CUET UG Physics Booster Test - 2 Material Properties and Power
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QUESTION 1 OF 20
A given electrical device exhibits a non-linear V-I deviation. This implies that its dynamic electrical resistance behaves such that:
QUESTION 2 OF 20
Identify the correct statements about sign-dependent relations in specialized electrical devices.
Statements:
1. Reversing the bias voltage yields a current of drastically different magnitude.
2. This property represents a limitation of Ohm's law.
3. Such behavior is typically observed in standard metallic conductors like copper.
QUESTION 3 OF 20
Analyzing the characteristic curve of a diode explicitly reveals that it does not follow Ohm's law. Which observation confirms this?
QUESTION 4 OF 20
In a material like Gallium Arsenide (GaAs), the relationship between applied voltage and resulting current is non-unique. This physically implies that:
QUESTION 5 OF 20
Identify the correct statements about conduction properties of metallic elements.
Statements:
1. They are generally classified as insulating materials at room temperature.
2. They possess very large free-electron densities at room temperature.
3. Their fundamental resistivities generally lie in the range 10^(-8)to 10^(-6) Ω m.
QUESTION 6 OF 20
Identify the incorrect statement regarding high resistivity materials.
QUESTION 7 OF 20
Identify the correct statements about the temperature coefficient of resistivity (α).
Statements:
1. It represents the fractional change in resistivity per unit change in temperature.
2. For metals, a positive αimplies that resistivity increases with temperature.
3. Its SI unit is ohm per kelvin (Ω/K).
QUESTION 8 OF 20
A heating wire has a resistance of 100 Ωat T_0=27^∘C. If its temperature coefficient is
α=2.0×10^(-4) ^∘C^(-1)
what will be the approximate temperature when its resistance becomes 102 Ω?
QUESTION 9 OF 20
Because typical metals possess a positive temperature coefficient of resistivity, actively heating a metallic wire connected to a fixed voltage source will result in:
QUESTION 10 OF 20
The resistivity graph of copper versus absolute temperature deviates from
ρ_T=ρ_0[1+α(T-T_0)]
at very low temperatures. In this region, the relationship becomes:
QUESTION 11 OF 20
When an alloy like nichrome is used as a heating element in a standard toaster, its weak dependence of resistivity on temperature ensures that:
QUESTION 12 OF 20
Identify the correct statements about wire-bound standard resistors.
Statements:
1. Materials like manganin and constantan are commonly used.
2. They require materials whose resistance changes drastically with temperature.
3. Their purpose is to maintain nearly constant resistance despite small temperature changes.
QUESTION 13 OF 20
Identify the incorrect statement regarding the behavior of charge carriers in a semiconductor as temperature increases.
QUESTION 14 OF 20
Why does the resistivity of an intrinsic semiconductor decrease with increasing temperature despite more frequent collisions?
QUESTION 15 OF 20
For a steady current I flowing from point A to point B through a uniform resistor of length l, the potential gradient magnitude is:
QUESTION 16 OF 20
Match List I with List II regarding macroscopic energy changes in a solid conductor carrying a steady electric current
| List I | List II |
|---|---|
| 1. ΔUₚₒₜ (Change in macroscopic potential energy) | a. Zero (drift velocity remains steady) |
| 2. Total internal electrical energy change (Ideal) | b. +IVΔt |
| 3. ΔK if mobile charges moved freely (no lattice collisions) | c. Unchanged (Energy Conservation) |
| 4. Actual kinetic energy change on average (with collisions) | d. −IVΔt |
QUESTION 17 OF 20
Identify the correct statements about kinetic energy transfer in a metallic conductor.
Statements:
1. Charges accelerate initially but lose energy during collisions with ions.
2. Energy lost by electrons increases lattice vibrational energy.
3. The macroscopic result is heating of the conductor.
QUESTION 18 OF 20
An electric heating element operates at 220 V and draws a steady current of 5 A. Calculate the energy dissipated as heat in 10 s.
QUESTION 19 OF 20
Identify the correct statements regarding electrical power.
Statements:
1. Power depends on both voltage and current.
2. Power is supplied by an external energy source such as a cell or battery.
3. Power equals the total change in potential energy over the entire lifetime of a circuit.
QUESTION 20 OF 20
To transmit power P through cables of resistance R_c at source voltage V, the power lost in the cables is:
Test Complete!
