Exercise Short Questions of Current Electricty, Chapter 13 of 2nd Year Physics

 Q # 1. A potential difference is applied across the ends of a copper wire. What is the effect on the drift velocity of free electrons by (i) Increasing the potential difference (ii) Decreasing the length and the temperature of the wire.

Ans. The drift velocity $V_{d}$ of electrons in a conductor is described by the formula:

$$V_{d} = \frac{\mathrm{\Delta}V}{ne\rho L}$$

Where $\mathrm{\Delta}V$ is the potential difference between the ends of conductor, $L$ is the length of conductor and $\rho$ is the resistivity of wire. From equation, it is clear that

Drift velocity of electron increases with increase in potential   difference

Drift velocity of electron also increases by decreasing the length and temperature of wire.

Q # 2. Do bends in a wire affect its electrical resistance? Explain.

Ans. The resistance of the conductor is described by the formula:

$$R = \rho\frac{L}{A}$$

Where L is the length and A is the cross-section area of conductor.$\rho$the electrical resistivity of the material which depends upon the nature of conductor.

Hence the resistance of conductor depends upon the geometry and nature of conductor. Hence the bends in conducting wires don't affect its electrical resistance.

Q # 3. What are the resistances of the resistors given in the figure A and B. What is the tolerance of each? Explain what is meant by the tolerance.

Tolerance:

Tolerance means the possible variation in the value of resistance from the marked value. For example, a 1000 Ω resistance with a tolerance of 10\% can have an actual resistance between 900 Ω and 1100 Ω.

Q # 4. Why does the resistance of conductor rise with temperature?

Ans. The resistance offered by a conductor to the flow of electric current is due to collisions which the free electrons encounter with atoms of the lattice. As the temperature of the conductor rises, the amplitude of vibration of atoms increases and hence the probability of their collision with free electrons also increases which result increase of resistance of conductor.

Q # 5. What are the difficulties in testing whether the filament of a lighted bulb obeys ohm's law?

Ans.  Ohm's law states that the current flowing through the conductor is directly proportional to the potential difference applied across its ends provided that the temperature of the conductor remains constant. In case of a lighted bulb, the temperature of the filament increases with the passage of current through it. Hence the Ohm's law can't be applied to filament bulb.

Thus the main difficulty in testing whether the filament of a lighted bulb obeys ohm's law is the change in temperature with the flow of current in it.

Q # 6. Is the filament resistance lower or higher in a 500 W, 220 V bulb than in a 100 W, 220 V bulb?}

Ans. We know that

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

The resistance of filament of 500 W, 220 V bulb is:

$$R = \frac{V^{2}}{P} = \frac{(220)^{2}}{500} = 98.6\ \Omega$$

The resistance of filament of 100 W, 220 V bulb is:

$$R = \frac{V^{2}}{P} = \frac{(220)^{2}}{100} = 484\ \Omega$$

It is clear that the filament resistance is lowered in a 500 W, 220 V bulb than 100 W, 220 V bulb.

Q # 7. Describe a circuit which will give a continuously varying potential.

Ans. To use rheostat as potential divider, potential difference $V$ is applied across the fixed ends A and B of rheostat with the help of a battery. If $R$ is the resistance of the wire AB, the current I passing through is given by:

$$I = \frac{V}{R}$$

The potential difference between the portion BC of the wire AB is given by:

$$V_{BC} = current \times resistance$$

$$V_{BC} = \frac{V}{R} \times r = \frac{r}{R}V$$

Where $r$ is the resistance of the portion BC of wire. The equation shows that this circuit can provide potential difference at output terminal varying from zero to the full potential difference of the battery depending on the position of sliding contact.

Q # 8. Explain why the terminal potential difference of a battery decreases when current drawn from it is increases.

Ans. The terminal potential difference $V_{t}$ of the battery of emf $\varepsilon$ is described by the formula:

$$V_{t} = \varepsilon - Ir$$

Where $r$ is the internal resistance of the battery and I is the current flowing through outer circuit.

It is clear from equation that when I is large, the factor $Ir\ $becomes large and $V_{t}$ becomes small. Hence terminal potential difference of a battery decreases when current drawn from it is increased.

Q # 9. What is Wheatstone bridge? How can it be used to determine an unknown resistance?

Ans. It is an electrical circuit which can be used to find the unknown resistance of a wire. The circuit of Wheatstone bridge is shown in the figure.

It consist of four resistance connected in the form of a mesh, galvanometer, battery and a switch. When the bridge is balanced, it satisfies the following relation:

$$\frac{R_{1}}{R_{2}} = \frac{R_{3}}{R_{4}} \Longrightarrow R_{4} = \frac{R_{2} \times R_{3}}{R_{1}}$$

If the values of $R_{1}$,$R_{2}$,$R_{3}$ are known, then $R_{4}$ can be calculated, provided the bridge is balanced.