# Questions OBJECTIVE - I and Answer Electric Current in Conductors HC Verma Part II

Q#1

A metallic resistor is connected across a battery. If the number of collisions of the free electrons with the lattice is somehow decreased in the resistor (for example, by cooling it), the current will

(a) increase

(b) decrease

(c) remain constant

(d) become zero.

Answer: (a)

If the number of collisions of free electrons with the lattice is decreased in the resistor, the average time between the successive collisions increases. Hence the drift speed increases. Thus the conductivity increases and so does the current. Hence the option (a) is true.

Q#2

Two resistors A and B have resistances Rₐ and Rᵦ respectively with Rₐ < Rᵦ. The resistivities of their materials are ρₐ and ρᵦ.

(a) ρₐ > ρᵦ

(b) ρₐ = ρᵦ

(c) ρₐ < ρᵦ

(d) The information is not sufficient to find the relation between ρₐ and ρᵦ.

Answer: (d)

The resistivity of a material is given as,

$\rho=\frac{RA}{l}$

From this, it is clear that the resistivity of the material of a given resistor is not only dependent on the resistance but on its cross-sectional area and length. Since the information of cross-sectional area and lengths of the given resistors are not given, the relation between ρₐ and ρᵦ can not be found. Hence the option (d) is true.

Q#3

The product of resistivity and conductivity of a cylindrical conductor depends on

(a) temperature

(b) material

(c) area of cross-section

(d) none of the above.

Answer: (d)

The resistivity of a given conductor is inverse of its conductivity. i.e. $\rho=\frac{1}{\sigma}$,

where ρ = resistivity and σ = conductivity.

From this relation, it follows that

σρ = 1.

So the product of resistivity and conductivity is always 1 without any condition. So the option (d) is true.

Q#4

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

(a) increases

(b) decreases

(c) remains same

(d) may increase or decrease

Answer: (c)

Same explanation as in above problem number 3. Their product is always constant and equal to 1. Hence the option (c) is correct.

Q#5

In an electric circuit containing a battery, the charge (assumed positive) inside the battery

(a) always goes from the positive terminal to the negative terminal

(b) may go from the positive terminal to the negative terminal

(c) always goes from the negative terminal to the positive terminal

(d) does not move.

Answer: (b)

In an electric circuit the battery is either in a discharging state or in a charging state. In a discharging state, the battery supplies charge (positive) in the circuit. The charge in this case goes from negative terminal to positive terminal inside the battery.

In the charging state, the charge (positive) goes from the positive terminal to the negative terminal inside the battery. Hence the option (b) is true.

Q#6

A resistor of resistance R is connected to an ideal battery. If the values of R is decreased, the power dissipated in the resistor will

(a) increase

(b) decrease

(c) remain unchanged

Answer: (a)

Since the battery is ideal, there is no internal resistance and the potential difference across the resistor is constant and equal to emf V of the battery.

The power dissipated in the resistor,

P = $\frac{V^2}{R}$

Since V is constant, the power dissipated P is inversely proportional to R. Here we are decreasing R, hence P will increase. Option (a) is true.

Q#7

A current passes through a resistor. Let K₁ and K₂ represent the average kinetic energy of the conduction electrons and the metal ions respectively.

(a) K₁ < K₂

(b) K₁ = K₂

(c) K₁ > K₂

(d) Any of these three may occur.

Answer: (c)

The conduction electrons are the free electrons that randomly collide with the metal ions of the conductor and drift opposite to the electric field. The metal ions simply vibrate at their mean positions depending on the temperature. So, the average velocity of a metal ion is much less than the average velocity of a conduction electron. On the other side, the mass of a metal ion is much more than an electron.

Now the kinetic energy is directly proportional to mass but it is directly proportional to the square of the velocity. Since an electron has a much greater velocity, the average kinetic energy of the conduction electrons is greater than the metal ions. Option (c) is correct.

Q#8

Two resistors R and 2R are connected in series in an electric circuit. The thermal energy developed in R and 2R are in the ratio

(a) 1 : 2

(b) 2 : 1

(c) 1 : 4

(d) 4 : 1

Answer: (a)

Since the two resistors are connected in series, the same current will go through them. Thermal energy developed per unit time is given as,

P = i²R, in the first resistor.

Thermal energy developed per unit time in the second resistor,

P' = i²(2R) = 2i²R.

The ratio of thermal energy developed,

P : P' =i²R : 2i²R = 1 : 2.

