Revision notes, practice numericals, a question bank, assertion and reason questions and answers to the NCERT questions. Tap a group to open it; opening one closes the others.
When electric charge flows through a conductor, such as a metal wire, there is an electric current. Current is the amount of charge passing through a cross-section in one unit of time, so it is the rate of flow of charge.
I = Q / tIn metal wires the moving charges are electrons, which are negative. Before electrons were known, current was thought of as a flow of positive charge. By convention, current flows from the positive terminal of the cell to the negative terminal, which is opposite to the direction of electron flow.
A continuous, closed path along which current flows is an electric circuit. A switch (plug key) makes or breaks the conducting link between the cell and the bulb. If the circuit is broken anywhere, current stops and the bulb does not glow.
An ammeter measures current. It is always connected in series in the part of the circuit where current is to be measured.
A circuit diagram uses standard symbols for the cell, battery, open and closed plug key, wire joint, wires crossing without joining, electric bulb, resistor, variable resistance (rheostat), ammeter and voltmeter. A circuit is made of a cell or battery, a plug key, the components and connecting wires.
Charges do not flow in a wire by themselves, just as water does not flow in a perfectly level tube. Water flows when there is a pressure difference between the two ends. In the same way, electrons move only when there is a difference of electric pressure, called the potential difference, along the conductor.
A cell or battery produces the potential difference between its terminals through chemical action, even when no current is drawn. To keep a current going, the cell has to spend the chemical energy stored in it.
The potential difference between two points is the work done to move a unit charge from one point to the other.
V = W / QThe SI unit is the volt (V). One volt is the potential difference when one joule of work is done to move one coulomb of charge between two points.
1 V = 1 J C⁻¹A voltmeter measures potential difference. It is always connected in parallel across the two points being measured.
Georg Simon Ohm found that the potential difference across the ends of a metallic wire is directly proportional to the current through it, provided the temperature stays the same.
V ∝ I V = IRA graph of V against I for such a wire is a straight line through the origin, so V/I is a constant.
The constant R in Ohm's law is the resistance of the conductor. It is the property of a conductor to resist the flow of charge. Electrons are not completely free inside a conductor; the atoms they move among hold them back.
R = V / I I = V / RA component of a given size that has low resistance is a good conductor. A conductor with appreciable resistance is called a resistor. A poor conductor has higher resistance, and an insulator of the same size has even higher resistance.
A component that changes the current without changing the voltage source is a variable resistance. A rheostat is the device commonly used to change the resistance in a circuit.
The resistance of a conductor depends on its length, its area of cross-section and the material it is made of.
Combining the two proportionalities gives the working formula, where ρ (rho) is the electrical resistivity of the material.
R = ρ l / AThe same current flows through every resistor, and the ammeter reads the same wherever it is placed in the circuit.
The total potential difference across the combination equals the sum of the potential differences across the individual resistors.
V = V₁ + V₂ + V₃Applying Ohm's law to each resistor and to the whole circuit gives the equivalent resistance.
Rs = R₁ + R₂ + R₃The combined resistance is the sum of the individual resistances, so it is greater than any single resistance.
Every resistor in a parallel combination has the same potential difference across it, equal to the voltage of the source.
The total current splits among the branches. It equals the sum of the currents through the separate resistors.
I = I₁ + I₂ + I₃The reciprocal of the equivalent resistance is the sum of the reciprocals of the individual resistances.
1/Rp = 1/R₁ + 1/R₂ + 1/R₃The total resistance of a parallel combination is less than the smallest resistance in it.
When series and parallel groups are combined, replace each parallel group by its equivalent resistance first, then add the results as a series circuit.
A source spends energy to maintain a current. Some of it may do useful work, such as turning a fan blade, and the rest becomes heat. In a purely resistive circuit, all of the source energy is dissipated as heat. This is the heating effect of electric current.
The work done in moving charge across a potential difference is the energy supplied, and in a resistor it all appears as heat. Using Ohm's law:
H = VIt H = I²RtHeat produced in a resistor is:
For an appliance on a known voltage, first find the current from I = V/R, then use the heat formula.
