Thick copper strips offer minimum resistance and hence avoid the error due to end resistance. Which have not been taken into account in the bridge formula.

\(I=\cfrac { E }{ R+r } \)

When resistors are connected in parallel there is an effective increase in area. Since \(R\alpha \cfrac { 1 }{ A } \) resistance is reduced.

When resistors are connected in series, there is an increase in the effective length. Since resistance varies directly proportional to the length the effective resistance increases \(R\ \alpha\ l\) .

(i) The current that flows due to the emf developed in thermocouple is called thermoelectric current.
(ii) The two dissimilar metals connected to form two junctions is known a thermocouple.

The electrical power P is the rate at which the electrical potential energy is delivered
\(P=\cfrac { dW }{ dt } =\cfrac { d }{ dt } \left( V.dQ \right) =V\cfrac { dQ }{ dt } \)
Since the electric current \(I=\cfrac { dQ }{ dt } \).
So the equation (1) can be rewritten as P = VI
This expression gives the power delivered by the battery to any electrical system, where I is the current passing through it and V is the potential difference across it. The SI unit of electrical power is a watt.

The resistanct is defined as the ratio of potential difference across the given conductor to the current passing through the conductor.
\(R=\cfrac { V }{ I } \)

Circuit Diagram:
Here, 2 cells are in series,
\(\therefore \varepsilon_{\mathrm{tot}} =\varepsilon+\varepsilon=2 \varepsilon  \)
\(\varepsilon_{\mathrm{tot}}  =10 \mathrm{~V}\)
Here 4, 6 and 12 are in parallel
\(\therefore \frac{1}{\mathrm{R}_{\mathrm{p}}}  =\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_1}=\frac{1}{4}+\frac{1}{6}+\frac{1}{12}  \)
\(\mathrm{R}_{\mathrm{p}}  =2 \Omega\)
Now, the circuit becomes,
(i) current drawn from the cell (through the circuit) is,
\(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{s}}}=\frac{10}{8+2}=1 \mathrm{~A}\)
Potential drop across the parallel combination of 3 resistors is \(\mathrm{V}^{\prime}=1 R_P=1 \times 2=2 \mathrm{~V}\)
(ii) Current through 8 resistor is I =1 A
\((\because 8 \Omega, 2 \Omega \text { in series) }\)
Current through \(\mathrm{R}=4 \Omega \ is, \mathrm{I}=\frac{\mathrm{V}^{\prime}}{\mathrm{R}}=\frac{2}{4}=0.5 \mathrm{~A}\)
Current through \(\mathrm{R}=6 \Omega \ is, \mathrm{I}=\frac{\mathrm{V}^1}{\mathrm{R}}=\frac{2}{6}=0.33 \mathrm{~A}\)
Current through \(\mathrm{R}=12 \Omega \ is, \mathrm{I}=\frac{\mathrm{V}^1}{\mathrm{R}}=\frac{2}{12}=0.17 \mathrm{~A}\)

Peltier discovered that, when an electric current is passed through a circuit of a thermocouple heat is evolved at one junction and absorbed at the other junction. This is known as Peltier effect.

Seebeck discovered that in a closed circuit consisting of two dissimilar metals, when the junctions are maintained at different temperature an emf is developed.

Temperature coefficient of resistance is defined as the ratio of increase in resistivity per degree rise in temperature to its resistivity at T_o.
\(\alpha=\frac{\rho_{T}-\rho_{0}}{\rho_{0}\left(T-T_{0}\right)}\)

The current (rate of flow of charge) in the wire is 
\(I=\frac { Q }{ t } =\frac { 120 }{ 60 } =2A\)

(i) When an electric current is passed through a circuit of a thermocouple, heat is evolved at one junction and absorbed at the other junction. This is known as the Peltier effect.
(ii) In the Cu-Fe thermocouple the junctions A and B are maintained at the same temperature.
(iii) Let a current from a battery flow through the thermocouple. At junction A, where the current flows from Cu to Fe, heat is absorbed and junction A becomes cold.
(iv) At junction B, where the current flows from Fe to Cu heat is liberated and it becomes hot.

Peltier effect: Cu - Fe thermocouple
(v) When the direction of current is reversed, junction A gets heated and junction B gets cooled as shown in Figure (b).
Hence Peltier effect is reversible.

