(i) T be the current flowing through a resistor of resistance 'R' and 'V' be the potential difference across the resistor. The charge flowing through the circuit for a time interval 't' is 'Q':
(ii) The work done in moving the charge Q across the ends of the resistor with a potential difference of V is VQ. This energy spent by the source gets dissipated in the resistor as heat. Thus, the heat produced in the resistor is:
H = W = VQ
(iii) You know that the relation between the charge and current is Q = I t. Using this, you get
H = V I t ------------------------ (A)
From Ohm's Law, V = I R. Hence, you have
H = I² R t ------------------------ (B)
This is known as Joule's law of heating.

A connection of a set of series resistors connected in a parallel circuit, is a parallel-series circuit. Let R₁ and R₂ be connected in series to give an effective resistance of R_s1. Similarly, let R₃ and R₄ be connected in series to give an effective resistance of R_s2. Then, both of these serial segments are connected in parallel (Figure).

Using the equation R_s = R₁ + R₂
R_s1 =R₁ + R₂'
R_S2 = R₃ + R₄
Finally, using equation \(\left( \frac { 1 }{ { R }_{ p } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } \right) \), the net effective resistance is given by \(\frac { 1 }{ { R }_{ total } } =\frac { 1 }{ { R }_{ S1 } } +\frac { 1 }{ { R }_{ S2 } } \)

(i) Resistor is used to fix the magnitude of the current through the circuit
(ii) Rheostat is used to fix the magnitude of the current through the circuit
(iii) Ammeter is used to measure the current.
(iv) Voltmeter is used to measure the potential difference.

40W. for the same applied voltage, power \(\alpha \frac{1}{resistance}\) If the power is less, higher will be its resistance.

(i) Melting point of tungsten is very.
(ii) Hence it is used in a filament bulb.
(iii) But a fuse wire should be made up of material which has low melting point.

(i) The SI unit of current is ampere (A).
(ii) The current flowing through a conductor is said to be one ampere, when a charge of one coulomb flows across any cross-section of a conductor  in one second. 
\(1\ ampere=\cfrac { 1\ coulomb }{ 1\ second } \)

Heat produced per unit time = V^2/R
Now slope of V-I graph = R (resistance of wire).

Since slope of V-I graph for wire A is greater than the slope of V-I graph for wire B, therefore, resistance of wire A is greater than the resistance of wire B, Hence, more heat will be produced per unit time in wire B than in wire A.

P = I² R is used when current flowing in every component of the circuit is constant. This is the - case of series combination of the devices in the circuit.
P = V²/R is used when potential difference (V) across every component of the circuit is constant. This expression is used in case of parallel combination in the circuit. In series combination, R is greater than the value of R in parallel combination.

Resistance of wire A, \({ R }_{ 1 }=\frac { \rho l }{ A } =\frac { \rho l }{ { \pi r }^{ 2 } } \)
Resistance of wire B, \({ R }_{ 2 }=\frac { \rho l' }{ A' } \)
\(=\frac { \rho \times 2l }{ \pi { (2r) }^{ 2 } } =\frac { \rho l }{ 2\pi { r }^{ 2 } } \)
Total resistance of the series combination, R = R₁ + R₂
or \(R=\frac { \rho l }{ { \pi r }^{ 2 } } +\frac { \rho l }{ { 2\pi r }^{ 2 } } \)
\(\\ =\frac { 3\rho l }{ { 2\pi r }^{ 2 } } \)
\(\therefore \frac { R }{ { R }_{ 1 } } =\frac { 3\rho l }{ { 2\pi r }^{ 2 } } \times \frac { { \pi r }^{ 2 } }{ \rho l } \)
\(=\frac { 3 }{ 2 } \)

(i) A material whose electrical resistivity is low is a good conductor of electricity. Since the electrical resistivity of silver is less than that of the copper, so silver is a better conductor than the copper.
(ii) For making the elements of heating devices, alloy is used instead of a pure metal This is because the resistivity of an alloy is more than that of a metal and alloy does not burn (or oxidise) even at higher temperature. Out of the given substances, nichrome is an alloy, so nichrome is used in electrical heating devices.