In many circuits, heat is a loss because it turns useful electrical energy into heat and can raise the temperature of components. Wires that connect an appliance have low resistance, so they stay cool while the heating element, which has high resistance, glows.
Power is the rate of doing work, which is also the rate at which energy is consumed. For an electric circuit:
P = VI P = I²R P = V²/RThe SI unit is the watt (W). One watt is the power of a device that carries 1 A of current at a potential difference of 1 V.
1 W = 1 V × 1 AThe watt is a small unit, so the kilowatt (1000 W) is used in practice.
Energy is power multiplied by time. The unit of electrical energy is the watt hour (W h), the energy used when 1 W of power runs for 1 hour.
The commercial unit is the kilowatt hour (kW h), commonly called a "unit" on electricity bills.
1 kW h = 3.6 × 10⁶ J| Quantity | Symbol | SI unit | Measured by |
|---|---|---|---|
| Charge | Q | coulomb (C) | – |
| Current | I | ampere (A) | Ammeter, in series |
| Potential difference | V | volt (V) | Voltmeter, in parallel |
| Resistance | R | ohm (Ω) | – |
| Resistivity | ρ | ohm metre (Ω m) | – |
| Power | P | watt (W) | – |
| Energy | E or H | joule (J); commercial unit kW h | – |
| Idea | Formula |
|---|---|
| Current | I = Q / t |
| Potential difference | V = W / Q |
| Ohm's law | V = IR |
| Resistance of a wire | R = ρ l / A |
| Series | Rs = R₁ + R₂ + R₃ |
| Parallel | 1/Rp = 1/R₁ + 1/R₂ + 1/R₃ |
| Joule's law | H = I²Rt |
| Power | P = VI = I²R = V²/R |
| Energy | E = P × t |
Try each problem first, then open the solution to check your working.
A bulb filament draws a current of 0.4 A for 15 minutes. How much charge flows through it?
Given: I = 0.4 A; t = 15 min = 15 × 60 = 900 s
Use: Q = I × t
Q = 0.4 A × 900 s = 360 C
Answer: 360 C
A charge of 900 C passes through a wire in 3 minutes. Find the current.
Given: Q = 900 C; t = 3 min = 180 s
Use: I = Q / t
I = 900 C ÷ 180 s = 5 A
Answer: 5 A
How many electrons make up a charge of 3.2 C? (Charge on one electron = 1.6 × 10⁻¹⁹ C)
Given: Q = 3.2 C; e = 1.6 × 10⁻¹⁹ C
Use: n = Q / e
n = 3.2 ÷ (1.6 × 10⁻¹⁹) = 2 × 10¹⁹
Answer: 2 × 10¹⁹ electrons
A current of 5 mA flows in a circuit for 2 minutes. Find the charge that flows.
Given: I = 5 mA = 5 × 10⁻³ A; t = 2 min = 120 s
Use: Q = I × t
Q = 5 × 10⁻³ × 120 = 0.6 C
Answer: 0.6 C
How much work is done in moving a charge of 5 C across two points that have a potential difference of 9 V?
Given: Q = 5 C; V = 9 V
Use: W = V × Q
W = 9 V × 5 C = 45 J
Answer: 45 J
A 12 V battery drives 4 C of charge round a circuit. How much energy does the battery give to this charge?
Given: V = 12 V; Q = 4 C
Use: W = V × Q
W = 12 × 4 = 48 J
Each coulomb receives 12 J.
Answer: 48 J
150 J of work is done to move 25 C of charge between two points. Find the potential difference.
Given: W = 150 J; Q = 25 C
Use: V = W / Q
V = 150 ÷ 25 = 6 V
Answer: 6 V
A 3 V cell delivers 90 J of energy. How much charge has passed through the cell?
Given: V = 3 V; W = 90 J
Use: Q = W / V
Q = 90 ÷ 3 = 30 C
Answer: 30 C
A heater draws 3 A when connected across 45 V. (a) Find its resistance. (b) What current will it draw at 75 V?