(i) Suppose n cells, each of emf \(\xi \) volts and internal resistance r ohms are connected in series with an external resistance R.
(ii) The total emf of the battery = \(n\xi \) The total resistance in the circuit = nr + R By Ohm's law, the current in the circuit is 
\(I=\cfrac { totalemf }{ taoal\ resistance } =\cfrac { n\xi }{ nr+R } \)
\(I=\cfrac { n\xi }{ R } =n{ l }_{ 1 }\)
(iii) where II is the current due to a single cell
\(\left( { I }_{ 1 }=\cfrac { \xi }{ R } \right) \)
Thus, if r is negligible when compared to R the current supplied by the battery is n times that supplied by a single cell.
Case (b) If r >> R,\(I=\cfrac { n\xi }{ nr } =\cfrac { \xi }{ r } \)
(iv) It is the current due to a single cell. That is, current due to the whole battery is the same as that due to a single cell and hence there is no advantage in connecting several cells.
(v) Thus series connection of cells is advantageous only when the effective internal resistance of the cells is negligibly small compared with R.

(i) Due to this internal resistance, the power delivered to the circuit is not equal to power rating mentioned in the battery.
(ii) For a battery of emf \({ \xi }_{ 1 }\) with an internal resistance r, the power delivered to the circuit of resistance R is given by
\(P=I\xi =I(V+Ir)\)
Here V is the voltage drop across the resistance R and it is equal to IR.
Therefore, P = I (IR +Ir)
P = I² R + I² r
(iii) Here Pr is the power delivered to the internal resistance and PR is the power delivered to the electrical device (here it is the resistance R). For a good battery, the internal resistance r is very small, then for P < r < P

(i) The emf of cell \(\xi \) is measured by connecting a high resistance voltmeter across it without connecting the external resistance R.

(ii) Since the voltmeter draws very little current for deflection, the circuit may be considered as open. Hence the voltmeter reading gives the emf of the cell.
(iii) Then, external resistance R is included in the circuit, and current I is established in the circuit. The potential difference across R is equal to the potential difference across the cell (V).
(iv) The potential drop across the resistor R is V + IR
(v) Due to internal resistance r of the cell, the voltmeter reads a value V, which is less than the emf of cell . It is because a certain amount of voltage (Ir) has dropped across the internal resistance r.
Then \(V=\xi -Ir\)
\(Ir=\xi -V\)
(vi) Dividing equation (2) by equation (1) we get
\(\cfrac { Ir }{ IR } =\cfrac { \xi -V }{ V } \)
\(r=\left| \cfrac { \xi -V }{ V } \right| R\)
Since \(\xi \) V and R are known, internal resistance r can be determined.

A battery or cell is called a source of electromotive force (emf).
(i) The emf of a battery or cell is the voltage provided by the battery when no current flows in the external circuit.
(ii) Electromotive force determines the amount of work a battery or cell does to move a certain amount of charge around the circuit. It is denoted by the symbol and to be pronounced as 'xi'. An ideal battery has zero internal resistance and the potential difference (terminal voltage) across the battery equals to its emf.
(iii) A real battery is made of electrodes and electrolyte, there is resistance to the flow of charges within the battery. This resistance is called internal resistance r. For a real battery, the terminal voltage is not equal to the emf of the battery. A freshly prepared cell has low internal resistance and it increases with ageing.

\(\frac { P }{ Q } =\frac { R }{ S } \)
\(S=\frac { Q }{ P } \times R\)
\(S=\frac { 1000 }{ 100 } \times 40S=400\Omega \)

We can denote the current that flows from 9V battery as I₁ and it splits up nto I₂ and (I₁ – I₂) at the junction E according Kirchoff’s current rule (KCR).
Now consider the loop EFCBE and apply KVR, we get
1I₂ + 3I₁ + 2I₁ = 9
5I₁ + I₂ = 9          (1)
Applying KVR to the loop EADFE, we get
3 (I₁ – I₂ ) – 1I₂ = 6
3I₁ – 4I₂ = 6           (2)
Solving equation (1) and (2), we get
I₁ = 1.83 A and I₂ = -0.13 A
It implies that the current in the 1 ohm resistor flows from F to E.

Thus applying Kirchoff’s second law to the closed loop EACE
I₁R₁ + I₂R₂ + I₃R₃ = ξ
and for the closed loop ABCA
I₄R₄ + I₅R₅ - I₂R₂ = 0

Equivalent emf of the combination
ξ_eq = nξ = 4 x 9 = 36 V
ii) Equivalent internal resistance r_eq = nr = 4 x 0.1 = 0.4 Ω
iii) Total current \(I=\frac { n\xi }{ R+nr } \)
\(=\frac { 4\times 9 }{ 10+(4\times 0.1) } \)
\(=\frac { 4\times 9 }{ 10+0.4 } =\frac { 36 }{ 10.4 } \)
I = 3.46 A
iv) Potential difference across external resistance V = IR = 3.46 x 10 = 34.6 V. The remaining 1.4 V is dropped across the internal resistance of cells.
v) Potential difference across each cell \(\frac { V }{ n } =\frac { 34.6 }{ 4 } =8.65V\)

