Class 12 Physics Electrical circuits Notes

Class 12 Physics Electrical circuits Notes
UNIT 4
CLASS 12 PHYSICS • ELECTRICITY AND MAGNETISM

Electrical Circuits

Chapter 14
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Kirchhoff’s First Law

It states, “In any network algebraic sum of currents at any junction is zero.”
ΣI = 0

Let ‘O’ be the junction point. Here we consider that the current coming towards the junction point is positive and the current goes away from ‘O’ as negative.

Kirchhoff current law at a junction showing incoming and outgoing currents I₁ I₂ I₃ I₄ I₅ I₆ O
Current flowing towards and away from the junction
Applying Kirchhoff’s first law at O:
I₁ + I₂ + I₃ − I₄ − I₅ − I₆ = 0
I₁ + I₂ + I₃ = I₄ + I₅ + I₆

i.e. sum of incoming current = sum of outgoing current.

Hence, the sum of current flowing towards the junction is equal to the sum of current flowing out of the junction. This law is also called Kirchhoff’s current law. This law follows from the principle of conservation of charge.

Kirchhoff’s Second Law

It states, “In a closed loop of an electric circuit, the algebraic sum of the product of currents and resistance is equal to the algebraic sum of emf of the circuit.”
ΣE = ΣIR

Application of Kirchhoff’s Laws in a Complex Electrical Circuit

Complex electrical circuit with three parallel branches and three sources I₁ I₂ I₃ R₁ R₂ R₃ E₁ E₂ E₃
Complex electrical circuit used for Kirchhoff’s laws

Consider that the direction of emf and current flows in anticlockwise direction is taken positive and that in clockwise direction as negative.

Applying Kirchhoff’s second law in closed loop ABCA:

(+E₁) + (−E₂) = (+I₁)R₁ + (−I₂)R₂
E₁ − E₂ = I₁R₁ − I₂R₂   — (i)

Similarly, in closed loop FCDEF:

(+E₂) + (−E₃) = (+I₂)R₂ + (+I₃)R₃
E₂ − E₃ = I₂R₂ + I₃R₃   — (ii)

At junction F, applying Kirchhoff’s first law:

(+I₁) + (+I₂) + (−I₃) = 0
I₁ + I₂ = I₃   — (iii)

Solving these three equations we can calculate the value of current I₁, I₂ and I₃. This law is also called Kirchhoff’s voltage law. This law follows from the principle of conservation of energy.

Application of Kirchhoff’s Law: Wheatstone Bridge

Wheatstone bridge is an electrical circuit which is used for the accurate measurement of resistance of a conductor. It is the arrangement of four resistance in which three of them are known and the unknown resistance is measured in terms of known resistance.

Wheatstone bridge consists of four resistance P, Q, R and X in the form of a quadrilateral and galvanometer is connected between two points B and D. The value of R is adjusted in such a way that there is no current through galvanometer. At this condition it is called balanced condition of Wheatstone bridge.

Balanced Wheatstone bridge with four arms P, Q, R and X G B A C D P Q X R
Wheatstone bridge circuit

Applying Kirchhoff’s voltage law in loop ABDA:

0 = −I₁P − IgG + I₂X
At balanced condition, Ig = 0
I₁P = I₂X   — (i)

Applying Kirchhoff’s voltage law in loop BCDB:

0 = −I₃Q + I₄R + IgG
At balanced condition, Ig = 0
I₃Q = I₄R   — (ii)

Using Kirchhoff’s first law at junction B:

I₁ − I₃ − Ig = 0
At balanced condition, Ig = 0
I₁ = I₃   — (iii)

Using Kirchhoff’s first law at junction D:

I₂ + Ig − I₄ = 0
At balanced condition, Ig = 0
I₂ = I₄   — (iv)

Dividing equation (i) by equation (ii) and using equations (iii) and (iv):

P/Q = X/R

which is the balanced condition of Wheatstone bridge.

Potentiometer

Potentiometer is an electrical device which is used to measure emf of cell, internal resistance of cell and to compare emf of two cells.

Principle

When a constant current is passed through a wire of uniform area of cross-section, potential difference across any portion of wire is directly proportional to length of that portion.
Simplified potentiometer wire AB with segment AC of length l A C B l I E₀
Simplified form of potentiometer

Let I be the current passing through potentiometer wire AB and V be the potential difference across the segment AC of given whose length is l.