Given: V = 45 V; I = 3 A
Use: R = V / I I = V / R
(a) R = 45 ÷ 3 = 15 Ω
(b) The resistance stays the same, so I = 75 ÷ 15 = 5 A
Answer: R = 15 Ω; I = 5 A
A 9 V battery is connected across an unknown resistor and a current of 3 mA flows. Find the resistance.
Given: V = 9 V; I = 3 mA = 3 × 10⁻³ A
Use: R = V / I
R = 9 ÷ (3 × 10⁻³) = 3000 Ω
Answer: 3000 Ω (3 kΩ)
A bulb filament of 440 Ω and a heater coil of 55 Ω are each connected to a 220 V supply. Find the current in each.
Given: V = 220 V; R₁ = 440 Ω; R₂ = 55 Ω
Use: I = V / R
Bulb: I = 220 ÷ 440 = 0.5 A
Heater: I = 220 ÷ 55 = 4 A
The lower resistance draws the larger current.
Answer: 0.5 A and 4 A
The readings for a resistor are: current 0.2, 0.4, 0.6, 0.8 A and potential difference 1.0, 2.1, 2.9, 4.0 V. Find its resistance.
Given: pairs of V and I as listed
Use: R = V / I
V/I for each pair: 5.0, 5.25, 4.83, 5.0 Ω
The average is about 5 Ω. The V–I graph is a straight line through the origin with slope about 5 Ω.
Answer: about 5 Ω
A wire has a resistance of 6 Ω. Another wire of the same material is twice as long and has half the area of cross-section. Find its resistance.
Given: R₁ = 6 Ω; l₂ = 2l; A₂ = A/2
Use: R = ρ l / A
R₂ = ρ (2l) / (A/2) = 4 ρ l / A = 4 R₁
R₂ = 4 × 6 = 24 Ω
Answer: 24 Ω
A wire 2 m long and 1 mm in diameter has a resistance of 125 Ω. Find the resistivity and identify the material from these values: constantan 49 × 10⁻⁶ Ω m, manganin 44 × 10⁻⁶ Ω m, nichrome 100 × 10⁻⁶ Ω m.
Given: R = 125 Ω; l = 2 m; d = 1 mm = 1 × 10⁻³ m
Use: A = π d² / 4 ρ = R A / l
A = 3.14 × (1 × 10⁻³)² ÷ 4 = 7.85 × 10⁻⁷ m²
ρ = 125 × 7.85 × 10⁻⁷ ÷ 2 = 4.9 × 10⁻⁵ Ω m = 49 × 10⁻⁶ Ω m
This matches constantan.
Answer: ρ ≈ 49 × 10⁻⁶ Ω m; the wire is constantan
Find the resistance of an aluminium wire 100 m long with area of cross-section 2 × 10⁻⁶ m². (ρ of aluminium = 2.63 × 10⁻⁸ Ω m)
Given: l = 100 m; A = 2 × 10⁻⁶ m²; ρ = 2.63 × 10⁻⁸ Ω m
Use: R = ρ l / A
R = 2.63 × 10⁻⁸ × 100 ÷ (2 × 10⁻⁶) = 1.315 Ω
Answer: about 1.3 Ω
Two wires of the same material and length have diameters in the ratio 1 : 3. Find the ratio of their resistances.
Given: d₁ : d₂ = 1 : 3; same l and same material
Use: R ∝ 1/A A ∝ d²
So R ∝ 1/d²
R₁ : R₂ = 1/1² : 1/3² = 1 : 1/9 = 9 : 1
Answer: 9 : 1 (the thinner wire has nine times the resistance)
A copper wire is 20 m long and 0.8 mm in diameter. Find its resistance. (ρ of copper = 1.62 × 10⁻⁸ Ω m)
Given: l = 20 m; d = 0.8 mm = 0.8 × 10⁻³ m; ρ = 1.62 × 10⁻⁸ Ω m
Use: A = π d² / 4 R = ρ l / A
A = 3.14 × (0.8 × 10⁻³)² ÷ 4 = 5.02 × 10⁻⁷ m²
R = 1.62 × 10⁻⁸ × 20 ÷ (5.02 × 10⁻⁷) = 0.645 Ω
Answer: about 0.64 Ω
A 15 Ω bulb and a 5 Ω resistor are connected in series to a 10 V battery. Find (a) the total resistance, (b) the current, (c) the potential difference across each.
