CBSE Class 10 Science: Chapter 12 Electricity LA Questions (with Answers)
Welcome to Part 4 of our 8-part series on Chapter 12, Electricity. This post focuses on Long Answer (LA) Questions, typically worth 4 or 5 marks. These questions involve detailed derivations, complex circuit calculations, and in-depth conceptual explanations.
Part 4: Long Answer (LA) Questions
(a) Define electrical resistivity. What is its SI unit?
(b) State the relationship between commercial unit and SI unit of electrical energy.
(c) Two resistors of resistance 10 Ω and 20 Ω are connected in series and then in parallel. In which case will the equivalent resistance be more? Calculate the equivalent resistance in each case.
SI Unit: The SI unit of resistivity is the Ohm-meter (Ω m).
(b) Relationship:
- The commercial unit of electrical energy is the kilowatt-hour (kWh).
- The SI unit of electrical energy is the Joule (J).
- 1 kWh = 3.6 × 10⁶ J
Given: R1 = 10 Ω, R2 = 20 Ω
Case 1: Series Connection
The equivalent resistance (Rs) is:
Rs = R1 + R2
Rs = 10 Ω + 20 Ω = 30 Ω
Case 2: Parallel Connection
The equivalent resistance (Rp) is:
1/Rp = 1/R1 + 1/R2
1/Rp = 1/10 + 1/20
1/Rp = (2 + 1) / 20 = 3/20
Rp = 20 / 3 Ω ≈ 6.67 Ω
Conclusion: The equivalent resistance is more in the series connection (30 Ω).
(a) Derive the expression for Joule's Law of Heating (H = I²Rt).
(b) An electric heater of resistance 8 Ω draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.
- From the definition of potential difference (V), the work done (W) to move a charge (Q) is: W = V × Q.
- The rate of work done is Power (P). So, P = W / t = (V × Q) / t.
- We know that electric current (I) is the rate of flow of charge: I = Q / t.
- Substituting (I) into the power equation: P = V × (Q/t) = P = V × I.
- The energy supplied by the source for time 't' is E = P × t. This energy is dissipated as heat (H) in the resistor. So, H = P × t = V × I × t.
- From Ohm's Law, we know V = I × R.
- Substituting V = IR into the heat equation: H = (I × R) × I × t.
- This gives the final expression: H = I²Rt. This is Joule's Law of Heating.
The question asks for the "rate at which heat is developed," which is another term for Electric Power (P).
Resistance (R) = 8 Ω
Current (I) = 15 A
Time (t) = 2 hours (This information is not needed to find the *rate* of heat)
The formula for power (rate of heat) is: P = I²R
P = (15 A)² × 8 Ω
P = 225 × 8
P = 1800 W (or 1800 J/s)
The rate at which heat is developed is 1800 Watts (W) or 1800 Joules per second (J/s).
(a) List the factors on which the resistance of a conductor depends.
(b) Why are metals good conductors of electricity while glass is a bad conductor?
(c) Why are alloys commonly used in electrical heating devices?
- Length of the conductor (L): Resistance is directly proportional to length (R ∝ L).
- Area of cross-section (A): Resistance is inversely proportional to the area of cross-section (R ∝ 1/A).
- Nature of the material (Resistivity, ρ): Different materials offer different resistance.
- Temperature: Resistance of pure metals generally increases with an increase in temperature.
- Metals: Metals are good conductors because they possess a large number of free electrons. These electrons are loosely bound and can move easily through the material, constituting an electric current.
- Glass: Glass is a bad conductor (an insulator) because it has almost no free electrons. Its electrons are tightly bound to the atoms and are not free to move and carry current.
- High Resistivity: Alloys have a much higher resistivity than their constituent metals, which causes them to produce a large amount of heat (H = I²Rt).
- High Melting Point & Low Oxidation: They have a very high melting point and do not oxidize (burn) easily, even when they are red-hot.
(a) What is the function of an electric fuse in a domestic circuit?