Answer Review
1 A given electrical device exhibits a non-linear V-I deviation. This implies that its dynamic electrical resistance behaves such that:
�� Non-linear V-I relation implies variable resistance. �� Resistance depends on operating conditions. �� Ohm's law is not strictly obeyed.
Ohm's law states that the potential difference across a conductor is directly proportional to the current flowing through it, provided physical conditions remain constant. Under these conditions, the ratio V/I remains constant and is called resistance. In a non-linear device, the V-I graph is not a straight line. Therefore, the ratio V/I changes with voltage or current. This means the resistance is not constant and varies depending on the operating point of the device. Semiconductor devices such as diodes commonly exhibit such behavior. Since the resistance changes with the applied voltage or current, the device does not obey Ohm's law over its entire operating range. Hence, the correct answer is B.
- �� Option A → Constant resistance is a property of Ohmic conductors.
- �� Option C → Non-linear behavior does not imply zero resistance.
- �� Option D → No general inverse-square dependence exists.
Used – Concept Application
- Application
- Relate the shape of the V-I curve to the definition of resistance.
- Final Logic
- A non-linear V-I graph means V/I is not constant, so resistance varies.
- Curved Line → Variable Resistance
2 Identify the correct statements about sign-dependent relations in specialized electrical devices.
Statements:
1. Reversing the bias voltage yields a current of drastically different magnitude.
2. This property represents a limitation of Ohm's law.
3. Such behavior is typically observed in standard metallic conductors like copper.
�� Current depends on voltage polarity. �� Such behavior violates simple Ohmic proportionality. �� Metals like copper do not show this effect.
Certain electrical devices, especially semiconductor diodes, behave differently when the direction of the applied voltage is reversed. In forward bias, a large current may flow, while in reverse bias only a very small current exists. Thus, reversing the voltage does not simply reverse the current with the same magnitude. This sign-dependent behavior is one of the important limitations of Ohm's law because Ohm's law assumes a linear and symmetric relationship between voltage and current. Ordinary metallic conductors such as copper exhibit nearly identical resistance for both directions of current and therefore do not display this sign-dependent behavior. Hence, statements 1 and 2 are correct, while statement 3 is incorrect.
- �� Option B → Statement 3 is incorrect.
- �� Option C → Statement 3 is incorrect.
- �� Option D → Statement 3 is incorrect.
Used – NCERT Recall
- Application
- Recall examples of non-Ohmic devices discussed in NCERT.
- Final Logic
- Polarity-dependent current is a limitation of Ohm's law and is characteristic of semiconductor devices.
- Diode → Direction Matters
3 Analyzing the characteristic curve of a diode explicitly reveals that it does not follow Ohm's law. Which observation confirms this?
�� Diode behavior depends on voltage polarity. �� Forward and reverse currents are very different. �� The V-I graph is non-linear.
A diode is a non-Ohmic device whose current-voltage characteristic is strongly dependent on the direction of the applied voltage. When forward biased, the diode allows significant current to flow after a threshold voltage is reached. When reverse biased, only a very small leakage current flows. Therefore, the current obtained for positive voltage is drastically different from the current obtained for negative voltage. This asymmetrical and non-linear behavior demonstrates that the diode does not obey Ohm's law. Hence, observation B directly confirms the non-Ohmic nature of the diode.
- �� Option A → Describes an ideal Ohmic conductor.
- �� Option C → Not a defining property of a diode.
- �� Option D → Resistance is not constant.
Used – Concept Application
- Application
- Analyze the V-I characteristic curve of a diode.
- Final Logic
- Different current magnitudes for opposite voltage polarities indicate non-Ohmic behavior.
- Diode → One-Way Preference
4 In a material like Gallium Arsenide (GaAs), the relationship between applied voltage and resulting current is non-unique. This physically implies that:
�� GaAs exhibits non-unique V-I characteristics. �� Same current may correspond to different voltages. �� It is a non-Ohmic material.
Gallium Arsenide (GaAs) is a semiconductor material that can exhibit non-linear and non-unique current-voltage characteristics. In certain operating regions, the V-I curve bends in such a way that the same current value corresponds to more than one voltage value. This behavior is impossible for a simple Ohmic conductor, where every current value corresponds to exactly one voltage value according to V=IR. Therefore, non-uniqueness means that a single current can be associated with multiple voltage values.
- �� Option A → Current is not necessarily zero.
- �� Option C → GaAs is a conducting semiconductor.
- �� Option D → GaAs is not perfectly Ohmic.