Given: R₁ = 15 Ω; R₂ = 5 Ω; V = 10 V
Use: Rs = R₁ + R₂ I = V / Rs V = I R
(a) Rs = 15 + 5 = 20 Ω
(b) I = 10 ÷ 20 = 0.5 A
(c) Bulb: 0.5 × 15 = 7.5 V; resistor: 0.5 × 5 = 2.5 V (together 10 V)
Answer: 20 Ω; 0.5 A; 7.5 V and 2.5 V
A battery of four 1.5 V cells is connected in series with resistors of 4 Ω, 6 Ω and 10 Ω. What will an ammeter in the circuit and a voltmeter across the 10 Ω resistor read?
Given: battery voltage = 4 × 1.5 = 6 V
Use: Rs = R₁ + R₂ + R₃ V = I R
Rs = 4 + 6 + 10 = 20 Ω
I = 6 ÷ 20 = 0.3 A (ammeter)
V = 0.3 × 10 = 3 V (voltmeter)
Answer: 0.3 A; 3 V
Resistors of 0.5 Ω, 1.5 Ω, 2 Ω and 6 Ω are connected in series with a 12 V battery. Find the current and the potential difference across the 6 Ω resistor.
Given: V = 12 V; resistors as listed
Use: Rs = sum of resistances I = V / Rs
Rs = 0.5 + 1.5 + 2 + 6 = 10 Ω
I = 12 ÷ 10 = 1.2 A
V across 6 Ω = 1.2 × 6 = 7.2 V
Answer: 1.2 A; 7.2 V
Resistors of 2 Ω and 3 Ω and an unknown resistor R are in series across a 6 V source, and the current is 0.5 A. Find R.
Given: V = 6 V; I = 0.5 A
Use: Rs = V / I Rs = 2 + 3 + R
Total resistance = 6 ÷ 0.5 = 12 Ω
R = 12 − (2 + 3) = 7 Ω
Answer: 7 Ω
Resistors of 6 Ω, 12 Ω and 4 Ω are connected in parallel to a 12 V battery. Find (a) the current in each, (b) the total current, (c) the equivalent resistance.
Given: V = 12 V across every branch
Use: I = V / R I = I₁ + I₂ + I₃
(a) I₁ = 12/6 = 2 A; I₂ = 12/12 = 1 A; I₃ = 12/4 = 3 A
(b) I = 2 + 1 + 3 = 6 A
(c) Rp = 12 ÷ 6 = 2 Ω. Check: 1/6 + 1/12 + 1/4 = 6/12, so Rp = 2 Ω.
Answer: 2 A, 1 A, 3 A; 6 A; 2 Ω
Find the equivalent resistance of 2 Ω and 10⁵ Ω connected in parallel.
Given: R₁ = 2 Ω; R₂ = 10⁵ Ω
Use: 1/Rp = 1/R₁ + 1/R₂
1/Rp = 1/2 + 1/10⁵ = 0.5 + 0.00001 = 0.50001
Rp ≈ 1.99992 Ω, which is about 2 Ω
The result is always a little below the smaller resistance, and the very large resistor hardly matters.
Answer: about 2 Ω
A lamp of 200 Ω, a toaster of 55 Ω and a kettle of 22 Ω are connected in parallel to a 220 V supply. Find the total current and the equivalent resistance.
Given: V = 220 V
Use: I = V / R Rp = V / I
Lamp: 220 ÷ 200 = 1.1 A; toaster: 220 ÷ 55 = 4 A; kettle: 220 ÷ 22 = 10 A
Total I = 1.1 + 4 + 10 = 15.1 A
Rp = 220 ÷ 15.1 ≈ 14.6 Ω
Answer: 15.1 A; about 14.6 Ω
How many 110 Ω resistors must be connected in parallel to carry 10 A on a 220 V line?
