Discussion Questions
Answer:- In electrostatics, there is no electric field inside a conductor. There no electric field in a superconductor, since a superconductor has zero resistance and zero resistivity. Ordinary conductors like silver and copper do have a little resistance and hence do have small electric fields inside. They are not perfect conductors.When there is current flow, there is an electric field inside the conductor.
Q25.2 A cylindrical rod has resistance R. If we triple its length and diameter, what is its resistance, in terms of R?
A25.3:- The resistivity of a material doesn’t depend on its physical dimensions like length and cross sectional area. So if the people its length and diameter then resistivity of cylindrical road doesn’t change i.e. it will remains same.
Note-(1.) If we tripled the length and diameter of cylinder rod then it’s resistance will be one third of its initial value. ( 2. )Resistivity of cylindrical road depends on two factors first nature of material and second temperature.
Q25.4 Two copper wires with different diameters are joined end to end. If a current flows in the wire combination, what happens to electrons when they move from the larger-diameter wire into the smaller-diameter wire? Does their drift speed increase, decrease, or stay the same? If the drift speed changes, what is the force that causes the change? Explain your reasoning.
Q25.5 When is a 1.5-V AAA battery not actually a 1.5-V battery?
That is, when do its terminals provide a potential difference of less than 1.5 V?
A25.5 When a battery is connected to external circuit then in the process of of working of battery its internal resistance start working and due to which its terminal potential becomes less less than its EMF bi factor into Ir.
Q25.6 Can the potential difference between the terminals of a battery ever be opposite in direction to the emf? If it can, give an example. If it cannot, explain why not.
Answer:- Yes, it’s possible when Cell/battery is charging or When a cell is connected to cell of higher emf. We know that V = E – IR, here current I is positive and E > V. Now if we make the current negative then the equation become V = E + IR . Now V> E If we connect a new cell with higher emf then the current flows through the opposite direction and it’s become negative.
Q25.7 A rule of thumb used to determine the internal resistance of
a source is that it is the open-circuit voltage divided by the short circuit current. Is this correct? Why or why not?
Q25.8 Batteries are always labeled with their emf; for instance, an
AA flashlight battery is labeled “1.5 volts.” Would it also be appropriate to put a label on batteries stating how much current they provide? Why or why not?
Answer:- No, It will not be appropriate to put a label on batteries stating how much current they provide. Because Current drawn from the batteries depends on the External resistance and Internal Resistance of batteries. Therefore (i) For different circuit of different , value of current will be different (ii) Internal resistance of batteries is also increase on its use. So current drawn from batteries will be decrease on use.
Q25.9 We have seen that a coulomb is an enormous amount of charge; it is virtually impossible to place a charge of 1 C on an object. Yet, a current of 10 A, is quite reasonable. Explain this apparent discrepancy.
Answer:- Current is defined as the charge flows through a conductor per unit time. I = dq/dt
(i) It means if a very small amount of charge flowing through a conductor in a very short time, it can provide large amount of current.
(ii) Current I=neAVd So due to large amount of larges ( larger number density of charges n) may provides large amount of current.
Q25.10 Electrons in an electric circuit pass through a resistor. The wire on either side of the resistor has the same diameter. (a) How does the drift speed of the electrons before entering the resistor compare to the speed after leaving the resistor? Explain your reasoning. (b) How does the potential energy for an electron before entering the resistor compare to the potential energy after leaving the resistor? Explain your reasoning
Answer:- According to above formula, if n (Number density) and q( Charge ) are equal , then Current density will be constant in entire conductor. As per question , diameter i.e. cross-sectional area of resistor and wire are equal So drift speed of the electrons will remain same before a entering the resistor and after leaving the resistor.
We Know that current is always flow from higher potential to lower potential.So Charge carriers also moves from higher potential to lower potential . As U= qV=eV. Therefore in a Conductor moves from higher potential energy region to lower potential energy region. So finally we can say that the potential energy of electron is higher before entering the resistor than leaving it.
Q25.12 Which of the graphs in Fig. Q25.12 best illustrates the current I in a real resistor as a function of the potential difference V across it? Explain. (Hint: See Discussion Question Q25.11.)
Answer:- As we Know that slope of I-V graph is 1/R Which is straight line as shown in Graph (a) for constant value of resistance R.
But due to dissipation of energy in form of thermal energy , Resistance increase
Therefore 1/R (Slope of I-V Graph) decrease with temperature. So Graph (d) will best illustration of I-V graph.
Q25.13 Why does an electric light bulb nearly always burn out just as you turn on the light, almost never while the light is shining?
Answer:- Due to thermal expansion. When a bulb is turned on the temperature of its filament suddenly increases in short time. Due to this rapid change in the temperature of bulb, its filament to expand. over the time that filament is weakened. So sometimes we turned on the bulb is burn out.
Filament’s temperature is almost constant during shining so it never burn out while shining .