Hence option (a) is correct.

Q#9

Two resistors R and 2R are connected in parallel in an electric circuit. The thermal energy developed in R and 2R are in the ratio

(a) 1 : 2

(b) 2 : 1

(c) 1 : 4

(d) 4 : 1

Answer: (b)

Since the resistors are connected in parallel, the potential difference across each resistor is the same. Let this potential difference be V.

The thermal energy developed per unit time in the first resistor,

P = $\frac{V^2}{R}$

The thermal energy developed per unit time in the second resistor,

P' = $\frac{V^2}{2R}$

Hence the ratio of these two energies,

P : P' = $\frac{V^2}{R}$ : $\frac{V^2}{2R}$ = 2 : 1.

Hence option (b) is true.

Q#10

A uniform wire of resistance 50 Ω is cut into 5 equal parts. These parts are now connected in parallel. The equivalent resistance of the combination is

(a) 2 Ω

(b) 10 Ω

(c) 250 Ω

(d) 6250 Ω

Answer: (a)

Since the wire is of uniform cross-section, its resistance per unit length will be constant. The resistance of each part,

r = $\frac{50}{5}$ = 10 Ω.

When these parts are all connected in parallel, the equivalent resistance R is given as,

$\frac{1}{R}=\frac{1}{r}+\frac{1}{r}+\frac{1}{r}+\frac{1}{r}=\frac{5}{r}$

$R=\frac{r}{5}=\frac{10}{5}$ = 2 Ω.

Hence option (a) is true.

Q#11

Consider the following two statements:

(A) Kirchhoff's junction law follows from conservation of charge

(b) Kirchhoff's loop law follows from the conservative nature of the electric field.

(a) Both A and B are correct

(b) A is correct but B is wrong

(c) B is correct but A is wrong

(d) Both A and B are wrong.

Answer: (a)

Kirchhoff's Junction law states that the sum of all the currents directed towards a point in a circuit is equal to the sum of all the currents directed away from the point. This law is based on the conservation of charge. The total charge going towards the junction is itself coming out of the junction. Only the charges pass through the point. Hence A is correct.

Kirchhoff's loop law states that the algebraic sum of all the potential differences along a closed loop in a circuit is zero. This law follows from the fact that electrostatic force (here the electric field) is a conservative force and the work done by it in any closed path is zero. Statement B is also true.

The given option (a) is correct.

Q#12

Two nonideal batteries are connected in series. Consider the following statements:

(A) The equivalent emf is larger than either of the two emfs.

(B) The equivalent internal resistance is smaller than either of the two internal resistances.

(a) Each of A and B are correct

(b) A is correct but B is wrong

(c) B is correct but A is wrong

(d) Each of A and B is wrong.

Answer: (b)

Since the batteries are connected in series, equivalent emf is the sum of both emf. Hence the equivalent emf will be larger than either of the two emfs. Statement A is correct.

The internal resistances of the batteries are also in series. Hence the equivalent internal resistance will also be larger than either of the two internal resistances. Statement B is wrong.

Hence option (b) is true.

Q#13

Two nonideal batteries are connected in parallel. Consider the following statements:

(A) The equivalent emf is smaller than either of the two emfs.

(B) The equivalent internal resistance is smaller than either of the two internal resistances.

(a) Both A and B are correct

(b) A is correct but B is wrong

(c) B is correct but A is wrong

(d) Both A and B is wrong.

Answer: (c)

Let the emf and internal resistances be Ɛ₁, r₁ and Ɛ₂, r₂. When connected in parallel the equivalent emf Ɛ is given as,

$\epsilon=\frac{\epsilon_1r_2+\epsilon_2r_1}{r_1+r_2}$

and equivalent internal resistance as,

$r=\frac{r_1r_2}{r_1+r_2}$

As we can see, r is smaller than either r₁ or r₂. So statement B is correct. Now let us check for Ɛ. We find out the expression for $\epsilon-\epsilon_1$,

= $\frac{r_1r_2}{r_1+r_2}-\epsilon_1$

= $\frac{(\epsilon_2-\epsilon_1)r_1}{r_1+r_2}$

If Ɛ₂ > Ɛ₁, then Ɛ is greater than Ɛ₁. But in case of Ɛ₁ > Ɛ₂, Ɛ is smaller than Ɛ₁. So, the statement A is not true.

Hence option (c) is correct.