Given: V = 220 V; I = 10 A; R = 110 Ω each
Use: R_total = V / I R_total = R / n
Required total resistance = 220 ÷ 10 = 22 Ω
n = 110 ÷ 22 = 5
Answer: 5 resistors
A wire of resistance 20 Ω is cut into four equal parts and the parts are connected in parallel. Find the new resistance R′ and the ratio R / R′.
Given: R = 20 Ω; four equal parts
Use: 1/R′ = sum of reciprocals
Each part = 20 ÷ 4 = 5 Ω
1/R′ = 4 × (1/5), so R′ = 5/4 = 1.25 Ω
R / R′ = 20 ÷ 1.25 = 16
Answer: R′ = 1.25 Ω; R / R′ = 16
Two resistors of 6 Ω and 3 Ω in parallel are joined in series with a 4 Ω resistor and a 12 V battery. Find the current from the battery and the potential difference across the parallel pair.
Given: V = 12 V
Use: reduce the parallel part first I = V / R
Parallel pair: 1/R = 1/6 + 1/3 = 1/2, so R = 2 Ω
Total resistance = 2 + 4 = 6 Ω
I = 12 ÷ 6 = 2 A
Across the pair: V = 2 × 2 = 4 V
Answer: 2 A; 4 V
Two 8 Ω resistors in parallel are joined in series with a parallel pair of 6 Ω and 12 Ω. A 16 V battery is connected. Find the total resistance and the current.
Given: V = 16 V
Use: reduce each parallel part then add in series
First pair: 8 ÷ 2 = 4 Ω
Second pair: (6 × 12) ÷ (6 + 12) = 4 Ω
Total = 4 + 4 = 8 Ω
I = 16 ÷ 8 = 2 A
Answer: 8 Ω; 2 A
Using three 3 Ω resistors, show how to get (a) 4.5 Ω and (b) 2 Ω.
Given: three resistors of 3 Ω each
Use: series: add parallel: reciprocals
(a) Join two in parallel (1.5 Ω) and put the third in series: 1.5 + 3 = 4.5 Ω
(b) Join two in series (6 Ω) and connect the third in parallel with them: (6 × 3) ÷ (6 + 3) = 2 Ω
Answer: (a) two in parallel plus one in series; (b) two in series with the third in parallel
What are the highest and lowest total resistances that can be made from four coils of 2 Ω, 3 Ω, 6 Ω and 12 Ω?
Given: coils of 2, 3, 6 and 12 Ω
Use: series gives the highest parallel gives the lowest
Highest (all in series): 2 + 3 + 6 + 12 = 23 Ω
Lowest (all in parallel): 1/R = 1/2 + 1/3 + 1/6 + 1/12 = 13/12, so R = 12/13 ≈ 0.92 Ω
Answer: 23 Ω (highest); about 0.92 Ω (lowest)
Find the heat produced in an 8 Ω resistor carrying 3 A for 1 minute.
Given: R = 8 Ω; I = 3 A; t = 60 s
Use: H = I² R t
H = 3² × 8 × 60 = 4320 J
Answer: 4320 J
A heater coil of 55 Ω is connected to a 220 V supply for 10 minutes. Find the current, the power and the heat produced.
Given: R = 55 Ω; V = 220 V; t = 10 min = 600 s
Use: I = V / R P = V I H = P × t
I = 220 ÷ 55 = 4 A
P = 220 × 4 = 880 W
H = 880 × 600 = 5.28 × 10⁵ J
Answer: 4 A; 880 W; 5.28 × 10⁵ J
450 J of heat is produced every second in a 2 Ω resistor. Find the potential difference across it.
Given: H = 450 J; R = 2 Ω; t = 1 s
Use: I = √(H / R t) V = I R
I = √(450 ÷ (2 × 1)) = √225 = 15 A
V = 15 × 2 = 30 V
Answer: 30 V
20000 C of charge passes through a potential difference of 25 V. How much electrical energy is converted to heat in a resistor?
