CUET UG Physics Booster Test 3- Current Loop Dipoles and Meters
📌 Answers are locked once submitted — results and explanations appear at the end.
QUESTION 1 OF 20
A long wire carries 1 A from east to west. The earth's horizontal magnetic field is 3.0×10^(-5) T from south to north. The force per unit length on the wire evaluates to:
QUESTION 2 OF 20
Conductor A carries steady current South to North. Conductor B carries steady current North to South. The resulting magnetic interaction is:
QUESTION 3 OF 20
Identify the correct statements concerning the physical laboratory standardization of the Ampere.
Statements:
1. The effect of Earth's local magnetic field must be eliminated.
2. Multiturn coils replace the theoretically idealized infinite straight wires.
3. A current balance measures the resultant mechanical force.
4. Electrostatic inverse-square laws are used to determine steady current.
QUESTION 4 OF 20
Incorrect statement about the precise Ampere force magnitude standard:
QUESTION 5 OF 20
For a rectangular loop oriented such that its normal area vector makes an angle θwith the uniform magnetic field, the net translational force and magnitude of torque respectively are:
QUESTION 6 OF 20
When a rectangular current loop is oriented such that its area vector A is completely parallel to the uniform magnetic field B, the torque is:
QUESTION 7 OF 20
Which statement is correct regarding intrinsic and elementary magnetic moments?
QUESTION 8 OF 20
A loop of irregular shape carrying current is placed in an external magnetic field. If the wire is perfectly flexible, it shapes into a circle fundamentally to:
QUESTION 9 OF 20
Identify the correct statements regarding alignment of a magnetic dipole.
Statements:
1. When m is exactly parallel to B, the equilibrium is stable.
2. When m is exactly antiparallel to B, the equilibrium is unstable.
3. External work is required to rotate the coil from stable to unstable equilibrium.
4. The magnetic moment naturally rests at 90^∘.
QUESTION 10 OF 20
For an object with magnetic moment m placed in a uniform field B, unstable equilibrium dictates:
QUESTION 11 OF 20
Match List I with List II for circular loop dipole distance metrics.
| List I | List II |
|---|---|
| 1. Field Bat large distance xxR | a. πR2 |
| 2. Field Bat exact center x0 | b. IπR2 |
| 3. Magnetic moment m | c. Proportional to 1/x3 |
| 4. Loop Area A | d. μ0I2R |
QUESTION 12 OF 20
Incorrect statement regarding magnetic dipole field formulations.
QUESTION 13 OF 20
The mechanical purpose of the internal core configuration in a moving coil galvanometer is:
QUESTION 14 OF 20
By integrating a carefully shaped cylindrical soft iron core, the magnetic field specifically becomes:
QUESTION 15 OF 20
Identify the correct statements concerning torsional counter-torque.
Statements:
1. The spring S provides a mechanical counter torque kϕ.
2. In equilibrium, kϕbalances the magnetic torque NIAB.
3. The constant k represents restoring torque per unit angular twist.
4. Torsional torque depends exponentially on the number of turns N.
QUESTION 16 OF 20
At steady angular deflection ϕ, the balancing equation
kϕ=NIAB
allows the circuit current I to be directly computed as:
QUESTION 17 OF 20
M₁ (N=30, A=3.6×10^(-3)m^2, B=0.25T)and M₂ (N=42, A=1.8×10^(-3)m^2, B=0.50T)have identical springs. The current sensitivity ratio M_2/M_1 is:
QUESTION 18 OF 20
Continuing from the previous question, if the respective internal coil resistances are R_1=10Ωand R_2=14Ω, the voltage sensitivity ratio M_2/M_1 is:
QUESTION 19 OF 20
Identify the correct statements regarding construction of an ammeter.
