Electrical circuits (Kirchhoff’s law)

To calculate the value of electric current and terminal PD in a simple electric circuit, Ohm’s law is applied. For a complex electric circuit, to calculate the EMF, electric current and resistance, instead of Ohm’s law, Kirchhoff’s law is applied. Hence, Kirchhoff’s law is called advanced form of Ohm’s law. Kirchhoff’s law is mainly divided into two types:

1. Kirchhoff’s current law.

2. Kirchhoff’s voltage law.

Kirchhoff’s Current Law (KCL)

The algebraic sum of electric current at a junction of a given electric circuit becomes zero i.e., [latex]\sum I[/latex] = 0. Let, I₁ and I₂ be the electric current with resistance R₁ and R₂ flowing toward the junction and I₃ be the electric current with resistance R₃ flowing from the junction. The current flowing toward the junction is positive and the current flowing away from the junction is negative. Then, we know that,

[latex]\sum I = 0[/latex]

Or, (+I₁) + (+I₂) + (-I₃) = 0.

Or, I₁ + I₂ – I₃ = 0.

[latex]\therefore[/latex] I₁ + I₂ = I₃.

Hence, sum of electric current flowing toward the junction is equal to sum of electric current flowing away from the junction. In this law, there is conservation of charge which is called Kirchhoff’s current law or Kirchhoff’s junction law.

Kirchhoff’s Voltage Law (KVL)

It states that, “In a closed loop (group) of an electric circuit, the sum of emf is equal to algebraic sum of product of electric current and resistance at various part of the loop.”

[latex]\sum E = \sum IR[/latex]

The complete electric circuit is shown in the figure where I₁, I₂, I₃ represent electric current. E₁, E₂, E₃ represent emf and R₁, R₂, R₃ represents resistor of that circuit. To explain complex electric circuit Kirchhoff’s law is applied. Let, emf and current flowing in an anti-clockwise direction consider as positive and emf and current flowing in a clockwise direction is considered as negative. Now, in a closed loop ABCFA,

For, [latex]\sum E = \sum IR[/latex], we get,

(+E₁) + (-E₂) = (+I₁)R₁ + (-I₂)R₂

[latex]\therefore[/latex] E₁ – E₂ = I₁R₁ – I₂R₂ ……….. (i)

Similarly, for closed loop FCDEF,

[latex]\sum E = \sum IR[/latex]

Or, (+E₂) + (-E₃) = (+I₂)  + (-I₃)R₃

[latex]\therefore[/latex] E₂ – E₃ = I₂R₂ – I₃R₃ …………… (ii)

Now, applying Kirchhoff’s current law,

At a junction ‘F” of an electronic circuit,

For, [latex]\sum I = 0[/latex]

Or, (+I₁) + (I₂) + (-I₃) = 0.

Or, I₁ + I₂ – I₃ = 0.

[latex]\therefore[/latex] I₁ + I₂ = I₃ ………………… (iii)

On solving these 3 equations, we get the value of I₁, I₂ and I₃ when E₁, E₂, E₃ and R₁, R₂, R₃ are known. In Kirchhoff’s 2^nd law, total potential of the circuit remains constant. So, it gives conservation of energy.

Wheatstone bridge

Wheatstone bridge is an electrical device which is used to measure the accurate value of unknown resistance. Wheatstone bridge circuit contains 4 resistances P, Q, R and X among them, 3 are known and 1 is unknown. The positive side of the battery is connected with A and negative side of battery is connected with galvanometer (G), is kept between points B and D of the circuits for the balanced condition of wheatstone bridge P/Q = X/R, which represent principle of wheatstone bridge.

Verification of Balanced condition of wheatstone bridge

It is supposed that the wheatstone bridge was not in balanced condition initially. I₁ and I₃ represent electrical current passed through resistor P and Q and I₂ and I₄ represent electric current pass-through resistor X and R.