Given: Q = 20000 C; V = 25 V
Use: H = V I t = V Q
H = 25 × 20000 = 5 × 10⁵ J
Answer: 5 × 10⁵ J
A geyser rated 1.5 kW works on a 220 V supply. Which fuse should be used from 1 A, 2 A, 3 A, 5 A and 10 A?
Given: P = 1500 W; V = 220 V
Use: I = P / V
I = 1500 ÷ 220 ≈ 6.8 A
The fuse rating must be just above the working current. A 5 A fuse would blow, so choose 10 A.
Answer: 10 A fuse
Two identical wires are connected first in series and then in parallel across the same supply. Compare the heat produced in the same time.
Given: each wire has resistance R; same V and t
Use: H = V² t / R
Series: R_s = 2R. Parallel: R_p = R/2.
H ∝ 1/R, so H(series) : H(parallel) = 1/(2R) : 2/R = 1 : 4
Answer: 1 : 4 (more heat in parallel)
A bulb draws 0.25 A from a 220 V supply. Find its power.
Given: V = 220 V; I = 0.25 A
Use: P = V I
P = 220 × 0.25 = 55 W
Answer: 55 W
An iron works on 220 V. On maximum heat it uses 1100 W and on minimum heat 440 W. Find the current and resistance in each setting.
Given: V = 220 V; P = 1100 W and 440 W
Use: I = P / V R = V / I
Maximum: I = 1100 ÷ 220 = 5 A; R = 220 ÷ 5 = 44 Ω
Minimum: I = 440 ÷ 220 = 2 A; R = 220 ÷ 2 = 110 Ω
Answer: Maximum: 5 A, 44 Ω; minimum: 2 A, 110 Ω
A bulb rated 220 V, 60 W is used on a 110 V supply. How much power does it consume?
Given: rated V = 220 V; rated P = 60 W; used on 110 V
Use: R = V² / P P′ = V′² / R
R = 220² ÷ 60 = 48400 ÷ 60 ≈ 807 Ω
P′ = 110² ÷ 807 = 12100 ÷ 807 ≈ 15 W
Halving the voltage cuts the power to one-fourth: 60 ÷ 4 = 15 W.
Answer: 15 W
Lamps rated 40 W and 100 W at 220 V are connected in parallel to a 220 V supply. Find the total current drawn.
Given: V = 220 V; P₁ = 40 W; P₂ = 100 W
Use: P = P₁ + P₂ I = P / V
Total power = 40 + 100 = 140 W
I = 140 ÷ 220 ≈ 0.64 A
Answer: about 0.64 A
Bulbs rated 20 W for 220 V are connected in parallel on a 220 V line that allows 5 A at most. How many bulbs can be used?
Given: V = 220 V; I max = 5 A; P = 20 W each
Use: P = V I
Maximum power = 220 × 5 = 1100 W
Number of bulbs = 1100 ÷ 20 = 55
Answer: 55 bulbs
A 150 W TV runs 4 hours a day for 30 days. At Rs 6 per kW h, find the energy used and the cost.
Given: P = 150 W = 0.15 kW; t = 4 h × 30 days
Use: E = P × t
E = 0.15 × 4 × 30 = 18 kW h
Cost = 18 × 6 = Rs 108
Answer: 18 kW h; Rs 108
Which uses more energy: a 100 W bulb for 3 hours, or a 1500 W heater for 15 minutes?
Given: bulb: 100 W, 3 h; heater: 1500 W, 0.25 h
Use: E = P × t
Bulb: 100 × 3 = 300 W h
Heater: 1500 × 0.25 = 375 W h
Answer: the heater (375 W h against 300 W h)
A motor takes 4 A from a 220 V line. Find its power and the energy it uses in 3 hours, in kW h and in joules.