Statements:
1. A shunt resistance r_s is connected in parallel.
2. The value of r_s is much smaller than R_G.
3. The effective resistance becomes slightly less than r_s.
4. It maximizes the voltage drop across the measuring point.
QUESTION 20 OF 20
Match List I with List II for ideal measuring instruments.
| List I | List II |
|---|---|
| 1. Ideal Ammeter Internal Resistance | a. Infinity |
| 2. Ideal Voltmeter Internal Resistance | b. Zero |
| 3. Ammeter Connection | c. In Series with Target Component |
| 4. Voltmeter Connection | d. In Parallel with Target Component |
Test Complete!
Answer Review
1 A long wire carries 1 A from east to west. The earth's horizontal magnetic field is 3.0×10^(-5) T from south to north. The force per unit length on the wire evaluates to:
�� Use F=ILBsinθ. �� Current and magnetic field are perpendicular. �� Direction is obtained using the right-hand rule.
The magnetic force on a current-carrying conductor is F=ILBsinθ Hence, force per unit length is F/L=IBsinθ Given: I=1AB=3.0×10^(-5)T The current flows from east to west, while the magnetic field points from south to north. These directions are perpendicular, so θ=90^∘ and sin90^∘=1 Thus, F/L=1×3.0×10^(-5)=3.0×10^(-5)N m^(-1) Using the right-hand rule for F=IL×B, the force is directed vertically downward. Therefore, the correct answer is 3.0×10^(-5)N m^(-1)downward.
- �� Option B → Correct magnitude but wrong direction.
- �� Option C → Force is not zero because current and field are perpendicular.
- �� Option D → Incorrect magnitude.
Substitution
- Application
- Apply
- F/L=IBsinθ
- and determine direction using the right-hand rule.
- Final Logic
- F/L=3.0×10^(-5)N m^(-1)
- Direction → Downward.
"Current Cross Field Gives Force"
2 Conductor A carries steady current South to North. Conductor B carries steady current North to South. The resulting magnetic interaction is:
�� Opposite currents repel. �� Same-direction currents attract. �� Force acts sideways between conductors.
The force per unit length between two long parallel conductors is F/L=μ_0I_1I_2/2πd The direction of the force depends on the relative directions of the currents. If the currents flow in the same direction, the conductors attract each other. If the currents flow in opposite directions, they repel each other. In this problem, conductor A carries current from south to north while conductor B carries current from north to south. Thus, the currents are antiparallel. Therefore, the conductors exert equal and opposite lateral forces on each other, resulting in repulsion.
- �� Option A → Attraction occurs for parallel currents in the same direction.
- �� Option C → Antiparallel currents produce a force.
- �� Option D → Force acts laterally, not along the wire length.
NCERT Recall
- Application
- Recall the interaction rule for parallel currents.
- Final Logic
- Same Direction → Attraction
- Opposite Direction → Repulsion
"Opposites Repel for Currents"
3 Identify the correct statements concerning the physical laboratory standardization of the Ampere.
Statements:
1. The effect of Earth's local magnetic field must be eliminated.
2. Multiturn coils replace the theoretically idealized infinite straight wires.
3. A current balance measures the resultant mechanical force.
4. Electrostatic inverse-square laws are used to determine steady current.
�� External magnetic fields must be minimized. �� Practical instruments replace idealized conductors. �� Current balance measures magnetic force.
The historical definition of the ampere is based on the magnetic force between current-carrying conductors. In practical measurements, the effect of Earth's magnetic field and other stray magnetic fields must be minimized to obtain accurate results. Since infinitely long straight conductors cannot be realized experimentally, multiturn coils are often used to generate measurable magnetic forces. A current balance is employed to measure these forces accurately. The definition is based on magnetic interactions between currents rather than electrostatic inverse-square laws. Therefore, Statements 1, 2 and 3 are correct, while Statement 4 is incorrect.
- �� Option A → Includes Statement 4.
- �� Option B → Includes Statement 4, which is incorrect.
- �� Option D → Includes Statement 4.
NCERT Recall
- Application
- Recall the experimental realization of the ampere.