Applying Kirchhoff’s voltage law in a closed loop ADBA with sign convention, we get,

[latex]\sum E = \sum IR[/latex]

Or, 0 = (+I₂)X + (-I_g)R_g + (-I₁)P

Or, 0 = I₂X -I_gR_g -I₁P

[latex]\therefore[/latex] I₁P + I_gR_g = I₂X ……… (i)

Applying Kirchhoff’s voltage law in a closed loop BDCB, we get,

[latex]\sum E = \sum IR[/latex]

Or, 0 = (I_g)R_g + (I₄)R + (-I₃)Q

I₃Q = I_gR_g + I₄R…………….. (ii)

Now, Applying Kirchhoff’s current at a junction B, we get,

(+I₁) + (-I_g) + (-I­₃) = 0.

Or, I₁ – I_g – I₃ = 0.

[latex]\therefore[/latex] I₁ = I_g + I₃ …………….. (iii)

Again, applying Kirchhoff’s current law at the junction D, we get,

[latex]\sum I = 0[/latex]

(+I₂) + (+I_g) + (-I₄) = 0.

I₂ + I_g – I₄ = 0.

I₂ + I_g = I₄ ………………. (iv)

At a balanced condition, P, Q and R are adjusted in such a way that the amount of electric current passed through the galvanometer becomes zero. i.e.

I_g = 0.

Now, above 4 equations can be written as,

I₁P = I₂X ………… (v)

I₃Q = I₄R ………….. (vi)

I₁ = I₃ ………… (vi)

I₂ = I₄ …………………. (vii)

Now, dividing eq^n. (v) by (vi), we get,

[latex]\frac{I_1P}{I_3Q} = \frac{I_2X}{I_4R}[/latex] ………………… (ix)

From eq^n. (ix), (vii) and (viii), we get,

[latex]\frac{I_1P}{I_1Q} = \frac{I_2X}{I_2R}[/latex]

Or, [latex]\frac{P}{Q} = \frac{X}{R}[/latex] ………………… (x)

Eq^n. (x) represents the balanced condition of wheatstone bridge at which potential at point B is equal to potential at point D.

Meter Bridge

It is an electrical device which is used to measure the accurate value of unknown resistance. It is based on the principle of balanced condition of wheatstone bridge.

The device which has conducting wire with length 1m having uniform cross-sectional area and low temperature coefficient represent meter bridge. The conducting wire AC is fixed over wooden board between 2 copper strips. Resistance box is kept on the left side of meter bridge between A and D and unknown resistance wire is kept on right gap of meter bridge between CD. The galvanometer is connected between point D and jockey (slider). Jockey help to find the null point on the wire AC. From the balanced condition of wheatstone bridge, we get,

P/Q = R/X.

Or, [latex]\frac{Resistance\ of\ AB}{Resistance\ of\ BC} = \frac{R}{X}[/latex] …………… (i)

The length of part AB and BC are represented by L and (100-L) respectively by from the definition of resistance of the wire, we can write,

[latex]P = \rho \frac{L}{A}[/latex]

[latex]Q = \rho \frac{(100 – L)}{A}[/latex]

for AB and BC respectively.

For the constant resistivity and uniform cross-sectional area of the wire, the above equations become,

P ∝ L

Q ∝ (100-L)

Now, Equation (i) can be written as

[latex]\frac{L}{(100 – L)} = \frac{R}{X}[/latex]

[latex]\therefore X = \frac{(100 – L)}{L}\times R[/latex] …………….. (ii)

Hence, equation (ii) represents expression for unknown resistance of the wire.

For the different values of R, we obtain different value of X. The average value of X gives accurate value of unknown resistance.

Potentiometer

Principle of Potentiometer

Potentiometer is an electrical instrument which is used to measure e.m.f. and internal resistance of the cell, to compare the emf of the cell and to measure potential difference between any 2 points. Also, it is used to measure the electric current and resistance of a wire.