Given: I = 4 A; V = 220 V; t = 3 h
Use: P = V I E = P × t 1 kW h = 3.6 × 10⁶ J
P = 220 × 4 = 880 W = 0.88 kW
E = 0.88 × 3 = 2.64 kW h
E = 2.64 × 3.6 × 10⁶ ≈ 9.5 × 10⁶ J
Answer: 880 W; 2.64 kW h ≈ 9.5 × 10⁶ J
Convert 2.5 kW h into joules.
Given: 1 kW h = 3.6 × 10⁶ J
Use: E (J) = E (kW h) × 3.6 × 10⁶
E = 2.5 × 3.6 × 10⁶ = 9 × 10⁶ J
Answer: 9 × 10⁶ J
A 2 kW geyser works for 30 minutes a day for 30 days. At Rs 5 per unit, find the cost.
Given: P = 2 kW; t = 0.5 h × 30 days
Use: E = P × t
E = 2 × 0.5 × 30 = 30 kW h
Cost = 30 × 5 = Rs 150
Answer: Rs 150
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Answers to every in-text and exercise question of the NCERT chapter. Try each one first, then tap Show answer.
A continuous and closed path of an electric current is called an electric circuit.
The unit of current is the ampere (A). One ampere is the current when one coulomb of charge flows through a cross-section in one second: 1 A = 1 C / 1 s.
Charge of one electron = 1.6 × 10−19 C.
Number of electrons = 1 C ÷ (1.6 × 10−19 C) = 6.25 × 1018.
Answer: 6.25 × 1018 electrons
A cell or a battery (or a generator).
The potential difference between two points is 1 V if 1 joule of work is done to move a charge of 1 coulomb from one point to the other.
Energy = V × Q = 6 V × 1 C = 6 J.
Answer: 6 J
On its length, its area of cross-section and the material it is made of (and also its temperature).
Through the thick wire. Resistance is inversely proportional to the area of cross-section, so a thick wire has less resistance and carries more current.
By Ohm's law I = V/R. With R constant and V halved, the current is also halved.
An alloy has a higher resistivity than the pure metals it is made of, and it does not oxidise (burn) readily even at high temperatures, so the coil lasts long and gives out more heat.
(a) Iron, because its resistivity (10.0 × 10−8 Ω m) is lower than that of mercury (94.0 × 10−8 Ω m).
(b) Silver, which has the lowest resistivity (1.60 × 10−8 Ω m).
Draw a battery of three cells in series (total 6 V), then the plug key, then the three resistors 5 Ω, 8 Ω and 12 Ω one after another in a single closed loop back to the battery.
The ammeter is put in series; the voltmeter is connected in parallel across the 12 Ω resistor.
Total resistance R = 5 + 8 + 12 = 25 Ω; battery voltage V = 3 × 2 = 6 V.
I = V/R = 6/25 = 0.24 A.
Voltmeter reading = I × 12 = 0.24 × 12 = 2.88 V.
Answer: Ammeter 0.24 A; voltmeter 2.88 V
In parallel the equivalent resistance is less than the smallest resistance.
(a) 1/R = 1 + 10−6, so R ≈ 1 Ω (just under 1 Ω).
(b) 1/R = 1 + 10−3 + 10−6, so R ≈ 1 Ω (about 0.999 Ω).
1/R = 1/100 + 1/50 + 1/500 = (5 + 10 + 1)/500 = 16/500, so R = 31.25 Ω.
Current through the iron = V/R = 220/31.25 = 7.04 A (the sum of 2.2 + 4.4 + 0.44 A).
Answer: 31.25 Ω and 7.04 A
(a) Connect 3 Ω and 6 Ω in parallel: 1/R = 1/3 + 1/6 = 1/2, so R = 2 Ω. Put this in series with the 2 Ω resistor: 2 + 2 = 4 Ω.
(b) Connect all three in parallel: 1/R = 1/2 + 1/3 + 1/6 = 1, so R = 1 Ω.
(a) Highest, all in series: 4 + 8 + 12 + 24 = 48 Ω.