- Final Logic
- Magnetic Force Measurement → Current Balance → Ampere Standardization.
"Balance Measures Current Force"
4 Incorrect statement about the precise Ampere force magnitude standard:
�� Ampere definition is based on force between currents. �� No changing magnetic flux is involved. �� The setup uses steady currents.
The classical definition of the ampere states that two infinitely long parallel conductors of negligible cross-section placed 1 metre apart in vacuum carrying equal currents of 1 ampere experience a force of 2×10^(-7)N/m between them. This definition is based on the magnetic interaction produced by steady currents. It does not involve electromagnetic induction or the rate of change of magnetic flux. Concepts involving changing flux belong to Faraday's law of electromagnetic induction, which is entirely different from the definition of the ampere. Therefore, Statement D is incorrect.
- �� Option A → Correct force standard.
- �� Option B → Correct theoretical requirement.
- �� Option C → Correct separation and medium specification.
Concept Differentiation
- Application
- Distinguish Ampere definition from electromagnetic induction.
- Final Logic
- Ampere Definition → Steady Current Force
- Faraday's Law → Changing Flux
"Force Defines Ampere, Flux Defines Induction"
5 For a rectangular loop oriented such that its normal area vector makes an angle θwith the uniform magnetic field, the net translational force and magnitude of torque respectively are:
�� Uniform magnetic field produces zero net force. �� Opposite forces form a couple. �� Torque depends on sinθ.
A current-carrying rectangular loop placed in a uniform magnetic field experiences forces on its sides. These forces are equal and opposite, so the resultant translational force on the loop is zero. However, the forces act at different points and form a couple that produces a torque. The magnetic dipole moment of the loop is m=NIA and the torque is given by τ=mBsinθ Therefore, τ=NIABsinθ where θis the angle between the magnetic moment vector (normal to the loop) and the magnetic field. Thus, the net force is zero while the torque is NIABsinθ.
- �� Option A → Torque depends on sinθ, not cosθ.
- �� Option C → Net translational force is zero.
- �� Option D → Torque is generally not zero.
Formula Recall
- Application
- Use the magnetic torque relation for a current loop.
- Final Logic
- F_(net)=0τ=NIABsinθ
"No Shift, Only Twist"
6 When a rectangular current loop is oriented such that its area vector A is completely parallel to the uniform magnetic field B, the torque is:
�� Torque depends on sinθ. �� Parallel vectors imply θ=0^∘. �� sin0^∘=0.
The torque on a current-carrying loop placed in a magnetic field is given by τ=NIABsinθ where θis the angle between the area vector (or magnetic moment vector) and the magnetic field. When the area vector is parallel to the magnetic field, θ=0^∘ Therefore, τ=NIABsin0^∘τ=0 This corresponds to a stable equilibrium position because the magnetic moment is aligned with the magnetic field. In this orientation, no turning effect acts on the loop. Hence, the torque is zero.
- �� Option A → Torque is maximum at θ=90^∘.
- �� Option C → NIAB is the maximum torque, not the torque at θ=0^∘.
- �� Option D → Torque cannot become infinite.
Formula Recall
- Application
- Substitute θ=0^∘into
- τ=NIABsinθ
- Final Logic
- τ=NIABsin0^∘=0
"Parallel Means Peaceful"
7 Which statement is correct regarding intrinsic and elementary magnetic moments?
�� Electrons possess intrinsic magnetic moments. �� Intrinsic moments arise from quantum properties. �� Magnetic monopoles have not been observed.
Classically, magnetic dipoles are associated with circulating currents. However, elementary particles such as electrons possess intrinsic magnetic moments that cannot be explained purely by classical current loops. These magnetic moments arise from quantum mechanical properties such as spin. Modern physics recognizes that electrons, protons, and several other particles possess intrinsic magnetic moments. Magnetic monopoles, on the other hand, have not been experimentally observed. Therefore, the statement regarding intrinsic magnetic moments of elementary particles is correct.
