The efficiency of a cell when connected to a resistance $R$ is \[60\% \] What will be the efficiency when external resistance is increased to six times:
A. \[80\% \]
B. \[90\% \]
C. \[55\% \]
D. \[85\% \]
Answer
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Hint:A cell is a single electrical energy source that converts chemical energy to electrical energy. Efficiency is the ratio of useful power to the total power used by the device, in this case, the cell. The value of the new efficiency can be found out by using the formula for efficiency.
Formula used:
The efficiency of a cell is given by,
\[\eta = \dfrac{\text{useful power}}{\text{total power used}}\]
Complete step by step answer:
The following data is given in the solution, the efficiency of the cell \[60\% \]. Let \[r\] be the internal resistance of the cell. \[R\] be the external resistance in the circuit (as in the question). \[I\] be the current through the circuit. The amount of useful power in the circuit is, \[{I^2}R\]. The amount of total power used, \[{I^2}r + {I^2}R\].
This gives us the value of the efficiency when substituted in the formula given above,
\[\eta = \dfrac{{{I^2}R}}{{{I^2}r + {I^2}R}}\]
All the \[{I^2}\] values are cancelled as they can be taken commonly and cancelled.
Then we arrive at, \[\eta = \dfrac{R}{{r + R}}\]
This gives us the relation between external resistance and efficiency.
Substituting the value of efficiency, we get \[0.6 = \dfrac{R}{{r + R}}\]
Cross multiplying, we get \[0.6R + 0.5r = R\]
\[R = 1.5r\]
Now taking the value of new external resistance as, \[6R\]. The value of efficiency becomes,
\[\eta = \dfrac{{6\left( {1.5r} \right)}}{{6\left( {1.5r} \right) + r}} \\
\therefore \eta= \dfrac{9}{{10}}\]
Converting this into percentage, we get \[90\% \]. In conclusion, be the efficiency when external resistance is increased to six times will be \[90\% \]
Therefore, option B is the correct answer.
Note:An electrochemical cell consists of electrodes namely cathode and anode.The non-spontaneous reaction is controlled by an external source of current. The decomposition of an electrolytic solution by passing current, which results in the liberation of corresponding gases or metals is known as electrolysis.
Formula used:
The efficiency of a cell is given by,
\[\eta = \dfrac{\text{useful power}}{\text{total power used}}\]
Complete step by step answer:
The following data is given in the solution, the efficiency of the cell \[60\% \]. Let \[r\] be the internal resistance of the cell. \[R\] be the external resistance in the circuit (as in the question). \[I\] be the current through the circuit. The amount of useful power in the circuit is, \[{I^2}R\]. The amount of total power used, \[{I^2}r + {I^2}R\].
This gives us the value of the efficiency when substituted in the formula given above,
\[\eta = \dfrac{{{I^2}R}}{{{I^2}r + {I^2}R}}\]
All the \[{I^2}\] values are cancelled as they can be taken commonly and cancelled.
Then we arrive at, \[\eta = \dfrac{R}{{r + R}}\]
This gives us the relation between external resistance and efficiency.
Substituting the value of efficiency, we get \[0.6 = \dfrac{R}{{r + R}}\]
Cross multiplying, we get \[0.6R + 0.5r = R\]
\[R = 1.5r\]
Now taking the value of new external resistance as, \[6R\]. The value of efficiency becomes,
\[\eta = \dfrac{{6\left( {1.5r} \right)}}{{6\left( {1.5r} \right) + r}} \\
\therefore \eta= \dfrac{9}{{10}}\]
Converting this into percentage, we get \[90\% \]. In conclusion, be the efficiency when external resistance is increased to six times will be \[90\% \]
Therefore, option B is the correct answer.
Note:An electrochemical cell consists of electrodes namely cathode and anode.The non-spontaneous reaction is controlled by an external source of current. The decomposition of an electrolytic solution by passing current, which results in the liberation of corresponding gases or metals is known as electrolysis.
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