In all the sections, the resistors are connected in parallel.
Section I
\(\frac { 1 }{ { R }_{ { P }_{ 1 } } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } \)
\(\frac { 1 }{ { R }_{ { P }_{ 1 } } } =\frac { 1 }{ 2 } +\frac { 1 }{ 2 } =\frac { 2 }{ 2 } \quad { R }_{ { p }_{ 1 } }=1\Omega \)

Section II
\(\frac { 1 }{ { R }_{ { P }_{ 1 } } } =\frac { 1 }{ 4 } +\frac { 1 }{ 4 } =\frac { 2 }{ 4 } ,\quad \frac { 1 }{ { R }_{ { P }_{ 2 } } } =\frac { 1 }{ 2 } ,{ R }_{ { p }_{ 2 } }=2\Omega \)

Section III
\(\frac { 1 }{ { R }_{ { P }_{ 3 } } } =\frac { 1 }{ 6 } +\frac { 1 }{ 6 } =\frac { 2 }{ 6 } \)
\(\frac { 1 }{ { R }_{ { P }_{ 3 } } } =\frac { 1 }{ 3 } ,{ R }_{ { p }_{ 3 } }=3\Omega \)
Equivalent resistance is given by
R = Rp₁ + Rp₂ + Rp₃
R = 1 Ω + 2 Ω + 3 Ω = 6 Ω
The circuit became,

Equivalent resistance between A and B is

Since the resistances are connected in parallel, therefore, the equivalent resistance in the circuit is
\(\frac { 1 }{ { R }_{ p } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } =\frac { 1 }{ 4 } +\frac { 1 }{ 6 } \)
\(\frac { 1 }{ { R }_{ p } } =\frac { 5 }{ 12 } \Omega \quad or\quad { R }_{ p }=\frac { 12 }{ 5 } \Omega \)
The resistors are connected in parallel, the potential difference (voltage) across them is the same.
\({ I }_{ 1 }=\frac { V }{ { R }_{ 1 } } =\frac { 24V }{ 4\Omega } =6A\)
\({ I }_{ 2 }=\frac { V }{ { R }_{ 2 } } =\frac { 24 }{ 6 } =4A\)
The current I is the sum of the currents in the two branches. Then,
I = I₁ + I₂ = 6 A + 4 A = 10 A

R_s = 3 + 2.5 + P = 5.5 + P
V = 9 V, I = 1.0 A
R_s = \(\frac{V}{I}=\frac{9}{1}= 9 \Omega\)
∴ 9 = 5.5 + P
∴ P = 9 - 5.5 = 3.5 Ω

Diameter d 0.24 cm = 0.24 x 10^-2 m
radius \(r=\cfrac { d }{ 2 } =0.12\times { 1 }^{ -2 }m\)
Current, I = 10A
Current density \(J=\cfrac { 1 }{ A } =\cfrac { 1 }{ { \pi r }^{ 2 } } \)
= \(\cfrac { 10 }{ 3.14\times \left( 0.12\times { 10 }^{ -2 } \right) ^{ 2 } } \)
= 2.2 x 106 Am^-2

(i) Suppose n cells, each of emf volts and internal resistance r ohms are connected in series with an external resistance R as shown in Figure

(ii) The total emf of the battery = nr
The total resistance in the circuit = nr + R
By Ohm's law, the current in the circuit is
\(I=\cfrac { total\ emf }{ total\ resistance } =\cfrac { n\xi }{ nr+5 } \)
Case (a) If r << R, then
\(I=\cfrac { n\xi }{ R } ={ nl }_{ 1 }\)
where II is the current due to a single cell
\(\left( { I }_{ 1 }=\cfrac { \xi }{ R } \right) \)
(iii) Thus, if r is negligible when compared to R the current supplied by the battery is n times that supplied by a single cell
Case (b) If >> R, \(I=\cfrac { n\xi }{ nr } =\cfrac { \xi }{ r } \)
(iv) It is the current due to a single cell. That is, current due to the whole battery is the same as that due to a single cell and hence there is no advantage in connecting several cells.
(v) Thus series connection of cells is advantageous only when the effective internal resistance of the cells is negligibly small compared with R. Cells in parallel
(i) In parallel connection all the positive terminals of the cells are connected to one point and all the negative terminals to a second point. These two points form the positive and negative terminals of the battery.
(ii) Let n cells be connected in parallel between the points A and B and a resistance R is connected between the points A and B as shown in Figure. Let be the emf and r the internal resistance of each cell.

