A continuous and closed path along which an electric current flow is called an electric circuit.

If one coulomb of charge flows through any section of a conductor in one second, then current through it is said to be one ampere.

1 A = 1 Cs^-1

Charge on one electron, e = 1.6 x 10^-19 C

Total charge, Q = 1 C

Number of electrons, n = \(\frac{Q}{e}\) = \(\frac{1}{1.6×10^{-19}}\) = 6.25 x 10¹⁸ electrons.

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

The potential difference between two points is 1 volt if one joule of work is done in moving a positive charge of one coulomb from one point to the other point.

Energy given by battery = Charge x Potential difference = 1 C x 6 V = 6 J.

The resistance of a conductor depends

• on its length,
• on its area of cross-section,
• its temperature
• on the nature of its material.

• The current will flow more easily through a thick wire than a thin wire of the same material.
• Larger the area of cross-section of a conductor, more is the ease with which the electrons can move through the conductor. Hence smaller is the resistance of the conductor.

When potential difference is halved, the current through the component also decreases to half of its initial value. This is in accordance with Ohm's law i.e., V ∝ I.

The coils of electric toasters and electric irons are made of alloys instead of pure metal due to the following reasons:

• Alloys have higher resistivity than that of their constituent metals.
• Alloys do not oxidise (or burn) readily at high temperatures.

(a) Resistivity of iron = 10.0 x 10^-8 Ω m, Resistivity of mercury = 94.0 x 10^-8 Ω m

Thus, iron is a better conductor because it has lower resistivity than mercury.

(b) As silver has the lowest resistivity (= 1.60 x 10^-8 Ω m), so silver is the best conductor.

The required circuit diagram is given below:

The required circuit diagram is given below:

Equivalent resistance of the circuit,

R = R₁ + R₂ + R₃ = 5 + 8 + 12 = 25 Ω    …[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.

∴ Current in each resistors, I = V/R = 6/25 = 0.24 A

∴ Ammeter reading = 0.24 A

Potential across 12 Ω resistance,

V= IR = 0.24 ×12 = 2.88V

∴ Voltmeter reading is 2.88 V.

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

Let the resistance of lamp, R₁ = 100 Ω

Resistance of toaster, R₂ = 50 Ω

Resistance of filter, R₃ = 500 Ω

Net resistance, \(\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\) = \(\frac{1}{100}+\frac{1}{50}+\frac{1}{500}\) = \(\frac{16}{500}\)

∴ R = \(\frac{500}{16}\) = 31.25 Ω

So, resistance of iron to take same current as much current drawn by all the appliances should be 31.25 Ω.

Current through circuit, I = \(\frac{V}{R}\) = \(\frac{220}{31.25}\) = 7.04 A

Thus, current through iron is 7.04 A.

Following are the advantages of connecting electrical devices in parallel with the battery

• Parallel circuit divides the current among the electrical devices, so that they can have necessary amount of current to operate properly.
• If one of the devices in a parallel combination fuses or fails, then the other devices keep working without being affected.

(i) If 3 Ω and 6 Ω are connected in parallel, thus equivalent resistance of parallel combination = \(\frac{1}{1/3+1/6}\) = \(\frac{6}{3}\) = 2 Ω

If this combination is connected in series with 2 Ω resistance, then total equivalent resistance R = 2 Ω + 2 Ω = 4 Ω.

The resistor connections are as shown below

(ii) Since, equivalent resistance is less than the least value of resistance (i.e. 2 Ω), it means that all three resistors are connected in parallel.

R = \(\frac{1}{1/2+1/3+1/6}\) = 1 Ω

(i) Resistance is maximum when resistors are connected in series.

R_max = 4 + 8 + 12 + 24 = 48 Ω

(ii) Resistance is minimum when resistors are connected in parallel.

R_min = \(1/[\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{24}]\) = \(1/[\frac{12}{24}]\) = \(\frac{24}{12}\) = 2 Ω

The cord of an electric heater has lesser resistance than its heating element, so more heat is produced in the heating element than the cord and it glows.

Here, Q = 96,000 C, t = 1 hour = 3600 s, V = 50 V

Heat generated, H = VQ = 50 V x 96,000 C = 48,00,000 J = 4800 kJ.

