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 41. When a motor car is started, its light becomes slightly dim, why?   The potential difference across battery having internal resistance r is gives as V =Ir. Hence, p.d. is large in battery & there is reduction in p.d. across battery when a motor is started as the starter takes high current from the source i.e. battery. Hence, when a motor car is started, its light becomes slightly dim.     42. A high voltage supply must have a very high internal resistance. Why?   If the high voltage source doesn't have high resistance then the source is short circuited accidentally many times due to the flow of high voltage current through the source. This high current may damage the source of emf and may also cause firing. If the source has high internal resistance, the circuit in the short circuit will be limited by the internal resistance of the source which saves the source from damaging and provides safety from firing.        43. Why is heat generated across a resistor when electric field is applied?   When the electric field is applied across a resistor, free electrons move through resistor which collides frequently with many atoms of resistor. At each collisions they lose their kinetic energy & give it to atoms. It increases the K.E. of vibration of metal atoms which generates heat in the wire. Hence, heat is generated across a resistor when electric field is applied.       45. The same current is passed through the line wire and filament of a bulb, the filament becomes quite hot but not the line wire. Why? As the bulbs contain filament wire which has high resistance and resistivity, so it has to do much more work to oppose the flow of the current which results in the production of more heat according to the Joule's law of heating. where, H is the heat produced due to resistor, I is the current flowing through the wire, R is the resistance of the wire and t is the time at which the current is flowing through the wire. But, the line wire has less resistance as well as resistivity, so it opposes less current which is negligible as compared to that of heating wire. As the resistance is almost negligible there is very much less production of heat or heat production is almost negligible. Hence, for the same current passed through the line wire and filament of a bulb, the filament becomes quite hot but not the line wire.   46. The element of heater is very hot while the wires carrying current are cold. Why? As the heater contain heating wire which has high resistance and resistivity, so it has to do much more work to oppose the flow of the current which results in the production of more heat according to the Joule's law of heating. where, H is the heat produced due to resistor, I is the current flowing through the wire, R is the resistance of the wire and t is the time at which the current is flowing through the wire. But, the current carrying conductor has less resistance as well as resistivity, so it opposes less current which is negligible as compared to that of heating wire. As the resistance is almost negligible there is very much less production of heat or heat production is almost negligible. Hence, the element of heater is very hot while the wires carrying current are cold.   47. What are the special characteristics of a heating wire and a fuse wire? The special characteristics of a heating wire are as follow: 1. It has very high melting and boiling point. 2. It has high resistance and high resistivity. The special characteristics of a fuse wire are as follow: 1. It has low melting and boiling point. 2. It has less resistance and less resistivity.   48. ?     49. Why is easier to start a car engine on a hot day than on a cold day?   In a hot day due to the high temperature, more electrolyte in a battery dissociates into ions and their kinetic energy also increases due to heat energy. The internal resistance also decreases due to thermal energy and more current flows through the circuit. Hence, it is easier to start a car engine on a hot day than on a cold day.     50. Can the terminal p.d. be greater than the emf of a cell? In case of charging of cell having internal resistance r, the terminal potential difference V of the cell is greater than its emf E i.e. V = E + Ir: where I is the current. Here, the terminal p.d. exceeds the emf of a cell by the term Ir. In case of discharging of cell, the terminal p.d. cannot be greater than the emf of a cell but in the absence of the internal resistance, the terminal potential difference becomes equal to the emf of a cell. If the internal resistance of the resistor of resistance R is r and the emf of a cell is E and the potential difference be V then If there is no internal resistance then r = 0, so E = V.    51. Why do electric power co-operation reduce the voltage during high demand time? What is saved in doing so? As the energy is loss in the form of heat energy according to the Joule's law of heating. where, H is the heat produced on the supply line, I is the current flowing through it, R is the resistance of supply line and t is the time at which the current is flowing through the wire. In order to avoid the wastes of energy in the form of heat energy, if the voltage is reduced then the energy loss as