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

.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