- Introduction
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- ElectroMotive Force(emf)
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- Internal Resistance of Battery (or cell)
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- Electric Energy and Power
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- Kirchoff's Rules
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- The junction Rule (or point rule)
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- The Loop Rule (or Kirchoff's Voltage Law)
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- Grouping of the cell's
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- Meterbridge (slide wire bridge)
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- Potentiometer
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- Comparison of EMF's of two cells using potentiometer
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- Determination of internal resistance of the cell

- The resistance offered by medium in between plates of battery (electrolytes and electrodes of the cell) to the flow of current within the battery is
called internal resistance of the battery.

- Internal resistance of a battery usually d branch containing batteryenoted by r and in electric circuit its representation is shown below in the figure

- Internal resistance of a battery depends on factors like seperation between plates, plate area, nature of material of plate etc. For an ideal
cell r=0 , but real batteries or sourcesof emf always has same finite internal resistance.

- If P and Q are two terminals of the battery shown below in the figure

then potential difference between terminals P and Q is

V_{P}=(V_{P}-V_{x}) - (V_{Q}- V_{x}) = E-Ir

let V_{P}-V_{Q}=V

V=E-IR

now for I=0 and V=EMF

and this potential difference V is called the terminal difference of the cell or battery and defined as the emf of the battery when no current drawn from it.

- For real battery equation(4) which gives V=E-Ir whereI is the current in the branch containing battery.

- From figure(4) potential difference across the external resistance R of the circuit would be equal to terminal potential difference of the cell. Thus

V=IR also V=E-Ir

or, IR=E-Ir

which gives

I=E/(R+r) =Net EMF/Net resistance - From equation(4) we can calculate that when current is drawn from the battery terminal potential difference is less than the EMF of the battery.

- To understand the process of energy transfer in a simple circuit consider a simple circuit as shown in the figure given below

- Positive terminal of the battery as we all know is always at higher potential.

- Let ΔQ amount of charge begin to flow in the circuit from point E through the battery and resistor and then back to point E.

- When Charge ΔQ moves from point E to point F through the battery electric potential potential energy of the system increased by the amount

ΔU=ΔQ V -(6) and the electric energy of the battery decreased by the same amount.

- When charge ΔQ moves from point G to S through resistoe R, there comes a decrease in electric potential energy.

- This loss in potential energy appears as the increased in thermal energy of the resistor.

- Thermal energy of the resistor increases because when the charge moves through the resistor they loss there electrical potential energy by colliding with the atom in the resistor. This way electrical energy is transformed internal energy crossesponding to increase in vibrational motion of the atom of the resistor and this cause increase in temprature of the resistor.

- The connecting wires are assumed to have negligible resistance and no energy transfer occur for the path FG and HE.

- In time Δt charge ΔQ moves through the resistor i.e. from G to H. The rate at which it loss potential energy

ΔU/Δt=(ΔQ/Δt)ΔV=IΔV where I is thecurrent in the resistor and ΔV is the potential difference across it.

- This charge ΔQ regain its energy when it passes through the battery at the cost of conservation of chemical energy of electrolyte to the electrical energy.
- This loss of potential energy as stated earlier appears as increased thermal energy of the resistor. If P represents the rate at which energy is delivered to the resistor then

P=IΔV - We know that ΔV =IR for a resistor hence alternative forms of equation(8) are

P=I^{2}R=ΔV^{2}/R where I is expressed in amperes, ΔV iv volts and resustance R in ohm(Ω)

- SI unites of power is watt such that

1watt=1volt*1ampare

Bigger unit of electric power are Kilowatt(KW) and Megawatt(MW).

Class 12 Maths Class 12 Physics