Q25.14 A light bulb glows because it has resistance. The brightness of a light bulb increases with the electrical power dissipated in the bulb. (a) In the circuit shown in Fig. Q25.14a, the two bulbs A and B are identical. Compared to bulb A, does bulb B glow more brightly, just as brightly, or less brightly? Explain your reasoning. (b) Bulb B is removed from the circuit and the circuit is completed as shown in Fig. Q25.14b. Compared to the brightness of bulb A in Fig. Q25.14a, does bulb A now glow more brightly, just as brightly, or less brightly? Explain your reasoning
Answer:-
Q25.15 (See Discussion Question Q25.14.) An ideal ammeter A is placed in a circuit with a battery and a light bulb as shown in Fig. Q25.15a, and the ammeter reading is noted. The circuit is then reconnected as in Fig. Q25.15b, so that the positions of the ammeter and light bulb are reversed. (a) How does the ammeter reading in the situation shown in Fig. Q25.15a compare to the reading in the situation shown in Fig. Q25.15b? Explain your reasoning. (b) In which situation does the light bulb glow more brightly? Explain your reasoning
Answer:- (a) As per question, Ammeter is an ideal ammeter so its resistance is zero. So resistance in the both circuit is only due to Light bulb . So current in the both circuits will same.
If Ammeter is not ideal Ammeter then also current in the both circuits will same because Ammeter is connected in series in both circuits so resistance add in the both will equal.
(b) Brightness of bulb depends on electric power dissipated in the bulb. So in both cases current and resistance are equal , therefore power dissipated will same P=I2R hence brightness of bulb in both circuits will same.
Q25.16 (See Discussion Question Q25.14.) Will a light bulb glow more brightly when it is connected to a battery as shown in Fig. Q25.16a, in which an ideal ammeter A is placed in the circuit, or when it is connected as shown in Fig. 25.16b, in which an ideal voltmeter V is placed in the circuit? Explain your reasoning.
Answer:- As we know the resistance of Ideal Ammeter is zero and resistance of Ideal Voltmeter is infinite.So we Ideal ammeter is connected in series with bulb as fig (a) , there will be no change in the current and no change in brightness.
But when Ideal Voltmeter is connected in series with the bulb as shown in fig ( B), resistance becomes infinite. so brightness of bulb will be Zero.
Q25.17 The energy that can be extracted from a storage battery is always less than the energy that goes into it while it is being charged. Why?
Answer:- It is universal truth that power gain is always less then 1. it is due to energy losses in the various reasons. In case of battery production of heat and internal resistance, some energy of battery is lost So, the energy can be extracted from a storage battery is always less than the energy into it while it is being charged.
Q25.18 Eight flashlight batteries in series have an emf of about 12 V, similar to that of a car battery. Could they be used to start a car with a dead battery? Why or why not?
Answer:- No. A dead battery has huge resistance due to its internal resistance. So When all eight batteries connected in series with a dead battery , internal resistances of all batteries will add which becomes very high due to interanal resistance of dead battery .Therefore this combination will not able to start a car.
Q25.19 Small aircraft often have 24-V electrical systems rather than the 12-V systems in automobiles, even though the electrical power requirements are roughly the same in both applications. The explanation given by aircraft designers is that a 24-V system weighs less than a 12-V system because thinner wires can be used. Explain why this is so.
Answer:-
Q25.20 Long-distance, electric-power, transmission lines always operate at very high voltage, sometimes as much as 750 kV. What are the advantages of such high voltages? What are the disadvantages?
Answer:-
The advantages of using high voltage are mainly to reduce losses due to Joule heating (Ohmic heating).
Power delivered from one end of the line to the other goes like P=I∗VP=I∗V, the product of current and voltage. Because most transmission conductors are aluminum, they have nonzero resistance (R). As current moves down the length of the line, energy is lost to Joule (resistive) heating in the conductor. This energy loss goes like Presistive=I2∗RPresistive=I2∗R. Resistive losses are undesirable because this is energy that you pay for on one end of the line that never comes out of the other end. Essentially, you are paying to heat up the air around the line.
In order to minimize resistive losses, one must minimize current (or resistance; hello superconductors). To maximize power transfer, one must maximize the product of voltage and current. So, the maximum possible voltage level is chosen.
Q25.21 Ordinary household electric lines in North America usually operate at 120 V. Why is this a desirable voltage, rather than a value considerably larger or smaller? On the other hand, automobiles usually have 12-V electrical systems. Why is this a desirable voltage?
Answer:- When electricity is supply to households via long wires, resistance of wire becomes very high . So in order to maintain constant power, the voltage must be taken high.
In a car, short wires are used so there resistance is also same. Therefore , No need to take high voltage in this case to maintain constant power.
Q25.22 A fuse is a device designed to break a circuit, usually by melting when the current exceeds a certain value. What characteristics should the material of the fuse have?
Answer:- The material of the fuse should have a low melting point and high resistivity. When current is high, due to low melting point fuse wire melt and break the circuit. while high resistivity provide fuse wire more capacity to consume high power P= I2R.
Q25.23 High-voltage power supplies are sometimes designed intentionally to have rather large internal resistance as a safety precaution.
Why is such a power supply with a large internal resistance safer than a supply with the same voltage but lower internal resistance?
Answer:- High Voltage power supplies are designed intentionally to have larger internal resistance to limit the current up to safety limit.
Q25.24 The text states that good thermal conductors are also good electrical conductors. If so, why don’t the cords used to connect toasters, irons, and similar heat-producing appliances get hot by conduction of heat from the heating element?
Because cords of toasters,irons etc is made of good conductor of electricity i.e. less resistive material while in toasters, irons etc high resistive material is used to produce large amount of heat.
Video Discussion
coming soon
Enjoy physics 