Used – NCERT Recall
- Application
- Recall the special V-I characteristics of GaAs.
- Final Logic
- Non-unique V-I relation means one current may correspond to multiple voltages.
- GaAs → Same Current, Different Voltages
5 Identify the correct statements about conduction properties of metallic elements.
Statements:
1. They are generally classified as insulating materials at room temperature.
2. They possess very large free-electron densities at room temperature.
3. Their fundamental resistivities generally lie in the range 10^(-8)to 10^(-6) Ω m.
�� Metals have very low resistivity. �� Metals contain a large number of free electrons. �� Metals are conductors, not insulators.
Metals are characterized by very low resistivity values, typically ranging from about 10^(-8)to 10^(-6) Ω m. This low resistivity arises because metals possess an enormous number of free electrons available for conduction. The number density of free electrons in metals is typically of the order of 10^(28)to 10^(29) m^(-3), which allows large currents to flow even when the drift velocity is very small. Since metals conduct electricity efficiently, they are classified as conductors rather than insulators. Therefore, statement 2 and statement 3 are correct, while statement 1 is incorrect.
- �� Option A → Metals are not insulators.
- �� Option B → Statement 1 is incorrect.
- �� Option C → Statement 1 is incorrect.
Used – NCERT Recall
- Application
- Recall the classification of materials based on resistivity.
- Final Logic
- Low resistivity and high free-electron density make metals excellent conductors.
- Metals → Low ρ, High n
6 Identify the incorrect statement regarding high resistivity materials.
�� Insulators have very few mobile charge carriers. �� Their electrons are tightly bound. �� Hence their resistivity is extremely high.
High resistivity materials, commonly called insulators, strongly oppose the flow of electric current. Examples include ceramic, rubber, plastics, glass, and mica. Their resistivities are enormously larger than those of metals, often by factors of 10^(18)or more. The fundamental reason for this behavior is that electrons in insulators are tightly bound to their parent atoms and cannot move freely under ordinary electric fields. Consequently, the number of mobile charge carriers available for conduction is extremely small. Statement C is incorrect because metals actually possess a much larger density of free charge carriers than insulators. The scarcity of mobile carriers is precisely why insulators exhibit such high resistivity.
- �� Option A → Correct examples of insulators.
- �� Option B → Correct description of their high resistivity.
- �� Option D → Correct explanation of insulating behavior.
Used – Concept Application
- Application
- Relate electrical conduction to the availability of free charge carriers.
- Final Logic
- Insulators have very few mobile charge carriers and therefore very high resistivity.
- Metal → Free Electrons
7 Identify the correct statements about the temperature coefficient of resistivity (α).
Statements:
1. It represents the fractional change in resistivity per unit change in temperature.
2. For metals, a positive αimplies that resistivity increases with temperature.
3. Its SI unit is ohm per kelvin (Ω/K).
�� αmeasures sensitivity of resistivity to temperature. �� Metals generally have positive α. �� Unit of αis K^(-1).
The temperature coefficient of resistivity αmeasures the fractional change in resistivity per unit change in temperature. It appears in the relation ρ_T=ρ_0[1+α(T-T_0)] For most metallic conductors, αis positive. Therefore, an increase in temperature leads to an increase in resistivity because lattice vibrations become stronger and hinder electron motion. Since αrepresents change per unit temperature, its unit is the reciprocal of temperature, namely K^(-1)or ^∘C^(-1). It is not measured in Ω/K. Thus, statements 1 and 2 are correct while statement 3 is incorrect.
- �� Option B → Statement 3 is incorrect.
- �� Option C → Statement 2 is correct and cannot be excluded.
- �� Option D → Statement 3 is incorrect.
Used – Formula Recall
- Application
- Use the temperature dependence equation for resistivity.
- Final Logic
- αmeasures fractional change in resistivity per unit temperature.
- Alpha → Change per Degree
8 A heating wire has a resistance of 100 Ωat T_0=27^∘C. If its temperature coefficient is
α=2.0×10^(-4) ^∘C^(-1)
what will be the approximate temperature when its resistance becomes 102 Ω?
�� Use the linear resistivity relation. �� Find temperature rise first. �� Add it to the reference temperature.
Using R=R_0[1+α(T-T_0)] Given: R_0=100ΩR=102Ωα=2×10^(-4) ^∘C^(-1) Substituting: 102=100[1+2×10^(-4)(T-27)]1.02=1+2×10^(-4)(T-27)0.02=2×10^(-4)(T-27)T-27=100T=127^∘C Hence the correct answer is 127^∘C.