- �� Option B → Intrinsic magnetic moments exist in elementary particles.
- �� Option C → Magnetic monopoles have not been confirmed experimentally.
- �� Option D → Protons possess measurable magnetic moments.
NCERT Recall
- Application
- Recall NCERT's discussion on magnetic moments of elementary particles.
- Final Logic
- Magnetic moments can arise from both current loops and intrinsic particle properties.
"Electrons Carry Their Own Magnet"
8 A loop of irregular shape carrying current is placed in an external magnetic field. If the wire is perfectly flexible, it shapes into a circle fundamentally to:
�� A circle encloses maximum area for a given perimeter. �� Larger area gives larger magnetic flux. �� The system tends toward minimum energy.
For a given length of wire, a circle encloses the maximum possible area. When a flexible current-carrying loop is placed in an external magnetic field, magnetic forces act on different parts of the wire. The system tends to arrange itself in a configuration that maximizes the magnetic flux linked with the loop. Since magnetic flux is Φ=BAcosθ increasing the enclosed area increases the flux. Because a circle provides the maximum area for a fixed perimeter, the loop tends to assume a circular shape. This configuration is energetically favorable and is analogous to the behavior of soap films seeking configurations of minimum energy.
- �� Option A → The loop tends to maximize, not minimize, enclosed flux.
- �� Option C → Internal and external fields do not perfectly cancel.
- �� Option D → Physical systems generally seek stable configurations.
Concept Application
- Application
- Use the relationship between enclosed area and magnetic flux.
- Final Logic
- Fixed Perimeter → Maximum Area → Circle → Maximum Flux.
"Circle Captures Maximum Area"
9 Identify the correct statements regarding alignment of a magnetic dipole.
Statements:
1. When m is exactly parallel to B, the equilibrium is stable.
2. When m is exactly antiparallel to B, the equilibrium is unstable.
3. External work is required to rotate the coil from stable to unstable equilibrium.
4. The magnetic moment naturally rests at 90^∘.
�� Parallel alignment is stable. �� Antiparallel alignment is unstable. �� Energy must be supplied to move against the field.
The potential energy of a magnetic dipole in a magnetic field is U=-mBcosθ At θ=0^∘ the potential energy is minimum, corresponding to stable equilibrium. At θ=180^∘ the potential energy is maximum, corresponding to unstable equilibrium. To rotate the dipole from the stable position to the unstable position, external work must be done against the magnetic torque. The magnetic moment does not naturally remain at 90^∘, because that position is neither stable nor unstable equilibrium.
- �� Option A → Includes Statement 4.
- �� Option B → Includes Statement 4, which is incorrect.
- �� Option D → Includes Statement 4.
Concept Application
- Application
- Analyze potential energy and equilibrium conditions.
- Final Logic
- U=-mBcosθ
- Minimum Energy → Stable
- Maximum Energy → Unstable
"Parallel Peace, Antiparallel Peak"
10 For an object with magnetic moment m placed in a uniform field B, unstable equilibrium dictates:
�� Unstable equilibrium occurs at 180^∘. �� Torque is initially zero. �� Small disturbances move the dipole away.
For a magnetic dipole in a magnetic field, τ=mBsinθ At the unstable equilibrium position, θ=180^∘ Therefore, τ=mBsin180^∘=0 Although the torque is zero at this exact orientation, the position is unstable because the potential energy is maximum. Any small angular displacement produces a torque that drives the dipole farther away from the equilibrium position rather than restoring it. Hence, the defining characteristic of unstable equilibrium is that the initial torque is zero, but the system moves away under slight disturbance.
- �� Option A → Potential energy is maximum, not minimum.
- �� Option B → Unstable equilibrium occurs at 180^∘, not 90^∘.
- �� Option C → Unstable equilibrium does not produce restoring torque.
