# Electrical Circuits | Electricity and Magnetism | Chapter 1 | Class 12 | New Curriculam 2078 | Notes | Nepal |

### Galvanometer

A galvanometer is an instrument used to detect small(feeble) current in electric circuit. Galvanometer may be moving magnet galvanometer or moving coil galvanometer. In laboratory, a moving coil galvanometer is used.

When a current is passed through the coil, it experience a torque and deflects. The deflection of coil is directly proportional to the strength of current. A pointer attached to the coil shows a deflection under measuring skill. It is denoted by

### Shunt

A low (small) resistance which is connected in parallel to a galvanometer is called shunt. It carries most of the current passing through the circuit and thus leaving a small amount of current to flow through galvanometer.

### Uses:

1. It is used to for converting galvanometer into ammeter

2. It is used to prevent damage of galvanometer

3. It reduces decreases the total resistance of the circuit.

### Conversion of Galvanometer into Ammeter

An ammeter is an instrument used to measure the current in electric circuit. It is connected in series. If large current is passed, the galvanometer may damaged. So when galvanometer is converted into ammeter, it measures current without any damage to it.

To convert galvanometer into ammeter of range 0 to IA, a small resistance S is connected in parallel to the galvanometer which is called shunt. Let G be the resistance of galvanometer and I be the maximum current measured by ammeter. Then I_g be the current flowing through galvanometer. Then (I-I_q) is the current passing through the shunts.

Since G & S are parallel combination,

P.d across S  =  P.d across G

This is the value of shunt. Knowing the value of I_g, I and G, the value of S can be calculated. By connecting shunt of this value, the given galvanometer can be converted into ammeter of the range 0 to IA.

The effective resistance R_a of the ammeter is given by

\frac{1}{R_a} = \frac{1}{G} + \frac{1}{S}

\frac{1}{R_a} = \frac{S+G}{GS}

R_a = \frac{GS}{G+S}

In practice, G is large as compared to S. Since S is low resistance, equivalent resistance R_a of ammeter is very low and when it is connected in series in circuit, it will not affect the current passing through circuit.

### Conversion of Galvanometer into an Voltmeter

A voltmeter is a device used to measure P.d. between two points in a circuit. It is connected in parallel. Voltmeter should have very high resistance ideally infinite resistance and must not change the current.

To convert a galvanometer into voltmeter of range 0 to V volts, a high resistance R is connected in series to the galvanometer which is called multiplier. The combination of galvanometer and multiplier for the conversion of voltmeter is as shown in fig.

Let G be the resistance of the galvanometer and I_g be the current flowing through galvanometer which produces the maximum deflection in the galvanometer. Since high resistance R is connected in series with galvanometer, same current I_g flows through Resistance R. The voltage across the voltmeter is

V = (I_g)(R+G)

V = (I_g)R + I_gG

I_gR = V – I_g G

R = \frac{V}{I_g} – G

This is the value of resistance R. Knowing the value of V, I_g and G, the value of R can be calculated. When the resistance of this value is connected in series to the galvanometer, it will work as a voltmeter of range 0 to V volt.

## Kirchoff’s Laws:

### First Law:

It states that the algebraic sum of currents at a junction of circuit is zero.

Symbolically

\sum I = 0

Current flowing towards the junction as positive and current flowing out from the junction as negative. Then applying Kirchoff’s 1st law at the junction A, we have

I_1 + I_2 + I_3 - I_4 - I_5 - I_6 = 0

I_1 + I_2 + I_3 = I_4 + I_5 + I_6

Hence, the sum of current flowing towards the junction is equal to sum of current flowing out from the junction.

### Second Law

It states that in a closed loop, the algebraic sum of emfs is equal to the algebraic sum of product  of current and resistance.

Symbolically,

\sum E = \sum IR

Explanation:

ABCA be a closed circuit whose sides AB, BC and CA carry currents I_1, I_2 and I_3 respectively. The resistances of the sides are R_1, R_2 and R_3 respectively. There is a cell of emf E in the circuit. In complete circuit ABCA, we can write

\sum E = \sum IR

I_1R_1+I_2R_2+I_3R_3=E

If no cell in the circuit, E=0. So, we have

I_1R_1 + I_2R_2+I_3R_3=0

## Wheatstone bridge

Wheatstone bridge is an electrical circuit which is used to find the unknown resistance. It consists of four resistance P,Q,R & X in which P,Q,R are known resistance and X is unknown resistance, when the battery is connected across B&D then at balanced condition i.e. I_g=0

\frac{P}{Q} =\frac{X}{R}

This is the balanced condition for wheatstone bridge and known as principle of wheatstone bridge.