V = IR   — (i)
R = ρl/A   — (ii)
Using equation (ii) in equation (i):
V = Iρl/A
Since I, ρ and A all are constant:
V ∝ l

which is working principle of potentiometer.

Use of Potentiometer: To Determine the EMF of a Cell

Potentiometer arrangement for determining the emf of a cell G A C B E E₀
Potentiometer used to determine the emf of a cell

In potentiometer wire AB constant current is maintained by standard cell (driving cell) of emf E₀. If null deflection of galvanometer is obtained by sliding jockey at C, such that AC = l.

By principle of potentiometer:
VAC ∝ l

Since there is no current drawn from cell of unknown emf towards the galvanometer, potential difference across length AC is equal to emf E of cell.

E ∝ l
E = kl
Also, E₀ = k(AB)
E = (E₀/AB)l

This gives the emf of unknown cell.

To Determine Internal Resistance of a Cell

Potentiometer arrangement to determine internal resistance of a cell G Cell E, r R l₁ l₂
Potentiometer used to determine internal resistance of a cell

A cell of emf E whose internal resistance r is to be determined is connected with potentiometer. A resistance box R is connected parallel with cell of unknown internal resistance. A galvanometer is connected to circuit to determine null-deflection point. In potentiometer wire a constant current is maintained by driving cell of emf E₀.

When key is open, suppose null-deflection point is obtained at C such that AC = l₁:

E ∝ l₁
E = kl₁   — (i)

When key is closed, null-deflection point is obtained at D such that AD = l₂:

V ∝ l₂
V = kl₂   — (ii)

Dividing equation (i) by equation (ii):

E/V = l₁/l₂   — (iii)

Also, for cell and external resistance:

E = I(R + r)
V = IR
E/V = (R + r)/R

Using equation (iii):

l₁/l₂ = (R + r)/R
r = [(l₁/l₂) − 1]R

which is required internal resistance of cell.

To Compare EMF of Two Cells

Potentiometer arrangement to compare the emf of two cells G E₁ E₂ l₁ l₂
Potentiometer used to compare emf of cells

Two cells of emf E₁ and E₂ whose emf is to be compared are connected in a circuit with potentiometer. In potentiometer wire AB constant current is maintained by driving cell of emf E₀. Galvanometer is connected in a circuit to determine null-deflection point on a potentiometer wire.

When key K₁ is closed and K₂ is open, null-deflection point is found at C such that AC = l₁:

E₁ ∝ l₁
E₁ = kl₁   — (i)

When key K₁ is open and K₂ is closed, null-deflection point is found at D such that AD = l₂:

E₂ ∝ l₂
E₂ = kl₂   — (ii)

Dividing equation (i) by equation (ii):

E₁/E₂ = l₁/l₂

which is used to compare emf of cells.

Shunt

A very low resistance connected parallel to the galvanometer in order to convert galvanometer into ammeter is called shunt.
Low resistance shunt connected in parallel with a galvanometer G Ig Is S
Shunt connected parallel with galvanometer

Let I = total current of the circuit, Ig = current through galvanometer, Is = current through shunt, G = resistance of galvanometer and S = resistance of shunt.

Equivalent resistance R of the parallel combination:

1/R = 1/G + 1/S
R = GS/(G + S)

Potential difference across A and B:

VAB = IR
VAB = IGS/(G + S)

Now:

Ig = VAB/G = IS/(G + S)
Is = VAB/S = IG/(G + S)

Use of Shunt

  1. It is used to convert galvanometer into ammeter.
  2. It is used to increase current in a circuit.
  3. It is used to increase the range of galvanometer.

Galvanometer

A sensitive device which is used to detect very small current flowing in an electric circuit is called galvanometer.

Conversion of Galvanometer into Ammeter

A device which is used to measure current through the electric circuit is called ammeter. To convert a galvanometer into ammeter, a low resistance (i.e. shunt) is connected parallel to galvanometer.
Galvanometer converted to ammeter using a parallel shunt G Ig I − Ig S
Conversion of galvanometer into ammeter

Let G = resistance of galvanometer, S = resistance of shunt, I = total current flowing through circuit and Ig = current flowing through galvanometer.

Potential difference across shunt = potential difference across galvanometer:

(I − Ig)S = IgG
S = IgG/(I − Ig)

This is required value of shunt to convert galvanometer into ammeter.