The potentiometer wire is made up of by manganian and constantan having uniform cross-sectional area with length 10 m. The potentiometer wire is fixed over wooden board between 2 copper stripes. The potentiometer wire is divided into 10 equal segments with length 1m each. Each wire is fixed between 2 sides of the potentiometer. Constant current I is flowed in the circuit with the help of driver cell E_o which is varied by rheostat ‘R_h‘. Ammeter and voltmeter are used to measure electric current and potential difference in the circuit.

One side of the voltmeter is connected with point A and next side is connected with Jockey (slider) which helps to find balance point on the potentiometer wire.

A scale (meter) is fixed on the wooden board near to the potentiometer wire.

Let, I be the electric current passed through the potentiometer wire AB. R be the resistance of a wire. Then, from Ohm’s law,

V = IR ………………………. (i)

For the part of potentiometer wire AC, the length of wire is L then, its resistance represented by:

R = [latex]\rho \frac{L}{A}[/latex] .

Now, eq^n. (i) can be written as:

[latex]V = I\rho \frac{L}{A}[/latex]

Or, V = [latex]\frac{I\rho}{A}.L[/latex]

Or, V = KL, where, K = [latex]\frac{I\rho}{A}[/latex] = proportionality constant for steady electric current, is of constant value of resistivity and uniform cross-sectional area of the wire.

[latex]\therefore[/latex] V ∝ L ………. (ii)

The eq^n. (ii) represents the fundamental principle of potentiometer which sate that the potential difference between any 2 points of the potentiometer wire is directly proportional to corresponding length of that part for steady electric current.

Application of potentiometer:

1. Determination of internal resistance of a cell:

Potentiometer is basically used to measure the potentiometer difference between any two points of the circuit. It is also used to determine the internal resistance of the cell which is offered by the source itself.

Let, ‘E’ be the e.m.f. of a cell and ‘r’ is internal resistance which is to be determined. The positive terminal of E is connected with point A where A is connected with positive terminal of driver cell E_o. ‘I’ is a steady amount of electric current flowing in the circuit. R represents resistance produced on resistance box which is kept parallelly with cell along key.

One point of the galvanometer connected with negative terminal of the cell and next point is connected with jockey (slider) which is used to find balanced point on the potentiometer wire.

When key is open there is no flow of electrons across resistance box, the emf. of the circuit balance the potential produced between 2 points A and D where D is the balanced point showed by the galvanometer in null point deflection i.e.

E = V_AD

The length of part AD is L₁ i.e. according to the principle of potentiometer.

V_AD ∝ L₁

Or, E ∝ L₁ …………….. (i)

When key ‘K’ is closed, the potential difference of the circuit becomes V. The galvanometer shows null point at D’ where the length of AD’ is L₂. The potential difference produced on the circuit balance the p.d. between A and D’. i.e.

V = V_AD’

According to principle of potentiometer,

V_AD’ ∝ L₂

[latex]\therefore[/latex] V ∝ L₂ ……………. (ii)

From electric circuit, we know that,

E = I(R+r) …………… (iii)

And, V = IR ………….. (iv)

Now, dividing eq^n. (i) and (ii),

We get,

[latex]\frac{E}{V} = \frac{L_1}{L_2}[/latex] ………………… (v)

Again, from (iii), (iv) and (v),

[latex]\frac{I(R+r)}{IR} = \frac{L_1}{L_2}[/latex]

Or, [latex]\frac{(R+r)}{R} = \frac{L_1}{L_2}[/latex]

Or, [latex]1+\frac{r}{R} = \frac{L_1}{L_2}[/latex]

Or, [latex]\frac{r}{R} = \frac{L_1}{L_2}-1[/latex]

Or, r = [latex](\frac{L_1}{L_2}-1)R[/latex] ………………………… (vi)

Hence, eq^n. (vi) represents the expression for internal resistance of the cell as the value of L₁, L₂ and R is known, we can determine the internal resistance of cell.