a heat energy reduces according to above relation. Hence, electric power co-operation reduce the voltage during high demand time.   52. Two electric bulbs of 60Wand 100W are joined (I) in series and (II) in parallel. Which one will be brighter in each case? As the resistance of 60W bulb is greater than that of 100W bulb and brightness depends on the power consumed. So,1. If two electric bulbs of 60W and 100W are joined in series with each other then power consumed P=  I2R, hence 60W bulb having high resistance will glow brighter than 100W bulb. 2. If two electric bulbs of 60W and 100W are joined in parallel with each other then power consumed P= V2/ R, so 100W bulb having smaller resistance will more glow brightly than 60W bulb.   53. Why is it more economical to transmit electric power at high voltage and low current than that at low voltage and high current? According to the Joule's law of heating. where, H is the heat produced due to resistor, I is the current flowing through the transmission line, R is the resistance of transmission line and t is the time at which the current is flowing through the wire. If V be the voltage supplied then the power generated is VI. The power lost as a heat according to Joule's law of heating becomes I2R. So, the net power available in the line is "VI - I2R". In order to provide maximum power on the line, the value of I2R must be reduced. Hence, if the value of I2R is minimum then there is no wastage of energy in the form of heat. So, it is more economical to transmit electric power at high voltage and low current than that at low voltage and high current.   54. On an electric bulb, it is written 100W and 220V; what does it mean?   If it is written 100W and 220V, then it means if the bulb is connected on 220V supply it will consume 100W power of electric energy to convert into light energy i.e. it consumes 100J of energy in one second.       55. If external resistance is very high. How will you arrange a number of cells to get maximum current?   The equivalent resistance of the resistors when connected in series is greater than that connected in parallel. According to Ohm's law V = IR, potential difference between two terminal remaining constant the current is inversely proportional to the resistance offered by the resistor. For maximum current, the value of resistance should be high. Among parallel & series combination of resistors, I will arrange a number of cells in series to get maximum current.       56. Does the thermoelectric effect obey the law of conservation of energy?   Yes, the thermoelectric effect obeys the law of conservation of energy. In Seebeck effect, heat energy is converted into electrical energy; in Thomson effect, electrical energy is converted into heat energy whereas in Peltier effect, one junction absorbs heat and other junction evolves heat. Hence, the thermoelectric effect obeys the law of conservation of energy.       57. You are asked to measure the emf of a cell. Which instrument will you use to measure a high resistance: a voltmeter or a potentiometer and why?   An ideal voltmeter must have infinite resistance but in practice, a voltmeter is high resistance instrument so, it always draws some current but a potentiometer does not draw any current from the cell whose emf is to be measured. For exact measurement of emf of cell, potentiometer is used rather than voltmeter because voltmeter measures a lesser value of emf than it actually ought to be. If I am asked to measure the emf of a cell I would prefer a potentiometer to voltmeter.       58. A potentiometer is equivalent to a voltmeter of infinite resistance. Explain.   When a potentiometer is used to measure the emf of a cell, it draws no current from the cell at the null point. A voltmeter of infinite resistance also does not draws current as the current flowing through it is zero. Hence, a potentiometer is equivalent to a voltmeter of infinite resistance.       59. Why is Wheatstone bridge not suitable for measuring very low resistance?   In Wheatstone bridge, all four resistors have nearly equal resistance. If the resistance of the resistor used in the Wheatstone bridge is very low then other three resistors must also be of less resistance and a galvanometer used in it must be also of less resistance. These arrangement give a lot of error and the galvanometer itself is not sensitive. Hence, to avoid the error of measurement of the resistance of resistor, Wheatstone bridge is not used for measuring very high resistance.       60. Why do we prefer a potentiometer to measure the emf of a cell to a voltmeter? The principle of potentiometer states that the potential difference V across a portion of the wire is directly proportional to the length l of the portion of the wire i.e. Potentiometer follows a null method. At null point, the current flowing through the cell is zero and the emf can be measured using potentiometer. Thus, potential difference in an open circuit i.e. emf of the cell, is measured accurately by potentiometer. So, we prefer a potentiometer to measure the emf of a cell to a voltmeter as potentiometer gives more accuracy than voltmeter reading.           If you want certain questions answered, please write to
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