Conversion of Galvanometer into Voltmeter

A device which is used to measure p.d. across two ends of resistor is called voltmeter. To convert galvanometer into voltmeter, a high resistance is connected in series with galvanometer.
Galvanometer converted to voltmeter with a high series resistance G R Ig
Conversion of galvanometer into voltmeter

Let G = resistance of galvanometer, R = resistance of high resistor and Ig = current through galvanometer.

Total resistance, RT = G + R
V = Ig(G + R)
V/Ig = G + R
R = V/Ig − G

This is required value of high resistance to convert galvanometer into voltmeter.

Joule’s Law of Heating

According to this law, the amount of heat produced ‘H’ in a conductor due to flow of current is:
  1. directly proportional to square of current: H ∝ I²
  2. directly proportional to resistance of conductor: H ∝ R
  3. directly proportional to time for which current is passed: H ∝ t
Combining all:
H ∝ I²Rt
H = I²Rt joule
or, H = I²Rt/J calorie
J = 4.18 or 4.2 J cal⁻¹ is called Joule’s mechanical equivalent of heat.

Experimental Verification of Joule’s Law of Heating

Experimental arrangement for verification of Joule’s law of heating A Calorimeter with water Resistance R Thermometer Rheostat
Experimental arrangements of Joule’s law of heating

The experimental arrangement of Joule’s law of heating consists of a calorimeter which contains water. A thermometer is used to measure temperature. A resistance wire connected with source is kept inside calorimeter. A rheostat is connected in the circuit to change the current.

(i) To Verify H ∝ I²

Change the value of current I₁, I₂, I₃ … and calculate the corresponding heat developed H₁, H₂, H₃ … for fixed value of resistance and time.

H₁/I₁² = H₂/I₂² = H₃/I₃² = … = constant
H ∝ I²   — (i)

(ii) To Verify H ∝ R

Change the value of resistance R₁, R₂, R₃ … and calculate the corresponding heat developed H₁, H₂, H₃ … for fixed value of current and time.

H₁/R₁ = H₂/R₂ = H₃/R₃ = … = constant
H ∝ R   — (ii)

(iii) To Verify H ∝ t

Change the time t₁, t₂, t₃ … and calculate the corresponding heat developed H₁, H₂, H₃ … for fixed value of resistance and current.

H₁/t₁ = H₂/t₂ = H₃/t₃ = … = constant
H ∝ t   — (iii)

Combining equations (i), (ii) and (iii):

H ∝ I²Rt

This expression verifies Joule’s law of heating.

Derivation of Joule’s Law (Heat Developed in a Wire)

Let R be the resistance and I be the current passing in the circuit. We know that p.d. is amount of work done during moving unit charge from one point to another point in circuit.

If W be amount of work done in moving q charge:
V = W/q
W = Vq   — (i)

From Ohm’s law and current definition:

V = IR   — (ii)
q = It   — (iii)

Using equations (ii) and (iii) in equation (i):

W = IR(It)
W = I²Rt

This amount of work done appears in the form of heat across the resistance R. Therefore H = W = I²Rt. It is also called electrical energy consumed.

Power

It is defined as the rate at which electrical energy consumed by resistor.
P = W/t
P = I²R
Either, P = I²R = IR·I = VI
P = VI
or, P = I²R = (V/R)²R
P = V²/R

Meter Bridge

It is an electrical device which is used to measure unknown resistance. Its operation is based upon the principle of Wheatstone bridge. The length of wire in meter bridge is 1 meter, so the device is named meter bridge.
Meter bridge with one meter wire AC, resistance gaps X and R, galvanometer and balance point B G X R l 100 − l A B C
Meter bridge

It consists a 1 m long wire AC having uniform cross-section which is stretched on wooden board. Thick copper strips having negligible resistance are fitted on the wooden board leaving the gaps where a resistance R and unknown resistance X are kept. This arrangement is connected with galvanometer and source (i.e. cell of emf E). The jockey is connected to one end of galvanometer in order to find null deflection point by sliding it.

Let B be the balance point such that AB = l cm and BC = (100 − l) cm. Also let P and Q be the resistance of wire for length AB and BC respectively.

P ∝ l
P = kl   — (i)
Q ∝ (100 − l)
Q = k(100 − l)   — (ii)

Using the principle of Wheatstone bridge:

P/Q = X/R
kl/[k(100 − l)] = X/R
X = [l/(100 − l)]R

Here quantities R and l are known. Hence we can calculate the unknown resistance.

Conductor

Super Conductor

The materials which have zero electrical resistance are known as super conductor.