(b) Lowest, all in parallel: 1/R = 1/4 + 1/8 + 1/12 + 1/24 = (6 + 3 + 2 + 1)/24 = 1/2, so R = 2 Ω.
Answer: 48 Ω and 2 Ω
The same current flows through both, but H = I²Rt. The element has a very high resistance so it gets very hot and glows; the copper cord has a very low resistance, so very little heat is produced.
H = V × Q = 50 × 96000 = 4.8 × 106 J (the time does not change the result).
Answer: 4.8 × 106 J
H = I²Rt = 5 × 5 × 20 × 30 = 15000 J.
Answer: 1.5 × 104 J
The electric power, P = VI, determines the rate at which electric energy is delivered or consumed.
P = VI = 220 × 5 = 1100 W = 1.1 kW.
Energy = P × t = 1.1 kW × 2 h = 2.2 kW h = 7.92 × 106 J.
Answer: 1100 W; 2.2 kW h
Each part has resistance R/5. Five of them in parallel: R′ = (R/5)/5 = R/25.
So R/R′ = 25.
Answer: (d) 25
Power is VI = I²R = V²/R. IR² is not power.
Answer: (b) IR²
R = V²/P = 220²/100 = 484 Ω.
P′ = V′²/R = 110²/484 = 25 W.
Answer: (d) 25 W
Let each wire have resistance R. Series: 2R; parallel: R/2.
H = V²t/R, so Hseries : Hparallel = (1/2R) : (2/R) = 1 : 4.
Answer: (c) 1:4
A voltmeter is connected in parallel across the two points.
Area A = π r² = 3.14 × (0.25 × 10−3)² = 1.96 × 10−7 m².
R = ρ l / A, so l = R A / ρ = (10 × 1.96 × 10−7) / (1.6 × 10−8) = 122.7 m.
If the diameter doubles, the area becomes 4 times, so the resistance becomes R/4 = 2.5 Ω, a decrease of 7.5 Ω.
Answer: About 122.7 m; resistance falls to one-fourth (2.5 Ω)
Plot I on the x-axis and V on the y-axis. The points lie close to a straight line through the origin, so Ohm's law holds.
Slope = V/I. For example, 13.2/4 = 3.3 Ω, and the best-fit line gives about 3.3 Ω.
Answer: About 3.3 Ω
I = 2.5 mA = 2.5 × 10−3 A.
R = V/I = 12 / (2.5 × 10−3) = 4800 Ω.
Answer: 4.8 kΩ
R = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω.
I = 9 / 13.4 = 0.67 A (the same through every resistor in series).
Answer: 0.67 A
Current in one resistor = 220/176 = 1.25 A.
Number = 5 / 1.25 = 4.
Answer: 4 resistors
(i) Two 6 Ω in parallel give 3 Ω; connect this in series with the third 6 Ω: 3 + 6 = 9 Ω.
(ii) Two 6 Ω in series give 12 Ω; connect this in parallel with the third 6 Ω: (12 × 6)/(12 + 6) = 4 Ω.
Current in one bulb = 10/220 = 1/22 A.
Number = 5 ÷ (1/22) = 110.
Answer: 110 bulbs
One coil: I = 220/24 = 9.17 A.
Series: R = 48 Ω, I = 220/48 = 4.58 A.
Parallel: R = 12 Ω, I = 220/12 = 18.33 A.
Answer: 9.17 A, 4.58 A and 18.33 A
(i) I = 6/(1 + 2) = 2 A. P = I²R = 4 × 2 = 8 W.
(ii) The 2 Ω resistor has the full 4 V across it. P = V²/R = 16/2 = 8 W.
Answer: The power is the same, 8 W, in both circuits
Total power = 100 + 60 = 160 W.
I = P/V = 160/220 = 0.73 A.
Answer: About 0.73 A
TV: 250 W × 1 h = 250 W h = 0.25 kW h.
Toaster: 1200 W × (1/6) h = 200 W h = 0.20 kW h.
Answer: The TV uses more energy
Rate of heat = power = I²R = 5 × 5 × 44 = 1100 J/s.
Answer: 1100 W