# Electric Current - 4

(b) __Parallel grouping __ :-

Applying Kirchoff’s Law through branch :

© __Series grouping with different batteries__ :-

__E _{eq} = E_{1} + E_{2} + E_{3} +……………….. + E_{n}__

R

_{eq}= r

_{1}+ r

_{2}+ ……………………………. + r

_{n}

(d)

__Parellel Grouping with different batteries__:-

Number of rows are m and numbers of cells - in each rows in n :-

So, it is equivalent -

Applying Kirchoff’s laws :-

__Dumb question__:- How is total emf mE ?

__Solution__:- ’m’ . ‘E’ rated batteries are in series so net emf becomes mE.

__Dumb Question__:- How is net resistance ?

__Solution__:- ’m’ batteries in each row means ‘mr’ resistance becomes .

__Illustration - 8.__Find the equivalent emf and resistance of a single battery which is equivalent to a combination of al these batteries in the figure.

__Ans__:- For parallel combination :-

Now, this is in series with 6V battery -

Thus, net e.m.f. = (6 -3)V

And internal resistance = 2

So, the equivalent circuit is -

a)

__Galvanometer__:- It can detect very small currents as it has negligible resistance.

The corresponding potential difference for full scale deflection is :-

V = ig.G

b)

__Ammeter__:-

IT IS ALWAYS CONNECTED IN SERIES WITH THE BRANCH ‘I’ HAS TO BE MEASURED !

It is a corrent measuring instrument. A galvanometer can be converted into a ammeter by connecting a small resistance S (called shunt) in parallel with it.

Thus, S(I - I

_{g}) = G X I

_{g}

__Dumb Questions__:- Why is S in parallel ?

<>Solution :- As S is in parallel the current (I - I

_{g}) will be very high while Ig will be lo. So that the potential difference V

_{A}- V

_{B}does not change much.

__Dumb Questions__:- Why is Ig ?

<>Ans :- I

_{g}is that current which gives full scale deflection in galvanometer.

__TO THE BRANCH WHOSE BE MEASURED !__

A voltage measuring device is called a voltmeter. It measures the potential difference between two points. A galvanometer can be connecting a high resistance ® in series with it.

__Dumb Questions__:- Why is R

_{V}high ?

__Solution__:- If a large register (R

_{V}) is applied in parallel to two points then it doesn’t draw any current as R

_{V}is high. So I

_{g}is very low. And hence the current I

_{AB}doesn’t change much.

__Dumb Questions__:- What is the ideal resistance for as ideal ammeter and ideal voltmeter ?

__Ans__:- For ideal ammeter R = 0

For ideal voltmeter R =

__Illustration - 9.__A galvanometer having a coil resistance of 50 gives a full scale deflection when a current of 0.5mA is passed through it. What is the value of the resistance which can convert this galvanometer into ammeter giving full scale deflection for a current of SA ?

__Ans__:- As we derived earlier :

Now, I

_{g}= .5 10

^{- 3}A.

I = 5 A.

G = 50

As expected it is coming out to be very low.

In ideal ammeters this shunt resistance is 0.

__Illustration - 10__:- How can we make a galvanometer with G = 20

and I

_{g}= 1mA in to a voltmeter with a maximum range of ISV ?

__Ans__:- Using

.

we have,

.

As expected it is very high.

In ideal voltmeter it tends to infinity.

__* Tip__:- The basic Aim of using such devices is that they don’t affect the current much when used. And hence the voltmeter is such that when it is used, it doesn’t draw much current so that potential difference remains the same and is such that it also doesn’t let the voltage to da between the two points where it is installed. (and hence doesn’t change the current).

__WHEATSTONE’S BRIDGE__

In this system suppose galvanometer shows null (zero current.

Then V

_{BC}= 0

V

_{B}- V

_{C}= 0

V

_{B}= V

_{C}

I

_{1}R

_{1}= I

_{2}R

_{3}…………………………………… (1)

Similarly form other ends

I

_{1}R

_{1}= I

_{2}R

_{4}…………………………………… (2)

Dividing (1) and (2) :

And, At this point the bridge is said to be balanced.

__Dumb Questions - 14.__Why on value of current through galvanometer to be zero V

_{BC}= 0 ?

__Ans__:- Current flows between two points only if there is any potential difference between those points. Here, in this case as current through BC is zero, this suggests that potential of B equals the potential of C. Hence V

_{B}= V

_{C}or V

_{B}- V

_{C}= 0 or, in other words V

_{BC}= 0.

So if between B and C, another resistance is placed, then no current flows through that resistance.