- �� Option B → Arithmetic error.
- �� Option C → No temperature change assumed.
- �� Option D → Incorrect substitution.
Used – Substitution
- Application
- Substitute the given values into the resistance-temperature relation.
- Final Logic
- T=127^∘C
- Resistance Up → Temperature Up
9 Because typical metals possess a positive temperature coefficient of resistivity, actively heating a metallic wire connected to a fixed voltage source will result in:
�� Heating increases metallic resistivity. �� Resistance rises. �� Current decreases for fixed voltage.
Metals possess a positive temperature coefficient of resistivity. Therefore, when the temperature of a metallic conductor rises, its resistivity increases due to enhanced lattice vibrations that obstruct electron motion. Since R=ρl/A an increase in resistivity causes resistance to increase. For a fixed applied voltage, I=V/R Thus, as resistance increases, the current decreases. This effect is commonly observed in filament lamps and heating elements during operation. Hence, the correct answer is that resistance increases while current decreases.
- �� Option A → Current decreases, not increases.
- �� Option B → Resistance increases.
- �� Option D → Electrical properties change with temperature.
Used – Concept Application
- Application
- Combine temperature dependence of resistivity with Ohm's law.
- Final Logic
- Higher temperature → Higher resistance → Lower current.
- Hot Metal → More Resistance
10 The resistivity graph of copper versus absolute temperature deviates from
ρ_T=ρ_0[1+α(T-T_0)]
at very low temperatures. In this region, the relationship becomes:
�� Linear approximation works only over a limited range. �� At low temperatures the graph bends. �� Constant αis no longer valid.
The expression ρ_T=ρ_0[1+α(T-T_0)] is an approximate linear relation valid only over a moderate temperature range. For metals such as copper, experimental observations show that resistivity does not continue to vary linearly at very low temperatures. As the temperature approaches low values, the resistivity-temperature curve deviates significantly from a straight line. In this region, a single constant value of αcannot accurately describe the variation of resistivity. Therefore, the low-temperature behavior is non-linear and requires more detailed physical models than the simple linear approximation.
- �� Option A → The curve is not strictly linear.
- �� Option C → Resistivity is not zero below 273 K.
- �� Option D → No inverse-square dependence exists.
Used – NCERT Recall
- Application
- Recall the limitations of the linear resistivity equation.
- Final Logic
- The constant-α approximation fails at low temperatures.
- Low Temperature → Linear Law Breaks
11 When an alloy like nichrome is used as a heating element in a standard toaster, its weak dependence of resistivity on temperature ensures that:
�� Nichrome has a small temperature coefficient. �� Resistance remains relatively stable. �� Ideal for heating applications.
Nichrome is an alloy of nickel, chromium, and iron that exhibits only a weak dependence of resistivity on temperature. As the heating element becomes hot during operation, its resistance changes only slightly compared to ordinary metallic conductors. This property is highly desirable because it allows the heating element to operate predictably over a wide temperature range. A large change in resistance would lead to significant fluctuations in current and power consumption. Therefore, nichrome is widely used in toasters, electric irons, and heating coils because its resistance remains relatively stable even at high temperatures.
- �� Option A → Resistance does not fall sharply.
- �� Option B → Nichrome is not a superconductor.
- �� Option D → Weak temperature dependence prevents such instability.
Used – NCERT Recall
- Application
- Recall the properties and applications of nichrome.
- Final Logic
- Nichrome maintains nearly constant resistance as temperature rises.
- Nichrome → Stable Resistance, Stable Heating
12 Identify the correct statements about wire-bound standard resistors.
Statements:
1. Materials like manganin and constantan are commonly used.
2. They require materials whose resistance changes drastically with temperature.
3. Their purpose is to maintain nearly constant resistance despite small temperature changes.
�� Standard resistors require stability. �� Manganin and constantan are commonly used. �� Resistance should remain nearly unchanged with temperature.
Wire-bound standard resistors are designed to provide highly stable and accurate resistance values. For this reason, materials such as manganin and constantan are used because they possess very small temperature coefficients of resistivity. A small temperature coefficient ensures that minor changes in environmental temperature produce negligible changes in resistance. This stability is essential in precision electrical measurements and calibration instruments. Statement 1 is therefore correct. Statement 3 is also correct because maintaining a constant resistance value is the primary objective of standard resistors. Statement 2 is incorrect because materials with large resistance variations would make accurate measurements impossible.
