An electric circuit is a continuous and closed conducting path through which electric current flows. It usually includes a source such as a cell or battery, connecting wires and circuit components.
The SI unit of current is the ampere. One ampere is the current flowing when one coulomb of charge passes through a conductor in one second: 1 A = 1 C s⁻¹.
Charge on one electron = 1.6 × 10⁻¹⁹ C. Number of electrons n = 1 C / (1.6 × 10⁻¹⁹ C) = 6.25 × 10¹⁸.
One coulomb of charge contains about 6.25 × 10¹⁸ electrons.
A cell or a battery helps maintain a potential difference across a conductor.
The potential difference between two points is 1 V if 1 joule of work is done to move 1 coulomb of charge from one point to the other: 1 V = 1 J C⁻¹.
Potential difference V = W/Q. Therefore W = VQ = 6 V × 1 C = 6 J.
Each coulomb of charge receives 6 J of energy.
Resistance depends on the length of the conductor, its area of cross-section, the nature of its material and its temperature. It increases with length and decreases with cross-sectional area.
Current flows more easily through a thick wire. For the same material and length, a thicker wire has larger area of cross-section and therefore lower resistance.
By Ohm's law, I = V/R. If resistance remains constant and potential difference becomes half, the current also becomes half.
Alloys have higher resistivity than pure metals and do not oxidise easily at high temperature. Therefore they can become hot and remain stable when used as heating elements.
(a) Iron is a better conductor than mercury because its resistivity is lower.
(b) Silver is the best conductor among the materials listed because it has the lowest resistivity.
The circuit should show three 2 V cells connected in series to make a 6 V battery, a plug key, and the 5 Ω, 8 Ω and 12 Ω resistors connected one after another in the same loop.
Total voltage = 3 × 2 V = 6 V. Total series resistance = 5 + 8 + 12 = 25 Ω. Current I = V/R = 6/25 = 0.24 A, so the ammeter reads 0.24 A. Potential difference across 12 Ω = IR = 0.24 × 12 = 2.88 V.
The ammeter reads 0.24 A and the voltmeter across the 12 Ω resistor reads 2.88 V.
(a) 1/Rp = 1/1 + 1/10⁶ = 1.000001, so Rp ≈ 0.999999 Ω.
(b) 1/Rp = 1/1 + 1/10³ + 1/10⁶ = 1.001001, so Rp ≈ 0.999 Ω. In a parallel combination, the equivalent resistance is less than the smallest resistance.
In both cases the equivalent resistance is slightly less than 1 Ω; practically, it is about 1 Ω.
Currents in parallel branches: lamp = 220/100 = 2.2 A; toaster = 220/50 = 4.4 A; water filter = 220/500 = 0.44 A. Total current = 2.2 + 4.4 + 0.44 = 7.04 A. An iron taking the same current at 220 V must have R = V/I = 220/7.04 = 31.25 Ω.
The electric iron should have resistance 31.25 Ω, and the current through it is 7.04 A.
In parallel, each device gets the full potential difference of the source and can be switched on or off independently. If one device fails, the others continue to work. Parallel connection also allows each device to draw current according to its own resistance.
(a) 3 Ω and 6 Ω in parallel give Rp = (3×6)/(3+6) = 2 Ω. Series with 2 Ω gives 4 Ω.
(b) For all in parallel, 1/Rp = 1/2 + 1/3 + 1/6 = 1, so Rp = 1 Ω.
(a) Connect 3 Ω and 6 Ω in parallel, then connect that combination in series with 2 Ω.
(b) Connect all three resistors in parallel.
Highest resistance is obtained by series connection: 4 + 8 + 12 + 24 = 48 Ω. Lowest resistance is obtained by parallel connection: 1/Rp = 1/4 + 1/8 + 1/12 + 1/24 = 12/24 = 1/2, so Rp = 2 Ω.
The highest resistance is 48 Ω and the lowest resistance is 2 Ω.
The cord has low resistance and is usually made of good conducting copper, so little heat is produced in it. The heating element has high resistance and becomes very hot due to Joule heating, so it glows.
Electrical work/heat H = VQ. Given V = 50 V and Q = 96000 C, H = 50 × 96000 = 4,800,000 J = 4.8 × 10⁶ J.
The heat generated is 4.8 × 10⁶ J.
By Joule's law, H = I²Rt. Given I = 5 A, R = 20 Ω and t = 30 s. H = 5² × 20 × 30 = 25 × 600 = 15000 J.
The heat developed is 15,000 J.
Electric power determines the rate at which electrical energy is delivered or consumed by a current. It is given by P = VI.
Power P = VI = 220 × 5 = 1100 W = 1.1 kW. Energy in 2 h = 1.1 kW × 2 h = 2.2 kWh. In joules, 2.2 × 3.6 × 10⁶ = 7.92 × 10⁶ J.
The power of the motor is 1100 W, or 1.1 kW. The energy consumed in 2 h is 2.2 kWh, equal to 7.92 × 10⁶ J.