Here, R = 20 Ω, I = 5 A, t = 30 s

Heat developed is H = I²Rt = 5² x 20 x 30 = 25 x 600 = 15,000 J.

Electric power determines the rate at which energy is delivered by a current.

Given : I = 5 A, V = 220 V, t = 2h

∴ Power of motor, P = VI = 220 × 5 = 1100 W = 1.1 kW

∴ Energy consumed = Pt =1.1 x 2 = 2.2 kW-h

Thus, the power of the motor is 1.1 kW and energy consumed is 2.2 kW-h.

(d) 25

Resistance of complete wire is R. If it is cut into 5 equal parts, then resistance of each part will be R/5

Five parts of resistance R/5 each are connected in parallel as shown in the figure

Equivalent resistance,

R’ = \(\frac{1}{\frac{R}{5}+\frac{R}{5}+\frac{R}{5}+\frac{R}{5}+\frac{R}{5}}\) = \(\frac{R}{25}\)

∴ Ratio R/R’ = \(\frac{R}{R/25}\) = 25

(b) IR²

Electric power = VI = IR x I = I²R     ..[∵V = IR]

= \((\frac{V}{R})^2\)R = \(\frac{V^2}{R}\)

So, IR² does not represent electric power.

(d)

Given : V = 220 V, P = 100W

∴ Resistance of bulb, R = V²/P = \(\frac{100^2}{220}\) = 484 Ω

Now, when V = 110 V, then power consumed,

P = V²/R = \(\frac{110^2}{484}\) = 25 W

(c) 1:4

Let R be the resistance of each wire. The resistance of both the wires will be same because they are of same material and have same length and same cross- sectional area.

Equivalent resistance in series = R + R = 2R

Heat produced, H = \(\frac{V^2t}{R}\)

If wires are connected in series, then H_s = \(\frac{V^2t}{2R}\)

Equivalent resistance in parallel = R/2

Heat produced, H_p = \(\frac{2V^2t}{R}\)

Ratio of heat produced, \(\frac{H_s}{H_p} = \frac{v^2t/2R}{2V^2t/R}\) = \(\frac{1}{4}\) = 1: 4

A voltmeter is connected in parallel to measure the potential difference between two points in a circuit.

Given : radius of wire, r = diameter/2 = 0.5/2 = 0.25 mm = 0.25 × 10^-3 m,

ρ = 1.6 x 10^-8 Ω-m and R = 10 Ω

(i) We know that, resistance, R = \(\frac{ρL}{A}\) = \(\frac{ρL}{πr^2}\)       …[∵ A = π r²]

∴ L = \(\frac{Rπr^2}{ρ}\) = \(\frac{10×3.14×(0.25×10^{-3})^2}{1.6×10^{-8}}\)  = 122.66 m

(ii) Resistance, R ∝ 1/d²

If diameter is doubled, then resistance becomes one-fourth of its original value.

The graph between V and I for the given data is shown below:

Resistance of the resistor, R = \(\frac{V_2-V_1}{I_2-I_1}\) = \(\frac{10.2-6.7}{3-2}\) = 3.5 Ω

Given : V = 12 V, I = 2.5 mA = 2.5 x 10^-3 A

Resistance, R = \(\frac{V}{I}=\frac{12}{2.5×10^{-3}}\) = 4800 Ω

Total resistance, R = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω

Potential difference, V = 9 V

Current through the series circuit is I = V/R = 9/13.4 = 0.67 A

Current through 12 Ω  resistor = 0.67 A.

Suppose n resistances of 176 Ω are connected in parallel. Then,

\(\frac{1}{R}=\frac{1}{176}+\frac{1}{176}+\)..... n factor = \(\frac{n}{176}\)

Or R = \(\frac{176}{n}\)

By Ohm's law,  R = \(\frac{V}{I}\)

\(\frac{176}{n}\) = \(\frac{220}{5}\)

n = \(\frac{176×5}{220}\) = 4 resistors.

Here R₁ = R₂ = R₃ = 6 n

(a) When we connect R₁ in series with the parallel combination of R₂ and R₃, as shown in Fig. (a).