Formula Recall
- Application
- Use
- τ=mBsinθ
- and evaluate at θ=180^∘.
- Final Logic
- τ=0
- but
- U=+mB
- which is maximum.
"Zero Torque, Maximum Energy"
11 Match List I with List II for circular loop dipole distance metrics.
| List I | List II |
|---|---|
| 1. Field Bat large distance xxR | a. πR2 |
| 2. Field Bat exact center x0 | b. IπR2 |
| 3. Magnetic moment m | c. Proportional to 1/x3 |
| 4. Loop Area A | d. μ0I2R |
�� Far-field behaves like a dipole. �� Center field depends on loop radius. �� Magnetic moment equals current × area.
For a circular current loop, the magnetic field at a very large distance behaves like a magnetic dipole field and decreases according to B∝1/x^3 The magnetic field at the center of a circular loop is B=μ_0I/2R The magnetic moment of a single-turn circular loop is m=IA Since A=πR^2 we obtain m=IπR^2 Thus, the loop area corresponds to πR^2, the magnetic moment corresponds to IπR^2, the center field corresponds to μ_0I/2R, and the far-field varies as 1/x^3. Hence the correct matching is 1-c, 2-d, 3-b and 4-a.
- �� Option B → Interchanges center field and far-field behavior.
- �� Option C → Incorrectly matches magnetic moment and area.
- �� Option D → Multiple mismatches occur.
NCERT Recall
- Application
- Recall standard formulas for circular current loops.
- Final Logic
- A=πR^2,m=IπR^2,B_(center)=μ_0I/2R
"Area First, Moment Next"
12 Incorrect statement regarding magnetic dipole field formulations.
�� Magnetic monopoles have not been observed. �� Magnetic dipoles arise from currents. �� Magnetic fields are not produced by isolated magnetic charges.
Electric fields originate from electric charges, which are scalar sources. In contrast, magnetic fields are generated by moving charges and current distributions. No isolated magnetic monopoles have been experimentally confirmed. Therefore, magnetic fields cannot be described as arising from scalar sources identical to electric charge points. At large distances, a current loop behaves like a magnetic dipole, and many mathematical analogies exist between electrostatics and magnetism, including the correspondence between 1/ε_0 and μ_0. Hence, Statement C is incorrect.
- �� Option A → Correct theoretical description.
- �� Option B → Correct far-field approximation.
- �� Option D → Valid electromagnetic analogy.
Concept Comparison
- Application
- Compare the physical origins of electric and magnetic fields.
- Final Logic
- Electric Charges Exist Individually; Magnetic Monopoles Do Not.
"Charges Stand Alone, Poles Do Not"
13 The mechanical purpose of the internal core configuration in a moving coil galvanometer is:
�� Soft iron has high permeability. �� It strengthens the magnetic field. �� It helps create a radial field.
The cylindrical soft iron core inside a moving coil galvanometer concentrates magnetic flux because of its high magnetic permeability. This increases the magnetic field strength inside the instrument. Along with the specially shaped pole pieces, the core helps establish a radial magnetic field. A radial field ensures that the angle between the magnetic field and the plane normal remains fixed at 90^∘, thereby maintaining maximum torque. Hence, the internal core both strengthens the field and makes it radial.
- �� Option A → The core does not weaken the field.
- �� Option C → The field is radial, not axial.
- �� Option D → The field becomes stronger, not weaker.
NCERT Recall
- Application
- Recall the role of the soft iron cylindrical core.
- Final Logic
- Soft Iron Core ⇒ Stronger Field + Radial Geometry.
"Core Creates Strong Circular Control"
14 By integrating a carefully shaped cylindrical soft iron core, the magnetic field specifically becomes:
�� Radial fields maintain constant torque. �� θremains 90^∘. �� Sensitivity increases.
