Physics, asked by Pranav44591, 11 months ago

Derive a relation between emf constant potential internal resistance of a cell and external resistance

Answers

Answered by Ankitakashyap2005
0

Answer:

Hi..

To drive a current through a ideal cell (no internal resistance) from its positive terminal to its negative terminal requires there to be an external voltage source which exceeds the emf of the cell EE by an infinitesimal amount.

However with a cell that has internal resistance RR an extra potential difference must be applied across the cell terminals to drive the current II through the internal resistance =IR=IR.

So the total applied potential difference across the cell must be V=E+IRV=E+IR.

It might be clearer if both sides of the equation are multiplied by the current VI=EI+I2RVI=EI+I2R.

Now VIVI is the power being delivered by an external source to the cell.

I2RI2R is the rate at which heat is produced due to the cell having an internal resistance.

EIEI is the rate at which energy is being supplied to the cell to reverse the chemical change which the cell uses when it is discharging (converting chemical energy into electrical energy.

Now doing the same when the terminal potential difference is less than the emf of the cell.

V=E−IR⇒VI=EI−I2R⇒EI=VI+I2RV=E−IR⇒VI=EI−I2R⇒EI=VI+I2R

EIEI is the electrical power supplied by the cell from the chemical reaction within the cell.

I2RI2R is again the rate at which heat is produced due to the cell having an internal resistance.

VIVI is the power delivered to the external circuit by the cell.

So in one case (discharging) the cell is producing the electrical energy from chemical energy and in the other case (recharging) the cell is consuming electrical energy which is being converted into chemical energy.

hope it'll help you..

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