Comparison of emf of two cells:

Let us consider, two cell having emf E’ and E” connected with potentiometer for comparison of emf as shown in figure.

Let E be the emf of a cell flowing steady current in potentiometer circuit. We connect the positive terminal of the E’ and E” are joint to point A, the positive terminal of E and negative terminal is connected to the key. A galvanometer is connected between the key and the jokey that move over the potentiometer wire.  Where E is greater than E’ and E”.

Let us consider one cell E’ is connected in the circuit by closing K’ and E” is open by K”. The key is slided along wire between AB to find the null point. When the jockey at P near to the A on the wire, the length of portion AP is smaller and hence the potential difference is also smaller. Since the emf E’ is grater then V_AP and current flow through the galvanometer and deflection occur in the opposite direction.

Again, when the jockey is placed at Q near the B. then the potential difference will grater V_AP of portion BQ and current flow through the galvanometer and deflection occur in the opposite direction. This shows that the circuit is correct. on moving jockey, we find a point R at which the galvanometer shows null deflection at this condition the potential difference at R is equal to E’ and no current flow and there is no develop p.d in the internal resistance of E’ and in galvanometer and the whole emf E’ is balanced by the p.d V_AR by the jokey moving length L’. So, at the balanced condition,

E’=V_A

According to principle of potentiometer E’ ∝ L’ ………………. (i)

Similarly, if the process is repeated for next cell E” and we have

E” ∝ L”………. (ii)

Now from (i) and (ii), we have,

[latex]\frac{E’}{E”} = \frac{l’}{l”}[/latex]

Or, [latex]E’ = \frac{l’}{l”}[/latex]E”

On knowing the potential E” and measuring the length of the L’ and L”, we can find E’ from that relation.

Shunt

It is a low resistance wire which is connected in parallel with the coil of the galvanometer to convert it into an ammeter.

Uses of Shunt

i.      It is used for converting galvanometer into an ammeter.

ii.     It reduces the total resistance of the ammeter.

iii.    It is used to prevent from the damage of galvanometer.

iv.    It increases the range of ammeter.

Conversion of a galvanometer into an ammeter

Let us consider a galvanometer of resistance G connected in parallel with a shunt of resistance S as shown in the figure. Let [latex]I, I_g[/latex] and [latex](I-I_g)[/latex] are the current flowing through ammeter, galvanometer and shunt respectively when the galvanometer shows the full-scale deflection.

Now, p.d. across shunt = [latex](I – I_g)S[/latex] and

p.d. across galvanometer = [latex]I_gG[/latex]

Since, galvanometer and shunt are in parallel combination,

Then, p.d. across shunt = p.d. across galvanometer

[latex](I-I_g)S = I_gG[/latex]

[latex]\therefore S = \frac{I_gG}{I – I_g}[/latex] …………. (i)

This is the value of shunt to be connected in parallel with a coil of the galvanometer to convert it into an ammeter of range [latex](0 – I)[/latex] Ampere.

Conversion of Galvanometer into Voltmeter

Let us consider a galvanometer of resistance ‘G’ connected in series with a multiplier resistance ‘R’ as shown in figure. Let , [latex]V_g, V_R[/latex] and [latex]V[/latex] are the potential difference across galvanometer, multiplier and voltmeter respectively and [latex]I_g[/latex] be the current flowing through them when the voltmeter shows the full-scale deflection. Here, p.d. across galvanometer = [latex]I_gG[/latex]

p.d. across multiplier = [latex]I_gR[/latex]

Since, galvanometer and multiplier are connected in series combination,

Total p.d. = p.d. across galvanometer + p.d. across multiplier

Or, [latex]V = V_g + V_R[/latex]

Or, [latex]V = I_gG + I_gR = I_g(G+R)[/latex]

Or, [latex]R = \frac{V}{I_g} – G[/latex]

Which is the required value of resistance of multiplier to be connected in series with the galvanometer to convert it into voltmeter of range (0 – V) volt.