The torque on the galvanometer coil is τ=NIABsinθ The soft iron core and curved pole pieces create a nearly uniform radial magnetic field. Because of this arrangement, the magnetic field remains perpendicular to the area vector of the coil throughout its rotation. Therefore, θ=90^∘ at every position and sinθ=1 Consequently, τ=NIAB which ensures that torque is directly proportional to current. This produces a linear scale and high sensitivity, making the galvanometer suitable for accurate current measurement.
- �� Option B → The field becomes stronger, not weaker.
- �� Option C → Parallel orientation would make torque zero.
- �� Option D → The field remains magnetic, not electrostatic.
Formula Recall
- Application
- Use the torque equation and the concept of radial magnetic fields.
- Final Logic
- Radial Field ⇒ θ=90^∘⇒ sinθ=1.
"Radial Field, Full Torque"
15 Identify the correct statements concerning torsional counter-torque.
Statements:
1. The spring S provides a mechanical counter torque kϕ.
2. In equilibrium, kϕbalances the magnetic torque NIAB.
3. The constant k represents restoring torque per unit angular twist.
4. Torsional torque depends exponentially on the number of turns N.
�� Spring produces restoring torque. �� Equilibrium occurs when torques balance. �� k measures spring stiffness.
In a moving coil galvanometer, the suspension spring provides a restoring torque proportional to the angular deflection: τ_r=kϕ where k is the torsional constant and ϕis the angular deflection. The magnetic torque acting on the coil is τ_m=NIAB Under equilibrium conditions, NIAB=kϕ The torsional constant k physically represents the restoring torque produced per unit angular twist of the spring. Torsional torque does not depend exponentially on the number of turns N; instead, N affects the magnetic torque linearly. Therefore, Statements 1, 2 and 3 are correct while Statement 4 is incorrect.
- �� Option B → Includes Statement 4, which is incorrect.
- �� Option C → Includes Statement 4.
- �� Option D → Includes Statement 4.
Formula Recall
- Application
- Apply the equilibrium condition
- NIAB=kϕ
- for the galvanometer.
- Final Logic
- Restoring Torque = Magnetic Torque
- kϕ=NIAB
"Spring Stops the Swing"
16 At steady angular deflection ϕ, the balancing equation
kϕ=NIAB
allows the circuit current I to be directly computed as:
�� Equilibrium occurs when magnetic torque equals restoring torque. �� Rearrange the galvanometer equation. �� Current is proportional to deflection.
For a moving coil galvanometer operating in equilibrium, NIAB=kϕ where N is the number of turns, A is the coil area, B is the magnetic field strength, k is the torsional constant and ϕis the angular deflection. Rearranging for current, I=kϕ/NAB or I=(k/NAB)ϕ This equation shows that current is directly proportional to deflection. The proportionality constant depends on the physical construction of the galvanometer. Hence, the correct expression for current is Option D.
- �� Option A → Incorrect algebraic rearrangement.
- �� Option B → Represents current sensitivity, not current.
- �� Option C → Incorrect dimensional form.
Formula Rearrangement
- Application
- Rearrange
- NIAB=kϕ
- to isolate I.
- Final Logic
- I=kϕ/NAB
"Current Comes by Dividing NAB"
17 M₁ (N=30, A=3.6×10^(-3)m^2, B=0.25T)and M₂ (N=42, A=1.8×10^(-3)m^2, B=0.50T)have identical springs. The current sensitivity ratio M_2/M_1 is:
�� Current sensitivity =NAB/k. �� Springs are identical, so k cancels. �� Compare NAB values.
Current sensitivity is S_i=NAB/k Since both galvanometers have identical springs, k is the same. Therefore, S_(i2)/S_(i1)=N_2A_2B_2/N_1A_1B_1 Substituting values, =42×1.8×10^(-3)×0.50/30×3.6×10^(-3)×0.25=37.8×10^(-3)/27×10^(-3)=1.4 Hence, M_2/M_1=1.4
- �� Option A → Reciprocal-type error.
- �� Option B → Ignores parameter differences.