Q#14

The net resistance of an ammeter should be small to ensure that

(a) it does not get overheated

(b) it does not draw excessive current

(c) it can measure large currents

(d) it does not appreciably change the current to be measured.

Answer: (d)

An ammeter is connected in series in a circuit to measure the current. So the total current to be measured goes through it. The equivalent resistance in the circuit will be sum of the net resistance of the ammeter and the resistance of the circuit. Now the current in the circuit after the connection of the ammeter,

i' = $\frac{V}{R+r}$, where R is the resistance of the circuit and r is the net resistance of the ammeter.

Current in the circuit without ammeter,

i = $\frac{V}{R}$.

Hence r is kept small so that i ≈ i'. Option (d) is correct.

Q#15

The net resistance of a voltmeter should be large to ensure that

(a) it does not get overheated

(b)it does not draw excessive current

(c) it can measure large potential differences

(d) it does not appreciably change the potential difference to be measured.

Answer: (d)

A voltmeter is used to measure the potential difference across a circuit element and it is connected in parallel to that circuit element. Suppose the resistance of the circuit element = R and current through it = i. The potential difference across it without the voltmeter, V = iR. When a voltmeter with resistance r is connected in parallel, the equivalent resistance of the these two

R' = $\frac{rR}{r+R}$. If the potential difference across the element now = V' then

V' =i'R, where i' is the current in the element after connection of the voltmeter. To keep V ≈ V', we need to keep i ≈ i'. It can be only be achieved if a very small part of i is diverted through the voltmeter by keeping its resistance high. R' is slightly smaller than R. And it does not appreciably change the potential difference across the element.

Option (d) is correct.

Q#16

Consider a capacitor charging circuit. Let Q₁ be the charge given to the capacitor in a time interval of 10 ms and Q₂ be the charge given in the next time interval of 10 ms. Let 10 µC charges be deposited in a time interval t₁ and the next 10 µC charges are deposited in the next time interval t₂.

(a) Q₁ > Q₂, t₁ > t₂

(b) Q₁ > Q₂, t₁ < t₂

(c) Q₁ < Q₂, t₁ > t₂

(d) Q₁ < Q₂, t₁ < t₂.

Answer: (b)

Q₁ = ƐC(1 – e – $\frac{10}{CR}$)

Q₁ + Q₂ = ƐC(1 – e – $\frac{20}{CR}$)

Subtracting former from later,

Q₂ =ƐC(e – $\frac{10}{CR}$ – e – $\frac{20}{R}$)

= ƐC(1 – e – $\frac{10}{CR}$)e – $\frac{10}{CR}$

=Q₁e – $\frac{10}{CR}$

Q₁ = $\frac{10Q_2e}{CR}$

Since $\frac{10}{CR}$ is positive, the value of $\frac{10e}{CR}$ > 1, Hence,

Q₁ > Q₂.

Now for the second case,

10 = ƐC(1 – e – $\frac{t_1}{CR}$ (i)

and 20 = ƐC(1 – e – $\frac{t_1+t_2}{CR}$) (ii)

From (i) 1 – e – $\frac{t_1}{CR}$ = $\frac{10}{\epsilon C}$

→e-t₁/CR =(ƐC-10)/ƐC

→-t₁/CR =ln{(ƐC-10)/ƐC}

→t₁/CR =ln{ƐC/(ƐC-10)} ----(iii)

From (ii)

→1-e-(t₁+t₂)/CR =20/ƐC

→e-(t₁+t₂)/CR =(ƐC-20)/ƐC

→-(t₁+t₂)/CR =ln{(ƐC-20)/ƐC}

→(t₁+t₂)/CR =ln{ƐC/(ƐC-20)} --(iV)

Subtract (iii) from (iv)

t₂/CR =ln{ƐC/(ƐC-20)}-ln{ƐC/(ƐC-10)}

→t₂/CR =ln{(ƐC-10)/(ƐC-20)}

Now,

(t₂-t₁)/CR =ln{(ƐC-10)/(ƐC-20)}-ln{ƐC/(ƐC-10)}

= ln{(ƐC-10)²/(ƐC-20)ƐC}

= ln{(Ɛ²C²-20ƐC+100)/(Ɛ²C²-20ƐC)}

= ln (A+100)/A

Where A =Ɛ²C²-20ƐC,

ln (A+100)/A >1 and positive

Hence, t₂-t₁ > positive since CR is positive

Thus, t₁ < t₂,

Option (b) is correct.

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