HC Verma Questions and Solutions: Chapter 32: Electric Current in Conductors- 1 | HC Verma Solutions - JEE PDF Download

If the number of collisions of the free electrons with the lattice is decreased, then the drift velocity of the electrons increases.
Current i is directly proportional to the drift velocity 'V_d' and is given by the following relation:
i = neAV_d , where 'n'  is the number density of electrons and 'A' is the area of the cross-section of the conductor.
So, we can easily see that current increases with increase in drift velocity.

The resistance R of a conductor depends on its resistivity ρ, length l and cross-sectional area A. Thus, 
From the given comparison of resistances, we cannot derive the correct relation between the resistances. We also need to know the cross-sectional areas and the lengths of both the conductors before concluding about their resistivities. Only then can the relation between the resistivities be found.
The information is not sufficient to find the relation between ρ_A and ρ_B.

The relation between the resistivity ρ and conductivity σ of a material is given by

The product of conductivity and resistivity is unity. So, this product does not depend on any of the given quantities.

As the temperature of a metallic resistor is increased, the product of its resistivity and conductivity remains constant.

The product of resistivity and conductivity is equal to a constant.

In the study of electric current, the direction opposite the flow of electrons is regarded as the direction of flow of positive charge. In a battery, positive charge flows from the negative terminal to the positive terminal when it is discharging (connected to external circuit). But when the battery is charged, the positive charge flows from the positive terminal to the negative terminal.

As the resistance is connected to an ideal battery, it provides a constant potential difference across the two terminals. The internal resistance of the battery is also zero.
The power dissipated in the resistor,

V is constant; hence 
Thus, if the value of the resistance is decreased, the power dissipated in the resistor will increase.

The metal ions are bound at their positions and vibrate due to collisions with electrons and due to thermal energy. The conduction electrons are free to move. They get energy from the electric field set inside the conductor by connection with a battery and due to thermal motion. The velocity of the electrons is high.
Thus, the kinetic energy of the electrons is greater than the kinetic energy of the metal ions.

Thermal energy developed across a resistor,
U = i²Rt ,
where i is the current flowing through the resistor of resistance R for time t. Since the resistors are connected in series, the current flowing through both the resistors is same and the time for which the current flows is also same.
Thus, the ratio of the thermal energy developed in R and 2 R is 1 : 2.

Thermal energy developed in the resistances,

Heat developed in the resistance R, 
Heat developed in the resistance 2R, 
Thus, Heat developed in the resistance R and 2R are in ratio 2 : 1.

Resistance of a wire is directly proportional to its length.
So, when we cut the wire into 5 equal parts, the resistance of each part will be 10 Ω.
On connecting these wires in parallel, the net resistance will be

According to Kirchhoff's junction law, the net charge coming towards a point should be equal to the net charge going away from that point in the same time. It follows from the principle of conservation of charge.
The loop law follows from the fact that electrostatic force is a conservative force and the work done by it in any closed path is zero.

Let the emfs of the batteries be e₁ and e₂, and their respective resistances be r₁ and r₂.
Since the batteries are connected in series, the equivalent emf will be the sum of the emf of the two batteries (e = e₁ + e₂).
Thus, e > e₁ and  e > e₂
Thus, the equivalent emf is larger than either of the two emfs. Hence, statement A is correct.
Since the batteries are connected in series, the equivalent internal resistance (r) of the combination will be the sum of the internal resistance of the two batteries ( r = r₁ + r₂).
r > r₁ and r > r₂
Thus, the equivalent internal resistance is greater that either of the two resistances. Hence, statement B is wrong.

The equivalent emf ∈₀, when two non-ideal batteries of emfs ∈₁ and ∈₂ and internal resistances r₁ and r₂ respectively are connected in parallel to each other, is given by:-

Thus, the equivalent emf is greater than either of the emfs. Thus, statement A is wrong.
For the parallel combination of the batteries of internal resistance r₁ and r₂, the equivalent internal resistance r is given as 
Thus, the value of the resultant resistance is even smaller than either resistance.

The ammeter is connected in series with the circuit to measure the net amount of current flowing through the circuit. If the net resistance of the ammeter is high, then the amount of current that the circuit draws will have error. Hence, we cannot accurately measure the amount of current that the circuit draws from the voltage source. Hence, the net resistance or the equivalent resistance of an ammeter should be small enough to ensure that it does not appreciably change the current to be measured.

To measure potential difference across any circuit element, the voltmeter is connected in parallel to that circuit element. Let R_eq be the equivalent resistance of the voltmeter and V be the potential difference across the voltmeter.
Then, the current through the voltmeter, i = V/R_eq
Hence, the deflection in the voltmeter is proportional to the current i and, hence, proportional to V. However, when the voltmeter is used in a circuit, its resistance R_eq is connected in parallel to some circuit element. This might change the overall resistance of the circuit and, hence, the current. Consequently, the potential difference to be measured is also changed. To minimise the error due to this, the equivalent resistance R_eq of the voltmeter should be large. When a large resistance is connected in parallel to a small resistance, the equivalent resistance is only slightly less than the smaller one.

