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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 2101. |
Estimate the average drift velocity of conduction electrons in a copper wire of cross-sectional area `2.5 xx 10^(-7) m^(2)`, carrying a current of 2.7 A. Assume the density of conduction electrons to be `9 xx 10^(28) m^(-3)`. |
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Answer» Correct Answer - `0.75 mms^(-1)` `upsilon_(d) = (I)/(nAe)` `(2.7)/((9 xx 10^(28)) xx (2.5 xx 10^(-7)) xx (1.6xx 10^(-19)))` `= 0.75 xx 10^(-3)ms^(-1) = 0.75 "mms"^(-1)` |
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| 2102. |
Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area `1.0 xx 10^(-7) m^(2)` carrying a current of 1.5 A. Assume that each copper atom contributes roughly one conduction electron. The density of copper is `9.0 xx 10^(3) kg//m^(3)` and its atomic mass is 63.5 u. |
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Answer» The direction of drift velocity of conduction electrons is opposite to the electric field direction, i.e., electrons drift in the direction of increasing potential. The drift speed `upsilon_(d)` is given by Eq. `I Delta t = n eA//V_(d)//Delta t` `V_(d) = (I//n eA)` Now, `e = 1.6 xx 10^(-19) C,A = 1.0 xx 10^(-7) m^(2), I = 1.5 A`. The density of conduction electrons n is equal to the number of atoms per cubic meter (assuming one conduction electron per Cu atom as is reasonable from its valence electron count of one). A cubic metre of copper has a mass of `9.0 xx 10^(3)` kg. Since `6.0 xx 10^(23)` copper atoms have a mass of 63.5 g `n = (6.0 xx 10^(23))/(63.5) xx 9.0 xx 10^(6)` `= 8.5 xx 10^(28) m^(-3)` Which gives, `upsilon_(d) = (1.5)/(8.5 xx 10^(28) xx 1.6 xx 10^(-19) xx 10 xx 10^(-7))` `= 1.1 xx 10^(-3) ms^(-1) = 1.1 m m s^(-1)` |
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| 2103. |
Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area `3.0 xx 10^(-7) m^(2)` carrying a current of 5A. Assume that each copper atom contributions roughly one conduction electron. The density of copper is `9.0 xx 10^(3) kg//m^(3)` and its atomic mass is 63.5 u. |
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Answer» Given corss-sectional area of copper wire, `A = 3 xx 10^(-7) m^(2)` carrying current of copper, I = 5A charge of electron, `e = 1.6 xx 10^(-19) C` Density of conduction electrons = No. of atoms per cubic meter, `n = (N_(A) ("Avogadro number") xx "mass of copper" (M) "per cubic meter")/("Atomic mass"(m))` `:. n = (6 xx 10^(23) xx 9 xx 10^(3))/(63.5) = 8.5 xx 0^(23) m^(-3)` `:.` Average drift speed of conduction electrons. `V_(d) = (I)/(neA) = (5)/(8.5 xx 10^(28) xx 1.6 xx 10^(-19) xxx 3 xx 10^(-7))` `implies V_(d) = (5)/(8.5 xx 1.6 xx 3 xx 10^(2)) = 0.1225 xx 10^(-2) m//s` `:. V_(d) = 1.225 mm//s` |
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| 2104. |
The potential difference applied to an X-ray tube is 5 kV and the current through it is 3.2 mA. Then the number of electros striking the target par second isA. `2xx10^(16)`B. `5xx10^(16)`C. `1xx10^(17)`D. `14xx10^(15)` |
| Answer» Correct Answer - A | |
| 2105. |
In metan and vacuume tubes charge carries areA. electronsB. protonsC. bothD. positrons |
| Answer» Correct Answer - A | |
| 2106. |
The drift velocity of electron in a metal conductor under effective of elecrtic field applied isA. `10^(2)` m/sB. `10^(-2)` m/sC. `10^(4)` m/sD. `10^(-4)` m/s |
| Answer» Correct Answer - D | |
| 2107. |
The drift velocity of electron in a metal conductor under effective of elecrtic field applied isA. `10^(-13)(m)/(s)`B. `10^(-3)(mm)/(s)`C. `10^(-4)(m)/(s)`D. `10^(-30)(m)/(s)` |
| Answer» Correct Answer - C | |
| 2108. |