Perfect Conductor

The electrical conductor with no resistivity is known as perfect conductor.

Ohmmeter

Ohmmeter is an arrangement which is used for measuring resistance.

Solved Numericals

Q.1 — Kirchhoff’s Rules: Find Current, Resistance and EMF

Using Kirchhoff’s rules in the circuit, find: (i) the current in resistor R, (ii) the resistance R, (iii) the unknown emf E, and (iv) if the circuit is broken at P, what is the current in resistor R?

(i) Let the current in resistor R be I.

At junction C, by Kirchhoff’s first law:
6 − 4 − I = 0
I = 2 A

(iii) At loop DCEFD, from Kirchhoff’s second law:

−E + (4×6) + (6×3) = 0
E = 42 V

(ii) At loop ABCDA, from Kirchhoff’s second law:

−28 + 2R − (4×6) + E = 0
2R − 24 + 42 − 28 = 0
R = 5 Ω

(iv) If the circuit is broken at P:

I = Enet/Rnet
= 28/(5 + 3)
I = 3.5 A

Q.2 — Potential Gradient of a Potentiometer Wire

A potentiometer is 10 m long. It has a resistance of 20 Ω. It is connected in series with a battery of 3 V and a resistance of 10 Ω. What is the potential gradient along the wire?

I = Enet/Rnet
= 3/(20 + 10)
I = 0.1 A
VAB = IRAB = 0.1×20
VAB = 2 V
VAB = klAB
2 = k(10)
k = 0.2 V/m

Q.3 — Find EMF When Current Through 7 Ω is 1.80 A

What must be the emf E in the circuit so that the current flowing through the 7 Ω resistor is 1.80 A? Each emf source has negligible internal resistance.

Applying Kirchhoff’s second law in closed loop (I):

−3I₁ + 24 − E − 2I₂ = 0
24 − E = 3I₁ + 2I₂   — (i)

Applying Kirchhoff’s second law in loop (II):

2I₂ + E − 7(1.8) = 0
E = 12.6 − 2I₂   — (ii)

Using (ii) in (i):

24 − 12.6 + 2I₂ = 3I₁ + 2I₂
11.4 = 3I₁
I₁ = 3.8 A
I₁ − I₂ = 1.8
3.8 − I₂ = 1.8
I₂ = 2 A
E = 12.6 − 2(2)
E = 8.6 V

Q.4 — Simple Potentiometer Circuit

A simple potentiometer circuit is set up using a uniform wire AB, 1.0 m long, which has a resistance of 2 Ω. The resistance of the 4 V battery is negligible. If the variable resistor R were given a value of 2.4 Ω, what could be the length AC for zero galvanometer deflection?

lAB = 1 m, RAB = 2 Ω
R = 2.4 Ω, E = 4 V
I = Enet/Rnet = 4/(2.4 + 2) = 4/4.4 A
RAC = 2lAC
VAC = IRAC
1.5 = (4/4.4)(2lAC)
lAC = 0.825 m

Q.5 — Two Batteries Joined in Parallel

A battery of 6 V and internal resistance 0.5 Ω is joined in parallel with another of 10 V and internal resistance 1 Ω. The combination sends a current through an external resistance of 12 Ω. Find the current through each battery.

In loop ABCA:

−6 + 0.5I₁ − I₂ + 10 = 0
0.5I₁ − I₂ = −4
480 = 120I₂ − 60I₁   — (i)

In loop FCDEF:

−10 + I₂ + (I₁ + I₂)12 = 0
13I₂ + 12I₁ = 10
50 = 65I₂ + 60I₁   — (ii)

Adding (i) and (ii):

530 = 185I₂
I₂ = 2.86 A

From (i):

0.5I₁ − 2.86 = −4
I₁ = −2.28 A

Where negative sign shows that I₁ has clockwise direction. Hence current in batteries 6 V and 10 V are 2.28 A and 2.86 A respectively.

Q.6 — Potentiometer Length, Unknown P.D. and Maximum P.D.

The total length of the wire of a potentiometer is 10 m. A potential gradient of 0.0015 V/cm is obtained when a steady current is passed through this wire. Calculate: (i) distance of null point on connecting a standard cell of 1.081 V, (ii) unknown p.d. if the null point is obtained at a distance of 940 cm, and (iii) maximum p.d. which can be measured by this instrument.

l = 10 m
k = 0.0015 V/cm = 0.15 V/m

(i)

E = kl
l = 1.081/0.15
l = 6.78 m   [as written in source]

(ii)

l = 940 cm = 9.4 m
E = kl = 0.15×9.4
E = 1.41 V

(iii)

E = kl = 0.15×10
E = 1.5 V
The handwritten source writes 6.78 m for part (i). This value is preserved exactly rather than silently corrected.