- a. 1/25
- b. 1/5
- c. 5
- d. 25
Each part has resistance R/5. Five such equal resistances in parallel have equivalent resistance (R/5)/5 = R/25. Therefore R/R' = 25.
(d) 25
- a. I²R
- b. IR²
- c. VI
- d. V²/R
Power can be written as P = VI = I²R = V²/R. IR² is not a correct expression for power.
(b) IR²
- a. 100 W
- b. 75 W
- c. 50 W
- d. 25 W
For the same bulb, resistance is constant, so P ∝ V². Halving voltage from 220 V to 110 V makes power one-fourth: 100/4 = 25 W.
(d) 25 W
- a. 1:2
- b. 2:1
- c. 1:4
- d. 4:1
Let each wire have resistance R. Series resistance = 2R and parallel resistance = R/2. Heat H = V²t/R_eq, so H_series/H_parallel = [V²t/(2R)]/[V²t/(R/2)] = 1/4.
(c) 1:4
A voltmeter is connected in parallel across the two points between which the potential difference is to be measured.
Diameter = 0.5 mm, so radius r = 0.25 mm = 2.5 × 10⁻⁴ m. Area A = πr² = 3.14 × (2.5 × 10⁻⁴)² = 1.96 × 10⁻⁷ m². R = ρl/A, so l = RA/ρ = 10 × 1.96 × 10⁻⁷ / (1.6 × 10⁻⁸) ≈ 1.23 × 10² m. If diameter doubles, area becomes four times, so resistance becomes one-fourth: 10/4 = 2.5 Ω.
The length of the wire is about 123 m. If the diameter is doubled, the resistance becomes one-fourth, i.e. 2.5 Ω.
Resistance is the slope of the V-I graph. Using the points, V/I values are 3.2, 3.4, 3.35, 3.4 and 3.3 Ω. The average is about 3.3 Ω.
The V-I graph is approximately a straight line through the origin, and the resistance is about 3.3 Ω.
I = 2.5 mA = 2.5 × 10⁻³ A. R = V/I = 12/(2.5 × 10⁻³) = 4800 Ω.
The resistance is 4800 Ω, or 4.8 kΩ.
In series, the same current flows through every resistor. Total resistance = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω. Current I = V/R = 9/13.4 = 0.67 A.
The current through the 12 Ω resistor is about 0.67 A.
Current through one 176 Ω resistor at 220 V is I = V/R = 220/176 = 1.25 A. To carry 5 A, number of resistors = 5/1.25 = 4.
Four 176 Ω resistors are required in parallel.
(i) Two 6 Ω in parallel give 3 Ω; 3 Ω + 6 Ω = 9 Ω.
(ii) Two 6 Ω in series give 12 Ω; 12 Ω in parallel with 6 Ω gives (12×6)/(12+6) = 4 Ω.
(i) To get 9 Ω, connect two 6 Ω resistors in parallel and then connect this combination in series with the third 6 Ω resistor.
(ii) To get 4 Ω, connect two 6 Ω resistors in series and then connect this 12 Ω combination in parallel with the third 6 Ω resistor.
Maximum power available = VI = 220 × 5 = 1100 W. Each lamp uses 10 W, so number of lamps = 1100/10 = 110.
110 lamps can be connected in parallel.
One coil: R = 24 Ω, I = 220/24 = 9.17 A. Series: R = 24 + 24 = 48 Ω, I = 220/48 = 4.58 A. Parallel: equivalent R = 12 Ω, I = 220/12 = 18.33 A.
If one coil is used, current = 9.17 A. If the two coils are in series, current = 4.58 A. If they are in parallel, current = 18.33 A.
(i) Series total resistance = 1 + 2 = 3 Ω. Current = 6/3 = 2 A. Power in 2 Ω resistor = I²R = 2² × 2 = 8 W.
(ii) In parallel, the 2 Ω resistor has 4 V across it. Power = V²/R = 4²/2 = 8 W.
The power used in the 2 Ω resistor is 8 W in both circuits.
Total power in parallel = 100 + 60 = 160 W. Current I = P/V = 160/220 = 0.727 A.
The total current drawn is about 0.73 A.
TV energy = 250 W × 1 h = 250 Wh. Toaster energy = 1200 W × (10/60) h = 200 Wh. Therefore the TV uses more energy.
The 250 W TV set in 1 hour uses more energy.
Rate of heat development is power: P = I²R = 5² × 44 = 25 × 44 = 1100 W.
The rate of heat development is 1100 W, or 1.1 kW.
(a) Tungsten has a very high melting point and can glow at high temperature without melting easily.
(b) Alloys have high resistivity and do not oxidise readily at high temperatures, so they are suitable heating elements.
(c) In series, all appliances would share the supply voltage and the same current; if one appliance failed, the whole circuit would break. Appliances also cannot be operated independently.
(d) Resistance is inversely proportional to area of cross-section: R ∝ 1/A. A thicker wire has lower resistance.
(e) Copper and aluminium have low resistivity, so they conduct electricity well and reduce energy loss in transmission.