(b) An electric fuse is rated at 2 A. What is meant by this statement?
(c) Find the equivalent resistance of the following circuit:
(b) Meaning of 2 A Rating: A fuse rating of 2 A means that the fuse wire will melt and break the circuit if the current flowing through it exceeds 2 Amperes. It is the maximum safe current the circuit can handle.
(c) Circuit Calculation:
The two 4 Ω resistors are in series.
R_top = 4 Ω + 4 Ω = 8 Ω.
Step 2: Calculate resistance of the bottom branch (R_bottom).
The two 4 Ω resistors are also in series.
R_bottom = 4 Ω + 4 Ω = 8 Ω.
Step 3: Calculate the total equivalent resistance (Req).
The top branch (R_top) and the bottom branch (R_bottom) are in parallel with each other.
1/Req = 1/R_top + 1/R_bottom
1/Req = 1/8 + 1/8
1/Req = 2/8 = 1/4
Req = 4 Ω
The equivalent resistance of the circuit between points A and B is 4 Ω.
Two resistors, R1 and R2, are connected in parallel.
(a) Derive an expression for the equivalent resistance (Rp) of the combination.
(b) Two lamps, rated 100 W at 220 V and 60 W at 220 V, are connected in parallel to a 220 V supply. Find the total current drawn from the supply.
- When resistors are in parallel, the potential difference (V) across each resistor is the same.
- The total current (I) from the source is the sum of the currents in each branch: I = I1 + I2.
- Using Ohm's Law (I = V/R): I1 = V/R1 and I2 = V/R2.
- Let the equivalent resistance be Rp. The total current I = V/Rp.
- Substitute these into the main equation: V/Rp = (V/R1) + (V/R2).
- Factor out V: V/Rp = V * (1/R1 + 1/R2).
- Cancel V from both sides: 1/Rp = 1/R1 + 1/R2.
In a parallel connection, the total power consumed is the sum of individual powers.
P_total = P1 + P2 = 100 W + 60 W = 160 W.
Given: V = 220 V.
Using the formula P_total = V × I_total:
I_total = P_total / V
I_total = 160 W / 220 V = 16/22 A = 8/11 A
I_total ≈ 0.73 A
Method 2: Calculate individual currents.
Current for Lamp 1 (I1) = P1 / V = 100 W / 220 V.
Current for Lamp 2 (I2) = P2 / V = 60 W / 220 V.
Total Current (I_total) = I1 + I2 = (100/220) + (60/220) = 160/220 A ≈ 0.73 A.
For the circuit shown in the diagram, calculate:
(a) The total resistance of the circuit.
(b) The total current flowing through the circuit.
(c) The potential difference across the parallel combination (R2 and R3).
R2 and R3 are in parallel.
1/Rp = 1/R2 + 1/R3 = 1/8 + 1/12
Common denominator is 24.
1/Rp = (3/24) + (2/24) = 5/24
Rp = 24 / 5 Ω = 4.8 Ω.
Step 2: Calculate total resistance.
R1 is in series with Rp.
Req = R1 + Rp
Req = 7.2 Ω + 4.8 Ω
Req = 12.0 Ω
(b) Total Current (I):
Given: Total Voltage (V) = 6 V
Using Ohm's Law: I = V / Req
I = 6 V / 12 Ω
I = 0.5 A
(This is the current read by the ammeter A).
(c) Potential Difference across parallel (Vp):
We need to find the voltage across the parallel combination (Rp). The total current (I = 0.5 A) flows through this combination.
Vp = I × Rp
Vp = 0.5 A × 4.8 Ω
Vp = 2.4 V
(a) Define electric power. State its SI unit.
(b) Derive the relationship P = I²R.
(c) An electric lamp is marked 100 W, 220 V. It is used for 5 hours daily. Calculate the energy consumed in 30 days.
It is defined as the rate at which electrical energy is consumed or dissipated in an electric circuit. (P = E/t or P = W/t).
SI Unit: The SI unit of electric power is the Watt (W).