Let, initially the circuit is not in balanced condition and suppose I_1 , I_2, I_3 and I_4 be the current passing through P,X,Q and R respectively. Then applying Kirchhoff’s first law at the junction B is

\sum I = 0

I_1 – I_g – I_3 =0

I_1=I_g +I_3

And at junction D,

I_2 + I_g – I_4 =0

I_2 + I_g = I_4

Applying Kirchhoff’s second law (voltage law) in loop ABDA

\sum E = \sum IR

0 = I_1 P + I_g G – I_2 X

I_1 P + I_g G = I_2 X

Similarly in closed loop BCDA,

0 = I_2 Q – I_g G – I_4 R

I_3 Q = I_g G + I_4 R

At balanced condition, the value of P, Q & R are so adjusted that no current passes through the galvanometer i.e. I_g= 0 then above eqn becomes

I_1 = I_3  and  I_2 = I_4

And

I_1 P = I_2 X

I_3 Q = I_4 R

Dividing eq^n 6 by 7,

\frac{I_1 P}{I_3 Q} = \frac{I_2 X}{I_4 R}

From equation 5 & 8

\frac{I_1 P}{I_1 Q} = \frac{I_2 X}{I_2 R}

\therefore \frac{P}{Q} = \frac{X}{R}

This is required balanced condition of wheatstone bridge.

### Meterbridge

Meterbridge is an electric device which is used to determine the unknown resistance. It works on the principle of wheatstone bridge.

It consists of three metal strips of copper which are fixed on wooden board. A constantan wire of uniform cross section and 1m length is stretched tight between two terminal of meter scale as shown in figure

A resistance box R is connected in one gap and unknown resistance ‘X’ is connected in another gap. A galvanometer is connected between point D & jockey over the wire then resistance of AB = P and BC = Q

When the bridge is balanced then, wheatstone principle gives

\therefore \frac{P}{Q} = \frac{X}{R}

\therefore (Resistance   of   AB)/(Resistance  of   BC) = \frac{X}{R}

Let length of wire (AC) = 100cm and AB = l cm & BC = (100-l) cm. Since the wire has uniform cross-section and \rho is constant then its resistance is directly proportional to the length i.e. P \alpha l and Q \alpha (100-l)

So,

\frac{l}{100-l} = \frac{X}{R}

X = (\frac{l}{100-l})R

By knowing the value of l, 100-l and R, the unknown resistance (X) can be calculated.

### Potentiometer

Potentiometer is an electric device which is used to measure the emf of cell and internal resistance, to compare the emf of two cell and p.d. between two points. It is an ideal instrument for measuring the p.d. between two points because it does not draw any current from circuit.

Potentiometer is a long wire AB and uniform cross sectional area stretched between copper strips fixed on wooden board. The wire used in potentiometer is manganin or constantan. The length of wire is 5-10m and each of the segment length is 1m. Driving cell is connected across point A and B. The balanced point on wire is obtained by sliding the jockey over a wire. At balanced point there is no deflection on galvanometer.

Principle: It is based on the principle that steady current is passed through a wire of uniform cross sectional area, the p.d. across any portion(segment) of wire is directly proportional to its length.

V \alpha l

## Application:

### 1.       To find the internal resistance of given test cell:

Experimental arrangement to find the internal resistance of given cell using potentiometer is as shown in fig.

At first the key K_1 is closed (with key K_2 open) and jockey is slided over the potentiometer wire till the galvanometer shows null deflection. Suppose galvanometer shows null deflection at point e such that AC = l_1. Since l_1 is the balancing length for emf of test cell so.

Emf (E) = V_AC = k l_1

Now a suitable value of resistance R is taken from the resistance box and k_2 key is closed. Then above process is repeated till the galvanometer shows null deflection. Let galvanometer shows null deflection at point D such that AD = l_2. Since l_2 is the balancing length for terminal p.d. of the test cell so terminal p.d.(V) =V_AD = k l_2

Dividing equation 1 by 2

\frac{E}{V} = \frac{k l_1}{k l_2} = \frac{l_1}{l_2}

Also expression for internal resistance of a source in terms of its emf and terminal p.d. is

r = R(\frac{E}{V} -1) = R(\frac{l_1}{l_2} -1)

\therefore r = R(\frac{l_1}{l_2} -1)

By knowing the value of R, l_1 and l_2, the value of internal resistance of given cell can be determined.

### 2.       Compare the Emf of given cells:

The experimental arrangement to compare the emf of given cells is as shown in fig.

At first the test cell of emf E_1 is connected in circuit through the key k_2 and jockey is slided over the potentiometer wire till the galvanometer shows the null deflection. Let the galvanometer shows deflection at point C such that AC = l_1. Then the emf of first cell  is given by

E_1 = V_AC = kl_1

Now first cell is disconnected and the second cell of Emf E_2 is connected in the circuit through the key k_2 and then repeat above process till the galvanometer shows null deflection. Suppose galvanometer shows null deflection at point D such that AD = l_2. Then emf of second cell is given by

E_2 = V_AD = kl_2

Dividing equation 1 by 2

\frac{E_1}{E_2} = \frac{k l_1}{k l_2}

\frac{E_1}{E_2} = \frac{l_1}{l_2}

By knowing the value of l_1 and l_2, the emfs of two cell can be compared.