Meter Bridge

Class XII/PHYSICSMETRE BRIDGE or SLIDE WIRE BRIDGE

Principle: It is constructed on the principle of Wheatstone bridge, i.e. when the bridge is balanced,

, where P and Q are known resistances,

R is a variable resistance and X is the unknown resistance.

Construction: A constantan resistive wire of uniform diameter is mounted on a wooden plank along with ameter rule. Copper strips are connected at the ends A and Bof the wire as shown in the figure. The terminalson this strip are connected to a battery. Anothercopper strip is fixed between these two strips forming two gaps. In one gap, a resistance box R is connected and in the other gap, an unknown resistor X is connected. One end of a galvanometer is connected to the mid-point D of this strip and the other end to a jockey which can slide on the wire AB. For a given value R, the jockey is slided on the wire in such a way that galvanometer shows zero deflection.

Applications of meter bridge:1. To measure the unknown resistance: To find the value of unknown resistance, the circuit is

arranged as shown in figure. The bridge can be balanced by moving the jockey over the wire. For a given value R, the jockey is slided on the wire in such a way that galvanometer shows zero deflection. If null point is obtained at C such that AC = l then, CB = 100-l. The resistance between A and C is taken as P and that between C and B is taken as Q. Since the bridge wire has uniform area of cross-section, and .

Then,

From the principle of Wheatstone bridge,

Or, ----------------------- (1)

Knowing the values of R and l, X can be found.

2. To compare two unknown resistances: For comparing two unknown resistances R1 and R2, X in the circuit is replaced by R1. The jockey is moved over the bridge wire gently and by adjusting R, the balancing length l1 is obtained when the galvanometer gives zero deflection.In such a case, if AC= l1, then, CB=100- l1. Since the bridge wire has uniform area of cross-section,

and .

Therefore,


Now, the experiment is repeated with the other unknown resistance R2 in the right gap. If l2 is the balancing length, then we get- and

Therefore,


------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------SRHSS/2010/ CURRENT ELECTRICITY of 5 Prepared by: Sandhya.K

Class XII/PHYSICS

Dividing equation (2) by equation (3), we get-

Knowing the values of l1 and l2 , the ratio can be found.

POTENTIOMETER

A potentiometer is used to compare e.m.fs of two cells or to measure the internal resistance of a cell.

Principle: The principle of potentiometer states that when a constant current is passed through a wire of uniform area of cross-section, the potential drop across any portion of the wire is directly proportional to the length of that portion.

Construction: A potentiometer consists of a number of segments of wire of uniform area of cross-section stretched over a wooden board between two copper strips. Each segment is 100cm long. A battery connected between the two terminals sends current through wire, which is kept constant by using a rheostat.

Theory: Let V be the potential difference across certain portionof the wire, whose resistance is R. If I is the current through the wire, then V=IR.

We know, , where l, A and ρ are length, area of

cross-section and resistivity of the material of the wirerespectively.

Therefore,

If constant current is passed through the wire of uniform area of cross-section, then I and A are constants.Since for a given wire, ρ is also constant, we have-

Or,

Hence, if a constant current flows through a wire of uniform area of cross-section, the potential drop across any portion of the wire is directly proportional to the length of that portion.

Applications of potentiometer

1. To compare e.m.fs of two cells: Let E1 and E2 be the e.m.fs of the two cells. The positive poles of both the cells are connected to terminal A of the potentiometer. The negative poles of the two cells are connected to the terminals 1 and 2 of a two way key while its common terminal is connected to a jockey J through the galvanometer G. In the primary circuit, an ammeter, a battery of e.m.f E, a rheostat and a one way key is included.