(b) Derivation of P = I²R:
- We know that power P = V × I. (Equation 1)
- According to Ohm's Law, V = I × R. (Equation 2)
- Substitute the value of V from Equation 2 into Equation 1:
- P = (I × R) × I
- P = I²R
Power (P) = 100 W
Time per day = 5 hours
Total days = 30
Total time (t) = 5 hours/day × 30 days = 150 hours.
Total Energy (E) = Power (P) × Total Time (t)
E = 100 W × 150 h = 15000 Wh
To convert to the commercial unit (kWh), divide by 1000:
E = 15000 / 1000 kWh
E = 15 kWh
The total energy consumed in 30 days is 15 kWh (or 15 "units").
(a) How is an ammeter connected in a circuit? Why?
(b) How is a voltmeter connected in a circuit? Why?
(c) A 100 W electric bulb is connected to a 220 V mains power supply. Calculate the resistance of the bulb.
- Connection: An ammeter is always connected in series.
- Reason: It measures the current flowing *through* the circuit. In a series connection, the total current flows through the ammeter. It must have very low resistance so that it does not significantly change the total resistance or current of the circuit.
- Connection: A voltmeter is always connected in parallel across the component whose potential difference is to be measured.
- Reason: It measures the potential *drop* between two points. It must have very high resistance so that it draws a negligible amount of current from the main circuit, thus not affecting the voltage it is trying to measure.
Power (P) = 100 W
Voltage (V) = 220 V
We use the power formula that relates P, V, and R: P = V²/R
Rearranging for R: R = V²/P
R = (220 V)² / 100 W
R = (220 × 220) / 100
R = 22 × 22
R = 484 Ω
The resistance of the bulb is 484 Ω.
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A wire has a resistance of 10 Ω. It is stretched to three times its original length. Assuming the volume and resistivity of the wire do not change, what will be its new resistance?
When a wire is stretched, its volume (V) remains constant.
Let original length = L, original area = A. So, V = L × A.
New length L' = 3L.
New volume V' = L' × A' = (3L) × A'.
Since V = V', we have L × A = (3L) × A', which means the new area A' = A/3.
Step 2: Calculate new resistance.
Original Resistance R = ρ(L/A) = 10 Ω.
New Resistance R' = ρ(L'/A')
R' = ρ(3L / (A/3))
R' = ρ(9L / A)
R' = 9 × [ρ(L/A)]
R' = 9 × R
R' = 9 × 10 Ω = 90 Ω.
Shortcut: For a stretched wire, the new resistance R' = n² × R, where 'n' is the number of times the length is increased. Here n=3, so R' = 3² × 10 = 9 × 10 = 90 Ω.
(a) Define the commercial unit of electrical energy.
(b) A 60 W electric bulb is lit for 10 hours a day. How many units of electrical energy are consumed in 30 days?
(c) Calculate the cost of energy consumed at the rate of ₹ 6.00 per unit.
The commercial unit of electrical energy is the kilowatt-hour (kWh), often simply called a "unit".
1 kWh is defined as the amount of electrical energy consumed by an appliance of 1000 Watts (1 kW) when operated for 1 hour.
(b) Energy Calculation:
Time per day = 10 hours
Total days = 30
Total time (t) = 10 hours/day × 30 days = 300 hours.
Total Energy (E) = Power (P) × Total Time (t)
E = 0.06 kW × 300 h
E = 18 kWh
The energy consumed is 18 units.
(c) Cost Calculation:
Cost per unit = ₹ 6.00
Total Cost = Total Energy (kWh) × Cost per unit
Total Cost = 18 × 6.00
Total Cost = ₹ 108
(a) What is meant by "electric circuit"?
(b) What is overloading and short-circuiting? How can they be prevented?
(b) Overloading and Short-Circuiting:
- Overloading: This occurs when too many appliances are connected to a single socket, or when a high-power appliance is used. This causes the circuit to draw an excessively large current, which can heat up the wires and cause a fire.