- �� Option D → Overestimates the ratio.
Substitution
- Application
- Use the current sensitivity formula and cancel common factors.
- Final Logic
- S_(i2)/S_(i1)=42×1.8×0.50/30×3.6×0.25=1.4
"Current Sensitivity Depends on NAB"
18 Continuing from the previous question, if the respective internal coil resistances are R_1=10Ωand R_2=14Ω, the voltage sensitivity ratio M_2/M_1 is:
�� Voltage sensitivity =NAB/kR. �� Use the previous current sensitivity ratio. �� Include resistance ratio.
Voltage sensitivity is S_v=ϕ/V=NAB/kR Thus, S_(v2)/S_(v1)=N_2A_2B_2/N_1A_1B_1×R_1/R_2 From the previous question, N_2A_2B_2/N_1A_1B_1=1.4 Therefore, S_(v2)/S_(v1)=1.4×10/14=1.4×0.714≈1.0 Hence, the voltage sensitivities are equal.
- �� Option A → Uses only resistance ratio.
- �� Option C → Uses only current sensitivity ratio.
- �� Option D → Overestimates the result.
Substitution
- Application
- Combine current sensitivity ratio with resistance ratio.
- Final Logic
- 1.4×10/14=1
"Voltage Sensitivity = Current Sensitivity ÷ Resistance"
19 Identify the correct statements regarding construction of an ammeter.
Statements:
1. A shunt resistance r_s is connected in parallel.
2. The value of r_s is much smaller than R_G.
3. The effective resistance becomes slightly less than r_s.
4. It maximizes the voltage drop across the measuring point.
�� Ammeters need low resistance. �� A small shunt is connected in parallel. �� Effective resistance becomes very small.
To convert a galvanometer into an ammeter, a very small resistance called a shunt resistance is connected in parallel with the galvanometer coil. Because the shunt resistance is much smaller than the galvanometer resistance, most of the current bypasses the galvanometer and flows through the shunt. The equivalent resistance of two parallel resistors is less than the smaller resistor, making the total resistance slightly less than r_s. This ensures that the ammeter introduces minimal resistance into the circuit. An ammeter should minimize voltage drop rather than maximize it. Therefore, Statements 1, 2 and 3 are correct, while Statement 4 is incorrect.
- �� Option B → Includes Statement 4.
- �� Option C → Includes Statement 4.
- �� Option D → Includes Statement 4.
Concept Application
- Application
- Recall the purpose of shunt resistance in an ammeter.
- Final Logic
- Small Parallel Shunt ⇒ Very Low Effective Resistance.
"Ammeter Adds a Tiny Alternate Path"
20 Match List I with List II for ideal measuring instruments.
| List I | List II |
|---|---|
| 1. Ideal Ammeter Internal Resistance | a. Infinity |
| 2. Ideal Voltmeter Internal Resistance | b. Zero |
| 3. Ammeter Connection | c. In Series with Target Component |
| 4. Voltmeter Connection | d. In Parallel with Target Component |
�� Ideal ammeter should not oppose current. �� Ideal voltmeter should draw no current. �� Connection methods differ.
An ideal ammeter should measure current without affecting the circuit. Therefore, its internal resistance should be zero. Since current must pass through the instrument, it is connected in series with the circuit element. An ideal voltmeter should measure potential difference without drawing current. Therefore, its internal resistance should be infinite. Since it measures voltage across a component, it is connected in parallel with that component. Thus the correct matching is: 1-b,2-a,3-c,4-d
- �� Option B → Reverses ammeter and voltmeter properties.
- �� Option C → Incorrect connection methods.
- �� Option D → Incorrect resistance assignments.
NCERT Recall
- Application
- Recall the characteristics of ideal measuring instruments.
- Final Logic
- Ammeter → Zero Resistance → Series
- Voltmeter → Infinite Resistance → Parallel
"Ammeter Along, Voltmeter Across"