The charge Q on the plates of a capacitor at time t after connecting it in a charging circuit, 
where
ε = emf of the battery connected in the charging circuit
C = capacitance of the given capacitor
R = resistance of the resistor connected in series with the capacitor
The charge developed on the plates of the capacitor in first 10 mili seconds is given by

The charge developed on the plates of the capacitor in first 20 mili seconds is given by

The charge developed at the plates of the capacitor in the interval t = 10 mili seconds to 20 mili seconds is given by

For second part of the question
10 μC charge be deposited in a time interval t₁ and the next 10 μC charge is deposited in the next time interval t₂.
The time taken for 10 μC charge to develop on the plates of the capacitor is t₁

The time taken for 20 μC charge to develop on the plates of the capacitor is t'

Dividing (2) by (1)

Taking natural log both side,

Let the potentials at A and B be V_A  and V_B.
As potential, E = -(dV/dr) potential increases in the direction opposite to the direction of the electric field.
Thus, V_A < V_B
Potential energy of the electrons at points A and B:-
U_A = -eV_A
U_B = -eV_B
Thus, U_A > U_B
Let the kinetic energy of an electron at points A and B be K_A and K_B respectively.
Applying the principle of conservation of mechanical energy, we get:-
U_A  + K_A  = U_B  + K_B
As, U_A > U_B,
K_A < K_B
Therefore, the speed of the electrons is more at B than at A.

As the capacitor is connected to the battery at t = 0, current flows through the wire up to the time the capacitor is charged. As the capacitor is completely charged, the potential difference across the capacitor is equal to the terminal potential of the battery. So, the current stops flowing and does not pass through the point between the plates of the capacitor, as there is no medium for the flow of charge.

When no current is passed through a conductor, the average of the velocities of all the free electrons at an instant is zero.

As the resistor connected to the battery is heated, the thermal energy of the electrons increases. Thus, the relaxation time of the electrons will decrease and so will the drift velocity. Resistivity is inversely proportional to the relaxation time; thus, resistivity will increase with decrease in relaxation time. Therefore, resistance will also change. The number of electrons in a resistor will remain same.

Resistivity (ρ) of a conductor is the reciprocal of the its conductivity(σ). Thus,

The ratio of the resistivity and the conductivity,

As resistivity increases with temperature, the square of resistivity will also increase. Hence, the ratio of resistivity and conductivity will increase with increase in temperature.

Drift speed and current density are inversely proportional to the area of cross-section of a wire. Thus, they are dependent on the cross-section. The charge crossing in a given time interval is independent of the area of cross-section of the wire. Free electron density is the total number of free electrons per unit volume of the wire. The density of free electrons depends on the distribution of the free electrons throughout the volume of the wire. It does not depend on the cross-section of the wire.

The ammeter is connected in series in the circuit whose current is to be measured. If the net resistance of the ammeter is high, then the amount of current that the circuit draws will have error . So, we won't be able to accurately measure the amount of current that the circuit draws from the voltage source. Thus, it should have a small resistance.
To measure the potential difference across any circuit element, the voltmeter is connected in parallel to that circuit element. If the resistance of the voltmeter is large, then maximum voltage drop occurs across the voltmeter and it will measure the correct value of the potential.

The charge (Q) on the capacitor at any instant t,

where
C = capacitance of the given capacitance
R =  resistance of the resistor connected in series with the capacitor
RC = (10 × 10³) × (500 × 10^-6) = 5 s
The charge on the capacitor in the first 5 seconds,

The charge on the capacitor in the first 10 seconds,

Charge developed in the next 5 seconds,
Q' = Q₁ - Q₀
Q' =  CV(0.864 - 0.632) = 0.232 CV
The charge on the capacitor in the first 55 seconds,

Charge developed in the next 50 seconds,
Q' = Q₂ - Q₀
Q' =  CV(0.99 - 0.632) = 0.358 CV
Charge developed in the first 505 seconds,

Charge developed in the next 500 seconds,
Q' = CV (1 - 0.632) = 0.368 CV
Thus, the charge developed on the capacitor in the first 5 seconds is greater than the charge developed in the next 5,50, 500 seconds.

Let the voltage of the battery connected to the capacitors be V. Both the capacitors will charge up to the same potential (V).
The charge on the capacitors C₁ is C₁V = (1 μF)×V
The charge on the capacitors C₂ is C₂V = (2 μF)×V
The charge on the discharging circuit at an instant t,

The current through the discharging circuit,

At t = 0, the current through the discharging circuit will be V/R for both the capacitors.
Let the time taken by the capacitor C₁ to lose 50% of the charge be t₁.

Taking natural log on both sides, we get:-

Similarly,
Time taken for capacitor C₂:-

As, C₁ < C₂, t₁ < t₂
Thus, we can say that C₁ loses 50% of its initial charge sooner than C₂ loses 50% of its initial charge.