(a) 1\(\Omega \)   (b) 1\(\Omega \) 
When resistors are connected in parallel, then the equivalent resistance is less than the least resistance connected in the combination. In both the above cases, the equivalent resistance is less than 1\(\Omega \) but is approximately 1\(\Omega \).

Equivalent resistance of the circuit,
R = R₁ + R₂ + R₃ = 5 + 8 + 12 = 25 \(\Omega\)
[\(\because\) R₁, R₂ and R₃ are connected in series]
In series combination, current flowing through all the resistances is same and equal to the total current flowing through the circuit.
\(\therefore\) Current in the resistors, \(I=\frac{V}{R}=\frac{6}{25}=0.24 \mathrm{~A}\)
\(\therefore\) Ammeter reading = 0.24 A
Potential across 12 \(\Omega\) resistance,
V = IR = 0.24 \(\times\) 12 = 2.88 V
\(\therefore\) Voltmeter reading is 288 V.

Alloys have a higher resistivity than their constituent metals. They do not oxidise or burn at higher temperatures as they have high melting point. Thus, they are used to make coils of electrical roasters and electric irons rather than pure metals.

Let resistance be R. Potential difference V across the two ends becomes V/2. Since, I = V / R
As V \(\rightarrow\) V/2, \(I=\frac{V}{2R}=\frac{1}{2}I\)
In other words, current through the component becomes half of its original value.

Resistance is inversely proportional to the area of cross-section of the wire. Since, thìck wire has a large area of cross-section, its resistance will be less. Thus, current will flow more easily through the thick wire.

The resistance of a conductor depends on following factors:
(i) Length of the conductor.
(ii) Area of cross-section of the conductor.
(iii) Nature of material of the conductor.

Given, charge, q = 1 C, potential, V = 6 V, W=?
As we know, W = q V = 1 \(\times\)6 = 6J
6J is given to each coulomb of charge passing through a 6V battery.

The potential difference between two points is said to be 1 V if 1 J of work is done in moving 1 Coulomb of electric charge from one point to other point.

Electric cell or battery is a device that helps to maintain a potential difference across a conductor.

We know that, charge on one electron = 1.6 \(\times\) 10^-19C
\(\Rightarrow\) 1.6 \(\times\) 10^-19 coulomb charge = 1 electron.
\(\therefore\) 1 coulomb charge=\(\frac{1}{1.6\times 10^{-19}}\simeq 6.25\times 10^{18}\) electrons

The SI unit of electric current is ampere (A).
The current flowing through a conductor is said to be 1A, if a charge of 1coulomb (C) flows through it in 1second (s)
or \(1 A=\frac{1 C}{1 s}\)

We are given, potential difference V = 60 V, current I = 4 A.
According to Ohm's law, \(R=\frac{V}{I}=\frac{60 \mathrm{~V}}{4 \mathrm{~A}}=15 \Omega\).
When the potential difference is increased to 120 V the current is given by current \(=\frac{V}{R}=\frac{120 \mathrm{~V}}{15 \Omega}=8 \mathrm{~A}\)
The current through the heater becomes 8 A .

Total resistance in series \({ R }_{ S }={ R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }=3+4+6\)
\({ R }_{ S }=13\Omega \)
V = 10 V
I = ?
\(\therefore\) I = \(\overset { V }{ \underset { R }{ \_ \_ } } =\overset { 10 }{ \underset { 13 }{ \_ \_ } } \)
I = 0.77A
The current flowing through 6\(\Omega \) resistor is i.e. 0.77 A.
In series connection same current flows through each resistor.

1. To observe that flow of current is different for different components. Take a nichrome wire, a torch bulb, a 10 W bulb and an ammeter (0-5 A range), a plug key and some connecting wires.
2. Set up the circuit by connecting four dry cells of 1.5 V each in series with the ammeter leaving a gap XY in the circuit.

3. Complete the circuit using nichrome wire in gap XY. Note ammeter reading.
4. Now use torch bulb in the circuit.
5. Repeat the same with 10 W bulb.
 

R₁ = 5 Ω, R2 = 10 Ω, and R3 = 30 Ω.
Potential difference across the battery, V = 12 V.
This is also the potential difference across each of the individual resistor; therefore, to calculate the current in the resistors, we use Ohm’s law.
The current I₁ , through R₁ = V/ R₁
I₁ = 12 V/5 Ω = 2.4 A
The current l₂, through R₂ =V / R₂
\(I_2=12 \mathrm{~V} / 10 \Omega=1.2 \mathrm{~A}\)
The current I₃ , through R₃ = V/R₃
\(I_3=12 \mathrm{~V} / 30 \Omega=0.4 \mathrm{~A}\)
The total current in the circuit,
I = I₁ + I₂ + I₃
= (2.4 + 1.2 + 0.4) A = 4 A
The total resistance R_p, is given by [Eq]
\(\frac{1}{R_p}=\frac{1}{5}+\frac{1}{10}+\frac{1}{30}=\frac{1}{3}\)
Thus, R_p = 3 Ω

The resistance of electric lamp, R₁ = 20 Ω,
The resistance of the conductor connected in series, R₂ = 4 Ω.
Then the total resistance in the circuit
R = R₁ + R₂
R_s = 20 Ω + 4 Ω = 24 Ω.
The total potential difference across the two terminals of the battery
V = 6 V.
Now by Ohm’s law, the current through the circuit is given by
I = V/R_s
= 6 V/24 Ω
= 0.25 A
Applying Ohm’s law to the electric lamp and conductor separately,
we get potential difference across the electric lamp,
V₁ = 20 Ω × 0.25 A = 5 V;
and, that across the conductor, V₂ = 4 Ω × 0.25 A = 1 V.
Suppose that we like to replace the series combination of electric lamp and conductor by a single and equivalent resistor. Its resistance must be such that a potential difference of 6 V across the battery terminals will cause a current of 0.25 A in the circuit. The resistance R of this equivalent resistor would be
R = V/I
= 6 V/ 0.25 A
= 24 Ω.
This is the total resistance of the series circuit; it is equal to the sum of the two resistances.