- Short-Circuiting: This occurs when the live wire (positive) and the neutral wire (negative) come into direct contact. This can happen if the insulation on the wires is damaged. When they touch, the resistance of the circuit becomes almost zero, causing a very large current to flow, which can lead to fire.
Both conditions can be prevented by using a safety device called an electric fuse or an MCB (Miniature Circuit Breaker) in the circuit. These devices automatically break the circuit when the current exceeds a safe limit.
Two resistors of resistance R and 2R are connected in parallel. This combination is then connected in series with a resistor of resistance 3R. If the total resistance of the circuit is 11 Ω, find the value of R.
Resistors R and 2R are in parallel.
1/Rp = 1/R + 1/(2R)
Common denominator is 2R.
1/Rp = (2/2R) + (1/2R) = 3/(2R)
So, Rp = 2R/3
Step 2: Calculate the total equivalent resistance (Req).
This parallel combination (Rp) is in series with the 3R resistor.
Req = Rp + 3R
Req = (2R/3) + 3R
Find a common denominator (3).
Req = (2R/3) + (9R/3)
Req = 11R/3
Step 3: Find the value of R.
We are given that the total resistance (Req) is 11 Ω.
11R/3 = 11 Ω
R/3 = 1 Ω
R = 3 Ω
The value of R is 3 Ω.
Study the following circuit diagram and find:
(a) The total resistance of the circuit.
(b) The total current flowing in the circuit.
(c) The current flowing through the 3 Ω resistor.
1/Req = 1/R1 + 1/R2 + 1/R3
1/Req = 1/1 + 1/3 + 1/6
Common denominator is 6.
1/Req = (6/6) + (2/6) + (1/6)
1/Req = (6 + 2 + 1) / 6 = 9/6 = 3/2
Req = 2/3 Ω ≈ 0.67 Ω
(b) Total Current (I_total):
Given: Total Voltage (V) = 3 V
Using Ohm's Law: I_total = V / Req
I_total = 3 V / (2/3 Ω)
I_total = 3 × (3/2) A = 9/2 A
I_total = 4.5 A
(c) Current through 3 Ω resistor (I_3Ω):
In a parallel circuit, the voltage across each resistor is the same as the source voltage (V = 3 V).
Using Ohm's Law for the 3 Ω resistor:
I_3Ω = V / R_3Ω
I_3Ω = 3 V / 3 Ω
I_3Ω = 1 A
Two bulbs are rated (i) P1=60 W, V=220 V and (ii) P2=40 W, V=220 V.
(a) Calculate the resistance of each bulb.
(b) Which bulb has the higher resistance?
(c) If these two bulbs are connected in series to a 220 V supply, which bulb will glow brighter and why?
We use the formula P = V²/R, so R = V²/P.
Bulb 1 (60 W):
R1 = (220)² / 60 = (48400) / 60 = 4840 / 6 = 806.67 Ω
Bulb 2 (40 W):
R2 = (220)² / 40 = (48400) / 40 = 4840 / 4 = 1210 Ω
(b) Higher Resistance:
The 40 W bulb (R2 = 1210 Ω) has the higher resistance. (Note: For the same voltage, lower power means higher resistance).
(c) Brightness in Series:
When connected in series, the 40 W bulb will glow brighter.
Reason: In a series circuit, the current (I) is the same through both bulbs. The power (brightness) is given by P = I²R. Since I is constant, power is directly proportional to resistance (P ∝ R). The 40 W bulb has higher resistance (1210 Ω) than the 60 W bulb (806.67 Ω), so it will dissipate more power and glow brighter.
(a) Define 1 Ohm.
(b) How does resistance of a wire vary with its (i) area of cross-section and (ii) length?
(c) A 4 Ω wire is folded in half. What will be its new resistance?
(b) Variation of Resistance:
- (i) Area of cross-section (A): Resistance is inversely proportional to the area of cross-section (R ∝ 1/A). A thicker wire has less resistance.
