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Describe the factors affecting the internal resistance of a cell connected to a Voltmeter ?

Posted- 2266 days ago

.M.F or Electro Motive Force is the opposite of potential difference,
in that; it is the situation where a voltage is gaining energy. This
seems unlikely, but it is required in order to allow an electric
circuit to function. An electric current is a flow of electric charge,
the charge flows around the circuit, transferring some of its energy
to areas of resistance along the way (resistors, filament lamps,
buzzers). At some point along the way the energy must be initially
supplied, otherwise the whole process couldn't function. At the
beginning of the process the power supply provides the charge with
energy to pass to the circuit. This is the electro-motive force. EMF
is a type of voltage along with potential difference and these are
defined as:

A Voltage where the charge is losing energy is a potential difference,
V.

A Voltage where the charge is gaining energy is an electromotive force,
E.
a cell
contains chemical electrolytes and electrodes, both however work to
the same effect. When the charge passes through the source of e.m.f it
gains lots of energy, but also instantly loses some of that that it
gains, this is due to internal resistance. Internal resistance is
resistance that is found only in the power source. Commonly written as
r and graphically represented by a resistor enclosed in a circle
surrounding both the resistor and the power source.

It is possible to work out the current that flows when a power source
is connected to an external resistor, R. R and r are in series and
given that the current flows through both, one after the other, their
combined resistance can be written:

E=IR + Ir or E=I(R + r)

E cannot be directly measured because a voltmeter can only be
connected across the entire cell, including the cells internal
resistance, r. When a voltmeter is connected across the cell the
result returned is the terminal p.d. V using this formula

V=IR

By examining the previous formula, this is less than the e.m.f by Ir,
referred to as lost volts, and by combining the last two equations I
get the following:

V=IR - Ir or V=E-Ir