# Electric Current - 5

d) __Potentiometer__ :-

It is a device used to measure potential difference with high accuracy, without drawing any current from the unknown source. [* Basic Figure of a potentiometer : ??]

__Principle of Potentiometer__

It is based on the principle, that a current carrying wire has potential difference across its ends proportional to its length with a uniform cross sectional area.

It can be used for various purpose :

(1) To find e.m.f. of an unknown battery :

First a known battery (E_{k}) is placed as :

Suppose the null point is at l_{1}

Then, E_{k} = i(l l_{1}) ………………………………………(1)

__Dumb Questions - 15.__ What is this null point ?

__Ans__ :- On a close observation we can see that potentiometer is similar to wheatston’s bridge.

At null point current through galvanometer equals 0.

Now, the known battery is replaced by unknown battery (E_{u}).

Now, E_{u} = I(l l_{2}) ………………………………………. (2)

From (1) and (2)

2. Find internal resistance® of an unknown battery.

, where R is a known resistance.

For any circuit for battery : V = E -ir = iR

………………………………….. (A)

Where R = external resistance.

r = internal resistance.

We connect a battery of e.m.f. E and internal resistance r as:

Now, E = ill_{1} …………………………………………. (1) Now a known resistance R is connect across the terminals as :

Now,

……………………………………… (2)

from (1) and (2)

……………………………………… (3)

Now, placing value of (3) in (A) :

__Question__ :- Find current in all branches

__Ans__ :- Assume the current distribution as shown :

Applying kirchoff’s loop law in the three loops we get :

2l - Si_{1} - 6(i_{1} + i_{2}) - i_{1} = 0 ………………………………….. (1)

S - 4i_{2} - 6(i_{1} + i_{2}) - 8(i_{1} + i_{3}) = 0 …………………………..(2)

2 - P(i_{2} + i_{3}) - 16i_{3} = 0 ……………………………………… (3)

On simultaneously solving there three equations we get :

__Dumb Question - 16.__ In branch AC which will we Lonsides i_{1}, i_{2} ?

__Ans__ :- We will Lonsides the current i_{1} + i_{2} in AC as we have Lonsidered that in those two loops different currents i_{1} and i_{2} are flowing now, their common edge will have the sum according to the direction of two currents. This is called principle of superposition.

__E2__ :- Two resistance have temprature coefficient _{1} and _{2} and R_{01} an R_{02} the resistance at 0^{0}C. Find _{eq} if they both are connect (a) in series and (b) in parallel ?

__Ans__ :- For series :

At 0_{0}C R_{eq} = R_{01} + R_{02} ………………………………………… (1)

At any temperature t_{0}C

R_{eq} = R_{01} + R_{02}

R_{eq} = R_{01}[1 + _{1}(t - 0)] + R_{02}[1 + _{2}(t - 0)]

[R_{01} + R_{02}][1 + _{eq}t] = R_{01}[1 + _{1}t] + R_{02}[1 + _{2}t]

In Parallel :

At t_{0}C

Using Binomial Expansion :

E3. A rod length L and cross sectional area A lies along the axis between x = 0 and x = L.

Its resistivity is given by : P(x) = P_{0}.

The end x = 0 is at a potential V_{0} and x = L at V = 0.

(a) Find net resis tance of the rod.

(b) The electric potential V (n) ?

__Ans__ :- (a)

At any distance n suppose an elementary particle dn be considered such that dn 0.

Its resistance

As R =

Thus, R =

(b) We know that,

Potential difference = i dR = dV

04. A battery of e.m.f. 1.4V and internal resistance 2 is connected to a 100 resistor through an ammeter. The resistance of te ammeter is .

If ammeter rods .02 A. What is the resistance of the Voltmeter ?

__Ans__ :- Circuit diagram is :

Now first finding total resistance of the circuit.

where G = Resistance of voltmeter.

Now, Current =

In the given circuit 5 resistor develops 45 J/s Power due to current flowing through it. Calculate the heat developed/second through 2 resistor ?

__Ans__ :- Assume the shown current distribution after writing equivalent circuit :