- (ii) Length (L): Resistance is directly proportional to the length of the wire (R ∝ L). A longer wire has more resistance.
1. The new length (L') becomes half the original length: L' = L/2.
2. The new area of cross-section (A') becomes double the original area: A' = 2A.
Let the original resistance be R = ρ(L/A) = 4 Ω.
The new resistance (R') will be:
R' = ρ(L'/A')
R' = ρ( (L/2) / (2A) )
R' = ρ( L / (4A) )
R' = (1/4) × [ρ(L/A)]
R' = (1/4) × R
R' = (1/4) × 4 Ω = 1 Ω.
(a) Three resistors R1, R2, and R3 are connected in series. Derive an expression for the equivalent resistance (Rs) of the combination.
(b) Why is the series arrangement not used for domestic circuits? (Give two reasons).
- When resistors are in series, the total potential difference (V) across the combination is the sum of the potential differences across each resistor: V = V1 + V2 + V3.
- The current (I) is the same through all resistors in the series path.
- Using Ohm's Law (V = IR): V1 = IR1, V2 = IR2, and V3 = IR3.
- Let the equivalent resistance be Rs. The total voltage V = I * Rs.
- Substitute these into the main equation: I*Rs = IR1 + IR2 + IR3.
- Factor out I: I*Rs = I * (R1 + R2 + R3).
- Cancel I from both sides: Rs = R1 + R2 + R3.
- Circuit Break: If any one appliance in the series fails or is switched off, the circuit breaks, and all other appliances in that circuit stop working.
- Voltage Division: The total voltage (e.g., 220 V) is divided among all the appliances. This means each appliance does not receive the required voltage to operate correctly. A bulb would glow dimly, a heater would not heat properly, etc.
A wire of uniform cross-section and 20 Ω resistance is cut into four equal pieces. These pieces are then connected in parallel. Calculate the equivalent resistance of the parallel combination.
Step 1: Find the resistance of each piece.
The wire is cut into four equal pieces. Since resistance is directly proportional to length (R ∝ L), the resistance of each piece (R_piece) will be one-fourth of the total resistance.
R_piece = R_total / 4 = 20 Ω / 4 = 5 Ω.
Step 2: Calculate the parallel combination.
Now we have four resistors (R1, R2, R3, R4), each with a resistance of 5 Ω, connected in parallel.
1/Req = 1/R1 + 1/R2 + 1/R3 + 1/R4
1/Req = 1/5 + 1/5 + 1/5 + 1/5
1/Req = 4/5
Req = 5 / 4 Ω
Req = 1.25 Ω
The equivalent resistance of the parallel combination is 1.25 Ω.
List two advantages and two disadvantages of connecting electrical appliances in a parallel arrangement in a domestic circuit.
- Independent Operation: Each appliance has its own switch and can be turned on or off independently without affecting any other appliance in the circuit.
- Full Voltage: Each appliance is connected directly to the mains supply and receives the full rated voltage (e.g., 220 V), allowing it to operate at its correct power.
- Higher Current Draw: As more appliances are added in parallel, the total resistance of the circuit decreases (1/Req = 1/R1 + 1/R2...). This causes the total current drawn from the supply to increase (I = V/Req), which can lead to overloading if it exceeds the circuit's safe limit.
- No Protection if one Fails (Shorts): While an open-circuit failure is fine, if one appliance develops a short-circuit, it creates a very low-resistance path, drawing a massive current and potentially causing a fire (though this is what fuses/MCBs are for).
An electric iron consumes energy at a rate of 880 W when heating is at the maximum and 330 W when heating is at the minimum. The voltage is 220 V. What are the values of current and resistance in each case?
