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Current Electricity

Question
CBSEENPH12039178

a) State Kirchhoff's rules and explain on what basis they are justified.

(b) Two cells of emfs E1 and E2 and internal resistances r1 and r2 are connected in parallel.

Derive the expression for the

(i) Emf and

(ii) internal resistance of a single equivalent cell which can replace this combination.


OR

 

(a) "The outward electric flux due to charge +Q is independent of the shape and size of the surface which encloses is." Give two reasons to justify this statement.

(b) Two identical circular loops '1' and '2' of radius R each have linear charge densities -straight lambdaand +straight lambda C/m respectively. The loops are placed coaxially with their centre distance apart. Find the:

magnitude and direction of the net electric field at the centre of loop '1'.

Solution

a)

i) Junction Rule: The algebraic sum of currents meeting at a point in an electrical circuit is always zero.
                                      sum I space equals space 0

This law is in accordance with law of conservation of charge.

ii) Loop Rule: In a closed loop, the algebraic sum of emfs is equal to the algebraic sum of the products of the resistances and the current flowing through them.
                                Error converting from MathML to accessible text.

This law is based on the conservation of energy.

b) Consider the circuit, 
 

Here, E1 and E2 are the emf of two cells,

r1 and r2 are the internal resistance of cell,

I1 and I2 current due to two cells.

Terminal potential difference across the first cell is given by, 
space space space straight V space equals space straight E subscript 1 space minus space straight I subscript 1 straight r subscript 1 space

rightwards double arrow space space straight I subscript 1 space equals space fraction numerator straight E subscript 1 space minus space straight V over denominator straight r subscript 1 end fraction

For the second cell, terminal potential difference will be equal to that across the forst cell.

So, 
V space equals space E subscript 2 space minus space I subscript 2 r subscript 2
rightwards double arrow I subscript italic 2 space equals space fraction numerator E subscript italic 2 space italic minus space V over denominator r subscript italic 2 end fraction 

Let E be the effective emf and r the resultant internal resistance.

Consider, I as the current flowing through the cell.

Therefore, 
I thin space equals space I subscript 1 space plus space I subscript 2 italic space

rightwards double arrow space I thin space equals space fraction numerator E subscript italic 1 italic minus V over denominator r subscript italic 1 end fraction plus space fraction numerator E subscript italic 2 italic minus V over denominator r subscript italic 2 end fraction italic space

rightwards double arrow space I thin space equals space fraction numerator r subscript italic 2 space italic left parenthesis E subscript italic 1 italic minus V italic right parenthesis italic plus r subscript italic 1 italic left parenthesis E subscript italic 2 italic minus V italic right parenthesis over denominator r subscript italic 1 r subscript italic 2 end fraction italic space

rightwards double arrow space I r subscript italic 1 r subscript italic 2 space equals space E subscript italic 1 r subscript italic 2 space plus space E subscript italic 2 r subscript italic 1 minus space left parenthesis r subscript italic 1 plus r subscript italic 2 right parenthesis V

rightwards double arrow space V space equals space fraction numerator E subscript italic 1 r subscript italic 2 italic plus space E subscript italic 2 r subscript italic 1 over denominator r subscript italic 1 space italic plus space r subscript italic 2 end fraction minus space fraction numerator I r subscript italic 1 r subscript italic 2 over denominator r subscript italic 1 space italic plus space r subscript italic 2 end fraction

Now, comparing the equation with V = E –Ir, we have
straight E space equals space fraction numerator E subscript 1 r subscript 2 plus space E subscript 2 r subscript 1 over denominator r subscript 1 space plus space r subscript 2 end fraction space and
Internal space resistance comma space straight r space equals space fraction numerator r subscript 1 r subscript 2 over denominator r subscript 1 space plus space r subscript 2 end fraction
 
                                               OR

a) The outward electric flux due to the charge enclosed inside a surface is the number of electric field lines coming out of the surface. Outward flux is independent of the shape and size of the surface because:

i) Number of electric field lines coming out from a closed surface is dependent on charge which does not change with the shape and size of the conductor.

ii) Number of electric lines is independent of the position of the charge inside the closed surface.
b) Magnitude of electric field at any point on the axis of a uniformly charged loop is given by, 
E space equals space fraction numerator lambda over denominator 2 epsilon subscript o end fraction fraction numerator r R over denominator left parenthesis r squared plus R squared right parenthesis to the power of bevelled 3 over 2 end exponent end fraction 
 

Electric field at the centre of loop 1 due to charge present on loop 1 = 0

Electric field at the centre of loop1 due to charge present on loop 2 is given by, 

straight E '  equals straight space fraction numerator plus straight lambda over denominator 2 straight epsilon subscript straight o end fraction fraction numerator straight R square root of 3 straight R over denominator left parenthesis left parenthesis straight R square root of 3 right parenthesis squared plus straight R squared right parenthesis to the power of bevelled 3 over 2 end exponent end fraction 
rightwards double arrow space E space apostrophe space equals space fraction numerator 3 over denominator 16 epsilon subscript o end fraction lambda over R, is the required electric field.