The equivalent resistance is

R = R₁ + \(\frac{R_2×R_3}{R_2+R_3}\) = 6 + \(\frac{6×6}{6+6}\) = 6 + 3 = 9 Ω

(b) When we connect a series combination of R₁ and R₂ in parallel with R₃, as shown

in Fig. (b), the equivalent resistance is

R = \(\frac{(R_1+R_2)×R_3}{R_1+R_2+R_3}\) = \(\frac{72}{18}\) = 4 Ω

Resistance of each bulb, R = \(\frac{V^2}{P}\) = \(\frac{220^2}{10}\) = 4840 Ω

Suppose n bulbs are needed to be connected in parallel with each other. Then their equivalent resistance R_p is given by

\(\frac{1}{R_p}=\frac{1}{4840}+\frac{1}{4840}+\)..... n factor = \(\frac{n}{4840}\)

Or R_p = \(\frac{4840}{n}\) Ω

Given : V = 220 V, I = 5 A

By Ohm's law R_p = \(\frac{V}{I}\)

Or  \(\frac{4840}{n}\) = \(\frac{220}{5}\)

n = \(\frac{4840×5}{220}\) = 110 lamps

(i) When the two coils A and B are used separately,

R = 24 Ω, V = 220 V

Current, I = \(\frac{V}{R}\) = \(\frac{220}{24}\) = 9.167 A.

(ii) When the two coils are connected in series,

R = 24 + 24 = 48 Ω, V = 220 V

Current, I = \(\frac{V}{R}\) = \(\frac{220}{48}\) = 4.58 A.

(iii) When the two coils are connected in parallel,

R =  = 12 Ω, V = 220 V

Current, I = \(\frac{V}{R}\) = \(\frac{220}{12}\) = 18.33 A,

(i) The circuit shown in the fig has resistance connected in series combination:

Current in the circuit, I = \(\frac{V}{R_1+R_2}\) = \(\frac{6}{1+2}\) = 2 A,

Power = I²R = 2² x 2 = 2 x 2 x 2 = 8 W

(ii) The circuit is shown in below fig has resistance connected in parallel combination.

In parallel combination, potential across each resistor is same and equal to the potential applied to the circuit.

Potential across 2 Ω resistor, V = 4 V

Power = V²/R = 4²/2 = 8 W

Power used in both the cases is same.

Total power consumed in the circuit = 100 + 60 = 160 W

Voltage, V = 220 V,

Power = VI

∴ Current, I = Power/V = 160/220 = 0.727 A.

Energy used by 250 W TV set in 1 hour = 250 W x 1 h = 250 W-h

Energy used by 1200 W toaster in 10 minutes

= 1200 W x 10 min = 1200 W x \(\frac{10}{60}\)h = 200 W-h

Thus, the TV set uses more energy than the toaster.

Given: resistance, R = 8 Ω, current, I = 15 A

Time, t = 2 h = 7200 s

∴ Heat developed, H = I²Rt = (15 × 15 × 8 × 7200) J

∴ Rate of heat developed, P = \(\frac{H}{t}\) = \(\frac{15 × 15 × 8 × 7200}{7200}\) = 1800 W or 1800 J/s

Thus, the rate at which heat is developed in the heater is 1800 joule per second.

(i) Tungsten has a high melting point (3380 ℃). It does not melt at high temperature. It retains as much of heat generated, so that it becomes very hot and emits light. That is the reason why tungsten is used as filament of electric lamps.

(ii) Conductors of electric heating devices are made of alloys because alloys do not oxidise (burn) readily at high temperature unlike metals. Also, alloys have a greater resistivity (generally) as compared to their constituent pure metals.

(iii) There are 2 reasons for not using series connections for domestic circuits.

• Devices of different current ratings cannot be connected as the current is constant in the series circuit.
• If one device fails, the circuit is broken and all devices stop working.

(iv) Resistance is inversely proportional to the area of cross-section of the wire. Thus, if the wire is thick (large area of cross-section), then resistance is less. If the wire is thin (less area of cross-section), then resistance is large.

(v) Copper and aluminium wires are used for transmission of electricity because they have low resistivity. So, they conduct the electric current without heavy heat losses. Also, they are quite cost effective, as compared to silver.