Case 1: Maximum Heating
Power (P_max) = 880 W
Current (I_max):
Using P = V × I
I_max = P_max / V = 880 W / 220 V
I_max = 4 A
Resistance (R_max):
Using Ohm's Law (V = IR) or P = V²/R
R_max = V / I_max = 220 V / 4 A = 55 Ω
(Alternatively: R_max = V²/P_max = (220*220) / 880 = 48400 / 880 = 55 Ω)
Case 2: Minimum Heating
Power (P_min) = 330 W
Current (I_min):
Using P = V × I
I_min = P_min / V = 330 W / 220 V
I_min = 1.5 A
Resistance (R_min):
Using Ohm's Law (V = IR) or P = V²/R
R_min = V / I_min = 220 V / 1.5 A = 146.67 Ω
(Alternatively: R_min = V²/P_min = (220*220) / 330 = 48400 / 330 = 146.67 Ω)
(a) What is the resistivity of a material?
(b) A copper wire has a diameter of 0.5 mm and resistivity of 1.6 × 10⁻⁸ Ω m. What will be the length of this wire to make its resistance 10 Ω?
(c) How much does the resistance change if the diameter is doubled?
(b) Length Calculation:
R = 10 Ω
ρ = 1.6 × 10⁻⁸ Ω m
Diameter (d) = 0.5 mm = 0.5 × 10⁻³ m
Radius (r) = d/2 = 0.25 × 10⁻³ m
Step 1: Calculate Area (A).
A = πr² = 3.14 × (0.25 × 10⁻³ m)²
A = 3.14 × 0.0625 × 10⁻⁶ m²
A ≈ 0.19625 × 10⁻⁶ m²
Step 2: Calculate Length (L).
From R = ρL/A, we get L = RA/ρ
L = (10 Ω × 0.19625 × 10⁻⁶ m²) / (1.6 × 10⁻⁸ Ω m)
L = (1.9625 / 1.6) × 10⁻⁶ / 10⁻⁸ m
L = 1.226 × 10² m
L ≈ 122.6 m
(c) Change in Resistance (Diameter Doubled):
Resistance is inversely proportional to the area (R ∝ 1/A).
Area is proportional to the square of the radius (A ∝ r²) or square of the diameter (A ∝ d²).
Therefore, R ∝ 1/d².
If the diameter is doubled (d' = 2d), the new resistance R' will be:
R' ∝ 1 / (2d)² = 1 / 4d²
R' ∝ (1/4) × (1/d²)
This means the new resistance will be one-fourth (1/4) of the original resistance.
R' = 10 Ω / 4 = 2.5 Ω.
(a) State Ohm's Law.
(b) Draw a schematic diagram of a circuit to verify Ohm's law.
(c) What is the nature of the V-I graph for an ohmic conductor? How is resistance calculated from it?
V ∝ I
V = IR (where R is a constant called resistance).
(b) Circuit Diagram to Verify Ohm's Law:
(c) V-I Graph and Resistance:
The nature of the V-I graph for an ohmic conductor is a straight line passing through the origin (0,0).
Resistance (R) can be calculated from the slope of the V-I graph.
Slope = ΔV / ΔI = R. (If V is on the y-axis and I is on the x-axis).
An electric iron is rated 2 kW at 220 V.
(a) What is the resistance of the iron?
(b) What is the maximum current it can draw?
(c) It is used for 2 hours daily. Find the cost of energy consumed for 30 days if the rate is ₹ 5.50 per unit (kWh).
Power (P) = 2 kW = 2000 W
Voltage (V) = 220 V
(a) Resistance (R):
Using P = V²/R
R = V²/P = (220 V)² / 2000 W
R = 48400 / 2000 = 484 / 20
R = 24.2 Ω
(b) Maximum Current (I):
Using P = V × I
I = P / V = 2000 W / 220 V
I = 200 / 22 = 100 / 11
I ≈ 9.09 A
(c) Cost of Energy:
Time per day = 2 hours
Total days = 30
Cost per unit = ₹ 5.50 / kWh
Step 1: Calculate total energy in kWh.
Total time (t) = 2 hours/day × 30 days = 60 hours
Total Energy (E) = Power (P) × Total Time (t)
E = 2 kW × 60 h
E = 120 kWh (or 120 units)
Step 2: Calculate total cost.