Q.7 — Kirchhoff’s Laws of Current and Voltage

Using Kirchhoff’s laws of current and voltage, find the current in the 2 Ω resistor in the given circuit.

In loop (I):

35 − 3I₁ − 2(I₁ + I₂) = 0
35 = 5I₁ + 2I₂
105 = 15I₁ + 6I₂   — (i)

In loop (II):

4I₂ − 40 + 2(I₁ + I₂) = 0
40 = 2I₁ + 6I₂   — (ii)

Subtracting equation (ii) from equation (i):

65 = 13I₁
I₁ = 5 A

From equation (ii):

40 = 10 + 6I₂
I₂ = 5 A
Current through 2 Ω = I₁ + I₂ = 5 + 5
= 10 A

Q.8 — Driver Cell, Series Resistance and Thermocouple EMF

The driver cell of a potentiometer has an emf of 2 V and negligible internal resistance. The potentiometer wire has a resistance of 3 Ω. Calculate the resistance needed in series with the wire if a p.d. 5 mV is required across the whole wire. The wire is 100 cm long and a balanced length of 60 cm is obtained for a thermocouple of emf E. What is the value of E?

E = 2 V, RAB = 3 Ω, VAB = 5 mV = 5×10−3 V
I = E/(R + RAB) = 2/(R + 3)
VAB = IRAB
5×10−3 = [2/(R + 3)]×3
R + 3 = 6/(5×10−3) = 1200
R = 1197 Ω

For the thermocouple:

100 cm → 5 mV
60 cm → (5×60)/100 mV
E = 3 mV

Q.9 — EMF from Balancing Lengths

The emf of a battery A is balanced by a length 75 cm on a potentiometer wire. The emf of a standard cell 1.02 volts is balanced by a length of 50.0 cm. What is the emf of A?

l₁ = 75 cm, E₂ = 1.02 V, l₂ = 50 cm
E₁/E₂ = l₁/l₂
E₁/1.02 = 75/50 = 3/2
E₁ = 1.53 V

Q.10 — Resistance Needed in Series with Potentiometer Wire

The driving cell of a potentiometer has an emf of 2 V and negligible internal resistance. The potentiometer wire has a resistance of 3 Ω. Calculate the resistance needed in series with the wire if a p.d. of 1.5 mV is required across the whole wire.

I = Enet/Rnet = 2/(R + 3)
VAB = IRAB
1.5×10−3 = [2/(R + 3)]×3
1.5×10−3 = 6/(R + 3)
R + 3 = 4000
R = 3997 Ω

Q.11 — Moving Coil Meter Converted to Voltmeter and Ammeter

A moving coiled meter has a resistance of 25 Ω and indicates full scale deflection when a current of 4 mA passed through it. How could this meter be converted: (i) to a voltmeter with 0–3 V range, and (ii) to an ammeter with 0–1 A range?

(i) Voltmeter:

Ig = 4 mA = 4×10−3 A
G = 25 Ω, V = 3 V
V = Ig(G + R)
3 = 4×10−3(25 + R)
3000/4 = 25 + R
R = 725 Ω

(ii) Ammeter:

IgG = IsS
S = IgG/Is
S = (4×10−3×25)/(1 − 4×10−3)
= 0.1/0.996
S ≈ 0.1 Ω   [as written in source]
The question line in the source states a 0–1 A ammeter range, while the preceding conversation has sometimes been read as 0–0.1 A. The handwritten calculation itself uses 1 A and gives approximately 0.1 Ω; this source calculation is preserved.

Q.12 — Shunt Required for a 2 A Ammeter

A voltmeter coil has resistance 50 Ω and a resistor of 1.15 kΩ is connected in series. It can read p.d. up to 12 V. If the same coil is used to construct an ammeter which can measure current up to 2 A, what should be the resistance of shunt used?

From voltmeter:

V = Ig(G + R)
12 = Ig(50 + 1150)
Ig = 0.01 A

For ammeter:

VG = VS
IgG = (2 − Ig)S
S = (0.01×50)/(2 − 0.01)
S ≈ 0.25 Ω
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