# Questions OBJECTIVE - I and Answer Thermal and Chemical Effects of Electric Current HC Verma Part II

Q#1.

Which of the following plots may represent the thermal energy produced in a resistor in a given time as a function of electric current?

The thermal energy produced in a resistor in a given time t is, U = i²Rt

Since the R and t are given i.e. constant, the thermal energy U is proportional to square of the current. Hence the curve between U and i is represented as,

U = a.i²   {a = Rt}

which is in the form y =ax² that is the equation of a parabola. In the given figure plot (a) best represents the given condition. Option (a) is correct.

Note: Students should not confuse between the curve (a) and (d) because (d) does not represent y = ax². The reason is, (d) intersects y = x (represented by b) only at origin while the curve y = ax² should intersect y = x at two points, at origin and at (1/a, 1/a). You can find these two points by solving y = x and y = ax².

Q#2.

A constant current is passed through a resistor. Taking the temperature coefficient of resistance into account, indicate which of the plots shown in figure (33-Q3) best represents the rate of production of thermal energy in the resistor.

Thermal energy dU produced in the resistor in time interval dt,

dU = i²R.dt

Taking the thermal coefficient of resistance into account,

dU = i²R(1 + αT)dt, {T is the temperature differce} .

Here i, R and α are constants.

The rate of production of thermal energy, dU/dt = i²R(1 + αT

dU/dt = i²Ri²RαT

dU/dt = m'T + C

{where m' = i² and C = i²R}

dU/dt = mt + C                       (i)

{Since T is proportional to t; It can be shown as → i²rt = MsT', where r is resistance, M = mass and s = specific heat of the resistor. m is another constant of proportionality}

(i) is an equation of straight line which has an intersept on the dU/dt axis = C.

So the plot is a straight line. The slope of the straight line m is positive hence the plot (d). Option (d) is correct.

Q#3.

Consider the following statements regarding a thermocouple.

(A) The neutral temperature does not depend on the temperature of the cold junction.

(B) The inversion temperature does not depend on the temperature of the cold junction.

(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.

It has been found that the neutral temperature does not depend on the temperature of the cold junction. But the inversion temperature decreases by x°C where the temperature of the cold junction = 0°C + x°C = x°C. So, A is correct and B is wrong. Option (b) is correct.

Q#4.

The heat developed in a system is proportional to the current through it.

(a) It cannot be Thomson heat.

(b) It cannot be Peltier heat.

(c) It cannot be Joule heat.

(d) It can be any of the three heats mentioned above.

The Joule heat is always proportional to the square of the current through the system. Hence it cannot be Joule heat. Option (c) is correct.

Q#5.

Consider the following two statements.

(A) Free-electron density is different in different metals.

(B) Free-electron density in a metal depends on temperature.

Seebeck effect is caused,

(a) due to both A and B

(b) due toA but not due to B

(c) due to B but not due to A

(d) neither due to A nor due to B.

Both statements are true to explain the Seebeck effect. Only the fact in A can not cause a current in a thermocouple because the direction of emf at both ends will be equal and opposite. It is also due to the fact in statement B that a net emf is available in the circuit to drive the charges. Option (a) is correct.

Q#6.

Consider statements A and B in the previous question. Peltier effect is caused

(a) due to both A and B

(b) due to A but not due to B

(c) due to B but not due to A

(d) neither due to A nor due to B.

It is due to statement A only. Since the free electron density is different in different materials, a charge has to pass from low density to high density at one and from high to low at the other end when a current is passed through a thermocouple. So, the work done by the electric field on the charge inside the conductors at one end is positive while on the other end the work done is negative. Hence one end is hot and the other is cold. Peltier effect is not caused by the fact in statement B.

Q#7.

Consider statements A and B in question 5. Thomson effect is caused

(a) due to both A and B

(b) due to A but not due to B

(c) due to B but not due to A

(d) neither due to A nor due to B.

The fact in statement A is for different metals while the Thomson effect is related to emf produced in a metal wire with the nonuniform temperature at different sections. Hence option (c) is true.

Q#8.

(a) depends on the amount of the electrolyte.

(b) depends on the current on the electrolyte

(c) is a universal constant

(d) depends on the amount of charge passed through the electrolyte.