Class XII/PHYSICSTo compare the e.m.fs of two cells, a constant current is passed through the potentiometer wire between A and B.When the plug is put between 1 and 3 of the two way key, the cell of e.m.f E1 is included in the circuit. Let the balancing length be l1. If x is the resistance per unit length of the potentiometer wire, and I, the constant current passing through it, then-

--------------------- (1)When key is put between 2 and 3 of the two-way key after removing it from the gap 1 and 3, the cell with e.m.f E2 is included in the circuit. Let the balancing length be l2. Then, --------------- (2)

Dividing equation (1) by (2),

---------------- (3)

Measuring l1and l2, the ratio of the e.m.fs can befound.

2. To measure the internal resistance of a cell: The circuit is arranged as shown. Plug in the key K and maintain a constant current I through the circuit. Find the balancing length l1.Then, ---------- (4)

Now, plug the key K2 to include S in the circuit. Let the balancing length be l2. Then,

--------- (5)Dividing equation (4) by (5), we get-

-------------- (6)

The internal resistance of a cell is given by:

Using equation (6), we have-

Knowing the values of l1, l2 and S, internal resistance can be calculated.

AMMETER

Ammeter is used for measuring current. Galvanometer is a low resistance device and cannot be used to measure current in a circuit. To convert a galvanometer to an ammeter, a low value resistance called shunt is connected in parallel to the galvanometer. As its value is much smaller than G, most of the current flows through it and the galvanometer does not get damaged. Moreover, the equivalent resistance of the modified galvanometer is lower than thatof the shunt which when connected in the circuit does not alter its resistance appreciably and hence the true value of the current is measured.


Class XII/PHYSICS

Formula of ShuntLet IG = maximum current capacity of the galvanometer,I = desired maximum current in the ammeterS = necessary value of the shuntThe value of the shunt should be so selected that the current IG flows

through the galvanometer and the balance I - IG = Is flows through the shunt as shown in the figure. Applying Kirchhoff’s second rule to the loop,

VOLTMETER

Voltmeter is used to measure the potential difference between any two points of a circuit.Galvanometer with suitable modification is used for this purpose. To convert a galvanometer to a voltmeter a high value resistance called series resistance is connected in series with the galvanometer.

Deriving formula for the Series Resistance

Let IG = maximum current capacity of the galvanometer,G = resistance of galvanometer,V = desired maximum voltage in the voltmeter

Rs = necessary value of the series resistanceApplying Ohm’s law to the modified galvanometer as shown in the figure,

GROUPING OF CAPACITORS

Capacitors can be connected either in parallel or in series. When the capacitors are connected in series, the net capacitance DECREASES. When the capacitors are connected in series, the net capacitance INCREASES.

CAPACITORS CONNECTED IN SERIESCapacitors are connected in series if the positive plate of one is connected to the negative plate, as shown in Figure.

Let V1, V2 and V3 be potential differences across the plates of three capacitors C1, C2 and C3. If V is the potential difference across the series combination, then


Class XII/PHYSICS

From the definition of capacitance, ; ;

Therefore, ---------------- (1)

Let C be the equivalent capacitance of the series combination. Therefore, when a potential difference V is applied across the a capacitor of capacitance C, it will store same amount of charge.

Hence, ----------------------- (2)

From equations (1) and (2), we get-

Or, ------------------------ (3)

Thus, the reciprocal of the resultant (equivalent) capacitance of the series combination of a number of capacitors is equal to the sum of the reciprocals of the capacitances of individual capacitors.

CAPACITORS CONNECTED IN PARALLELTwo capacitors are in parallel if the negative plates are connected and the positive plates are connected, as shown in Figure

All the capacitors will have same potential difference across their plates. As the capacitors have different capacitances, they will draw different amount of charge from the source of e.m.f. If q1, q2, q3

are the values of the charges on the capacitors C1, C2 and C3 respectively and V is the potential difference across the parallel combination, then-

The total charge on the parallel combination of capacitors,

-------------------------- (4)If C is the equivalent capacitance of the parallel combination, then-

------------------- (5)From equations (4) and (5), we get-

----------------- (6)

Thus, the resultant (equivalent) capacitance of the parallel combination of a number of capacitors is equal to the sum of the capacitances of individual capacitors.


Meter Bridge

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