- �� Option A → Statement 2 is incorrect.
- �� Option C → Statement 2 is incorrect.
- �� Option D → Statement 2 is incorrect.
Used – Concept Application
- Application
- Relate the purpose of standard resistors to material selection.
- Final Logic
- Precision resistors require materials with minimal temperature dependence.
- Standard Resistor → Standard Resistance
13 Identify the incorrect statement regarding the behavior of charge carriers in a semiconductor as temperature increases.
�� Temperature generates additional charge carriers. �� Carrier density increases rapidly. �� Resistivity decreases.
In semiconductors, increasing temperature excites more electrons from the valence band into the conduction band. This creates a much larger number of free charge carriers. As a result, the carrier density n increases dramatically with temperature. Although the relaxation time τmay decrease because collisions become more frequent, the increase in carrier density is much greater. Consequently, conductivity increases and resistivity decreases. Therefore, statement D is incorrect because intrinsic carrier density is strongly temperature dependent and certainly does not remain constant.
- �� Option A → Correct semiconductor behavior.
- �� Option B → Correct explanation of conductivity increase.
- �� Option C → Correct consequence of increasing carrier density.
Used – Concept Application
- Application
- Use semiconductor conductivity concepts.
- Final Logic
- Higher temperature produces more free charge carriers.
- Semiconductor Hotter → More Carriers → Less Resistance
14 Why does the resistivity of an intrinsic semiconductor decrease with increasing temperature despite more frequent collisions?
�� Conductivity depends on both n and τ. �� Carrier density rises sharply with temperature. �� This dominates the reduction in τ.
The conductivity of a semiconductor is given by σ=ne^2τ/m As temperature increases, lattice vibrations become stronger, causing more collisions and reducing the relaxation time τ. However, temperature also generates a very large number of additional charge carriers. The increase in carrier density n is far greater than the decrease in τ. Consequently, the overall conductivity increases and the resistivity decreases. Thus, the dominant factor responsible for the reduction in resistivity is the dramatic rise in carrier density.
- �� Option A → Electron mass remains constant.
- �� Option B → Relaxation time decreases rather than increases.
- �� Option D → Electronic charge is constant.
Used – Formula Recall
- Application
- Analyze the conductivity equation.
- Final Logic
- The increase in n dominates the decrease in τ.
- Semiconductor Heating → Carrier Explosion
15 For a steady current I flowing from point A to point B through a uniform resistor of length l, the potential gradient magnitude is:
�� Potential gradient is potential drop per unit length. �� Use Ohm's law V=IR. �� Divide by conductor length.
The potential gradient is defined as the rate at which potential decreases along the length of a conductor. Mathematically, Potential Gradient=V/l For a resistor obeying Ohm's law, V=IR Substituting into the expression for potential gradient, V/l=IR/l This quantity represents the potential drop per unit length of the conductor and is directly related to the electric field inside the conductor.
- �� Option B → Incorrect dimensions.
- �� Option C → Not the definition of potential gradient.
- �� Option D → Physically incorrect relation.
Used – Formula Recall
- Application
- Use the definitions of potential gradient and Ohm's law.
- Final Logic
- Potential Gradient=V/l=IR/l
- Gradient = Potential Drop ÷ Length
16 Match List I with List II regarding macroscopic energy changes in a solid conductor carrying a steady electric current
| List I | List II |
|---|---|
| 1. ΔUₚₒₜ (Change in macroscopic potential energy) | a. Zero (drift velocity remains steady) |
| 2. Total internal electrical energy change (Ideal) | b. +IVΔt |
| 3. ΔK if mobile charges moved freely (no lattice collisions) | c. Unchanged (Energy Conservation) |
| 4. Actual kinetic energy change on average (with collisions) | d. −IVΔt |
�� Charges lose potential energy while moving. �� Energy is conserved in an ideal system. �� Average drift kinetic energy remains constant.
When charge moves through a conductor, electrical potential energy decreases. The energy lost during time Δt is ΔU_(pot)=-IVΔt Therefore, 1 → d. In an ideal energy accounting process, total energy remains conserved, so the total internal electrical energy change is unchanged, giving 2 → c. If charges could move without collisions, the electric field would continuously accelerate them and their kinetic energy would increase by +IVΔt Thus, 3 → b. In a real conductor, electrons undergo frequent collisions with lattice ions. As a result, the average drift velocity remains constant and there is no net increase in kinetic energy. Therefore, 4 → a. Hence the correct matching is: 1-d, 2-c, 3-b, 4-a
- �� Option B → Potential energy and kinetic energy terms are mismatched.