Total Cost = Total Energy (kWh) × Cost per unit
Total Cost = 120 × 5.50
Total Cost = ₹ 660
In the given circuit, if the current flowing through the 6 Ω resistor is 2 A, find:
(a) The potential difference across the parallel combination (R1 and R2).
(b) The current flowing through the 12 Ω resistor.
(c) The total current in the circuit.
(d) The potential difference of the battery (V).
(a) Potential difference across parallel (Vp):
R1 and R2 are in parallel, so the voltage across them is the same.
Vp = V1 = V2
We can find V1 using Ohm's Law:
Vp = I1 × R1 = 2 A × 6 Ω
Vp = 12 V
(b) Current through 12 Ω (I2):
The voltage across R2 is also Vp = 12 V.
I2 = Vp / R2 = 12 V / 12 Ω
I2 = 1 A
(c) Total current in the circuit (I_total):
The total current (I_total) leaves the battery, flows through R3, and then splits into I1 and I2.
I_total = I1 + I2
I_total = 2 A + 1 A
I_total = 3 A
(d) Potential difference of the battery (V):
The total voltage of the battery (V) is the sum of the voltage drop across R3 and the voltage drop across the parallel part (Vp).
V = V3 + Vp
First, find V3: V3 = I_total × R3 = 3 A × 4 Ω = 12 V.
V = 12 V + 12 V
V = 24 V
(a) What are the two main types of materials in terms of resistivity? Give two examples of each.
(b) Explain how the resistance of (i) a pure metal, (ii) an alloy, and (iii) a semiconductor, changes with an increase in temperature.
- Conductors: Materials with very low resistivity that allow electric current to pass easily.
Examples: Copper (Cu), Aluminum (Al), Silver (Ag). - Insulators: Materials with very high resistivity that do not allow electric current to pass through them.
Examples: Rubber, Glass, Wood, PVC.
(b) Effect of Temperature on Resistance:
- (i) Pure Metal: The resistance of a pure metal conductor increases with an increase in temperature. This is because higher temperature causes the metal ions to vibrate more, leading to more frequent collisions with the flowing electrons.
- (ii) Alloy: The resistance of an alloy (like Nichrome) also increases with temperature, but the change is much less significant than in pure metals. Their resistance is high but relatively stable.
- (iii) Semiconductor: The resistance of a semiconductor (like Silicon) decreases with an increase in temperature. This is because the higher temperature provides energy to break more covalent bonds, releasing more free electrons and charge carriers.
In the circuit diagram shown, R1 = R2 = R3 = 10 Ω. The battery is 12 V. Calculate:
(a) Current through each resistor.
(b) Total current in the circuit.
(c) Total equivalent resistance of the circuit.
(It's easiest to calculate this first).
Step 1: R2 and R3 are in parallel.
1/Rp = 1/R2 + 1/R3 = 1/10 + 1/10 = 2/10 = 1/5
Rp = 5 Ω
Step 2: R1 is in series with Rp.
Req = R1 + Rp = 10 Ω + 5 Ω
Req = 15 Ω
(b) Total Current (I_total):
Using Ohm's Law: I_total = V / Req
I_total = 12 V / 15 Ω
I_total = 0.8 A
(a) Current through each resistor:
Current through R1 (I1): The total current flows through R1.
I1 = I_total = 0.8 A
Current through R2 (I2) and R3 (I3):
First, find the voltage across the parallel combination (Vp).
Vp = I_total × Rp = 0.8 A × 5 Ω = 4 V
(Alternatively: V1 = I1*R1 = 0.8*10 = 8V. So Vp = V_total - V1 = 12V - 8V = 4V)
Now, use Vp to find I2 and I3:
I2 = Vp / R2 = 4 V / 10 Ω = 0.4 A
I3 = Vp / R3 = 4 V / 10 Ω = 0.4 A
(Check: I2 + I3 = 0.4 + 0.4 = 0.8 A, which matches I_total).