(iii) The equivalent internal resistance of the battery is \(\cfrac { 1 }{ { { r }_{ eq } } } =\cfrac { 1 }{ r } +\cfrac { 1 }{ r } +...\cfrac { 1 }{ r } (netrms)=\cfrac { n }{ r } \)
So \(\cfrac { 1 }{ { r }_{ eq } } =\cfrac { r }{ n } \) and the total resistance in the circuit = \(R+\cfrac { r }{ n } \) The total emf is the potential difference between the points A and B, which is equal to \(\xi \) The current in the circuit is given by
\(I=\cfrac { \xi }{ \frac { r }{ n } +R } \)
\(I=\cfrac { n\xi }{ r+nR } \)
Case (a) If >> R,\(I=\cfrac { n\xi }{ r } ={ nl }_{ 1 }\)
Case (b) If < \(I=\cfrac { \xi }{ R } \)
where II is the current due to a single cell and is equal to \(\cfrac { \xi }{ r } \) when R is negligible. Thus, the current through the external resistance due to the whole battery is n times the current due to a single cell.
  

(i) The emf of cell ε is measured by connecting a high resistance voltmeter across it without connecting the external resistance R.
(ii) Since the voltmeter draws very little current for deflection, the circuit may be considered as open.
(iii) Hence the voltmeter reading gives the emf of the cell.
(iv) Then, external resistance R is included in the circuit and current I is established in the circuit.
(v) The potential difference across R is equal to the potential difference across the cell V.

(vi) The potential drop across the resistor R is,
v = IR .........(1)
(vii) Due to internal resistance r of the cell, the voltmeter reads a value V, which is less than the emf of cell ε. It is because, certain amount of voltage (Ir) has dropped across the internal resistance r.
Then V = \(\varepsilon\) - Ir
Ir = \(\varepsilon\) - v  ......(2)
(viii) Dividing equation (2) by equation (1), we get
\(\frac{I r}{I R}=\frac{\varepsilon-V}{V} \)
\(r=\left|\frac{\varepsilon-V}{V}\right| R \)
(ix) Since \(\varepsilon\), V and R are known, internal resistance r can be determined.

Resistors in series:
(i) When two or more resistors are connected end to end, they are said to be in series. The resistors could be simple resistors or bulbs or heating elements or other devices. Figure (a) shows three resistors R₁, R₂ and R₃ connected in series.
(ii) The amount of charge passing through resistor R₁ must also pass through resistors R₂ and R₃ since the charges cannot accumulate anywhere in the circuit. Due to this reason, the current I passing through all the three resistors are the same.

(iii) According to Ohm's law, if same current pass through different resistors of different values, then the potential difference across each resistor must be different. Let V_1, V₂ and V₃ be the potential difference (voltage)across each of the resistors R₁, R₂ and R₃ respectively, then we can write V₁ = IR₁, V₂= RI₂ and V₃ = IR₃. But the total voltage V is equal to the sum of voltages across each resistor.
V = V₁ + V₂ + V₃ = IR₁+ IR₂ + IR₃ 
V = I (R₁ + R₂ + R₃)
V = IR_S 
where R_s is the equivalent resistance,
R_S = R₁ + R₂ + R₃
(iv) When several resistances are connected in series, the total or equivalent resistance is the sum of the individual resistances.
Note: The value of equivalent resistance in series connection will be greater than each individual resistance.
Resistors in parallel:
(i) Resistors are in parallel when they are connected across the same potential difference as shown in figure (a).
(ii) In this case, the total current I that leave the battery split into three separate components.
Let I₁, I₂ and I₃ be the current through the resistors R₁, R₂, and R₃ respectively. Due to the conservation of charge, total current in the circuit I is equal to sum of the currents through each of the three resistors.
I = I₁ + I₂ + I₃  .....(1)
(iii) Since the voltage across each resistor is the same, applying Ohm's law to each resistor, we have
\({ I }_{ 1 }=\cfrac { V }{ { R }_{ 1 } } ,{ I }_{ 2 }=\cfrac { V }{ { R }_{ 2 } } ,{ I }_{ 3 }=\cfrac { V }{ { R }_{ 3 } } \) 
Substituting these values in equation (1), we get,
\({ I }_{ 1 }=\cfrac { V }{ { R }_{ 1 } } +\cfrac { V }{ { R }_{ 2 } } +\cfrac { V }{ { R }_{ 3 } } =V\left[ \cfrac { 1 }{ { R }_{ 1 } } +\cfrac { 1 }{ { R }_{ 2 } } +\cfrac { 1 }{ { R }_{ 3 } } \right] \)
\(I=\cfrac { V }{ { R }_{ p } } \)
\(\cfrac { 1 }{ R_{ P } } =\cfrac { 1 }{ { R }_{ 1 } } +\cfrac { 1 }{ { R_{ 2 } } } +\cfrac { 1 }{ R_{ 3 } } \)  

Here R_P is the equivalent resistance of the parallel combination of the resistors. Thus, when a number of resistors are connected in parallel, the sum of the reciprocal of the values of resistance of the individual resistor is equal to the reciprocal of the effective resistance of the combination.
Note: The value of equivalent resistance in parallel connection will be lesser than each individual resistance.