- �� Option C → Energy conservation incorrectly assigned.
- �� Option D → Multiple physical quantities are interchanged.
Used – Concept Application
- Application
- Apply conservation of energy and the microscopic model of conduction.
- Final Logic
- Potential energy decreases, energy is conserved, and collisions prevent continuous kinetic energy growth.
- Collisions → Constant Drift Speed
17 Identify the correct statements about kinetic energy transfer in a metallic conductor.
Statements:
1. Charges accelerate initially but lose energy during collisions with ions.
2. Energy lost by electrons increases lattice vibrational energy.
3. The macroscopic result is heating of the conductor.
�� Electrons gain energy from the field. �� Collisions transfer energy to the lattice. �� The conductor heats up.
In a conductor carrying current, free electrons are accelerated by the electric field between successive collisions. During collisions with lattice ions, a portion of the gained kinetic energy is transferred to the lattice. This transfer increases the vibrational energy of the ions. As lattice vibrations become more intense, the temperature of the conductor rises. This phenomenon is observed as Joule heating. Therefore, electrons continuously gain energy from the electric field and transfer it to the lattice through collisions. The net result is the conversion of electrical energy into thermal energy. Hence statements 1, 2, and 3 are all correct.
- �� Option A → Statement 3 is also correct.
- �� Option B → Statement 1 is also correct.
- �� Option C → Statement 2 is also correct.
Used – NCERT Recall
- Application
- Recall the microscopic explanation of Joule heating.
- Final Logic
- Electron-lattice collisions convert electrical energy into heat.
- Electron Energy → Lattice Energy → Heat
18 An electric heating element operates at 220 V and draws a steady current of 5 A. Calculate the energy dissipated as heat in 10 s.
�� Use the heat dissipation formula. �� Energy = Power × Time. �� Power = VI.
Electrical energy dissipated as heat is given by W=VIΔt Given: V=220VI=5AΔt=10s Substituting, W=220×5×10W=11000J Thus the heating element converts 11,000 joules of electrical energy into thermal energy during the 10-second interval.
- �� Option B → Time factor not applied correctly.
- �� Option C → Arithmetic error.
- �� Option D → Extra factor of 10 introduced.
Used – Substitution
- Application
- Substitute values directly into W=VIΔt.
- Final Logic
- W=11000J
- Energy = Voltage × Current × Time
19 Identify the correct statements regarding electrical power.
Statements:
1. Power depends on both voltage and current.
2. Power is supplied by an external energy source such as a cell or battery.
3. Power equals the total change in potential energy over the entire lifetime of a circuit.
�� Power is energy per unit time. �� Batteries supply electrical energy. �� Power is a rate, not total energy.
Electrical power is defined as the rate at which electrical energy is supplied or dissipated. Mathematically, P=VI Therefore, power depends directly on both voltage and current. The energy required to maintain current flow is supplied by an external source such as a battery, generator, or power supply. Statement 3 is incorrect because power represents the rate of energy transfer per unit time, not the total energy transferred during the entire operation of a circuit. Hence statements 1 and 2 are correct.
- �� Option A → Statement 3 is incorrect.
- �� Option B → Statement 3 is incorrect.
- �� Option D → Statement 3 is incorrect.
Used – Formula Recall
- Application
- Recall the definition of electrical power.
- Final Logic
- P=Energy/Time=VI
- Power = Rate of Energy Transfer
20 To transmit power P through cables of resistance R_c at source voltage V, the power lost in the cables is:
�� Cable loss is I^2R_c. �� Express current using P=VI. �� Substitute into the loss formula.
The power transmitted is P=VI Therefore, I=P/V The power dissipated in the transmission cable is P_c=I^2R_c Substituting the value of current, P_c=(P/V)^2R_cP_c=P^2R_c/V^2 This equation explains why electrical power is transmitted at very high voltages. Increasing V reduces current and therefore greatly reduces transmission losses.
- �� Option B → Ignores transmitted power.
- �� Option C → Incorrect dimensional form.
- �� Option D → Wrong dependence on voltage and resistance.
Used – Substitution
- Application
- Substitute I=P/V into P_c=I^2R_c.
- Final Logic
- P_c=P^2R_c/V^2
- High Voltage → Low Current → Low Loss
