is also wasted in the cell. The efficiency of such an arrangement

A cell of internal resistances r is connected to a load of resistance R. Energy is dissipated in the load, but some thermal energy is also wasted in the cell. The efficiency of such an arrangement is found from the expression

\frac{\text{Energy dissipated in the load}}{ \text{Energy dissipated in the complete circuit}}

Which of the following gives the efficiency in this case?
 

Option: 1

\mathrm{\frac{r}{R}}


Option: 2

\mathrm{ \frac{R}{r}}


Option: 3

\mathrm{ \frac{r}{R+r}}


Option: 4

\mathrm{ \frac{R}{R+r}}


Answers (1)

best_answer

Let current I flow through the circuit

Energy dissipated in load = \mathrm{I^2 R}
Energy dissipated in the complete circuit =\mathrm{I^2(r+R)}
Therefore, the efficiency = \mathrm{\frac{I^2 R}{I^2(R+r)}=\frac{R}{R+r}}

Posted by

Suraj Bhandari

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