The ratio of output power to input power of an elecrtic device is called……. |
| Answer» Correct Answer - effeciency of electric device | |
| 2109. |
The drift velocity of electron in a metal conductor under effective of elecrtic field applied isA. `10^(-1)m//s`B. `10^(-2)m//s`C. `10^(-3)m//s`D. `10^(-4)m//s` |
| Answer» Correct Answer - D | |
| 2110. |
The drift velocity of electron in a metal conductor under effective of elecrtic field applied isA. `10^(29)cm^(-3)`B. `10^(29)m^(-3)`C. `10^(23)m^(-3)`D. `10^(26)m^(-3)` |
| Answer» Correct Answer - B | |
| 2111. |
Calculate the amount of charge flowing in 2 minutes in a wire of resistance `10 Omega` when a potential difference of 20 V is applied between its endsA. `120C`B. `240 C`C. `20 C`D. `4C` |
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Answer» Correct Answer - B |
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| 2112. |
If an observer is moving with respect to a stationary electron, then he observesA. Only magnetic fieldB. Only electric fieldC. Both (a) and (b)D. None of the above |
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Answer» Correct Answer - C |
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| 2113. |
If potential `V = 100 +- 0.5` Volt and current `I = 10 +- 0.2` amp are given to us. Then what will be the value of resistanceA. `10 +- 0.7` ohmB. `5 +- 2` ohmC. `0.1 +- 0.2` ohmD. None of these |
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Answer» Correct Answer - D |
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| 2114. |
What is the length of a nichrome wire of radius 0.32 mm, resistance `9.3 Omega` and resistivity `15 xx 10^(-6)Omega m` ? If a potnetial differences of 10 V is applied across the wire, what will be the current in the wire ? |
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Answer» Here, `l =?, r = 0.32xx10^(-3) xx 10^(-3) m, R =9.3 Omega , rho = 15xx10^(-6)Omega m, V= 10V` `R=(rho l) /(pi r^(2))` or `l = ((9.3)xx3.14 xx (0.32 xx 10^(-3))^(2))/ ((15xx10^(-6))) = 0.199 m ~= 0.20m` Current `I=V/R = 10/9.3 = 1.075= 1.08 A` |
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| 2115. |
Arrange the following materials in increasing order of their resistivity, Nichrome, Copper, Germanium,SiliconA. `"Copper" lt "Nichrome" lt "Germanium" lt "Silicon"`B. `"Germanium" lt "Copper"lt "Nichrome" lt "Silicon"`C. `"Nichrome" lt "Copper" lt "Germanium" lt "Silicon"`D. `"Silicon" lt "Nichrome" lt "Germanium" lt "Copper"` |
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Answer» Correct Answer - A `{:("Materials","Resistivity" (rho)(Omega "m at 0"^(@)C)),("Copper (Cu)",1.7 xx 10^(-8)),("Nichrome",100 xx 10^(-8)),("(alloy of Ni, Fe, Cr)",),("Germanium",0.46),("Silicon",2300):}` |
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| 2116. |
A nichrome wire 50 cm long and one square millimetre cross- section carries a current of 4 A when connected to a 2 V battery. The resistivity of nichrome wire in ohm metre isA. `1 xx 10^(-6)`B. `4 xx 10^(-7)`C. `3 xx 10^(-7)`D. `2 xx 10^(-7)` |
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Answer» Correct Answer - A |
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| 2117. |
A nichrome wire of length 100 cm and area of cross-section `0.5 m m^(2)` has a resistance of `2.2 Omega`. The resistivity of nichrome isA. `110Omega`B. `0.11Omega m`C. `121Omega m`D. `11xx10^(-7) Omega m` |
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Answer» Correct Answer - D `rho=(RA)/(l)=(2.2xx0.5xx10^(-6))/(1)=11xx10^(_7)Omegam` |
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| 2118. |
Are the paths of free electrons straight lines between successive collisions (with the positive ions of metal wire) in the- (i) Absence of electric field (ii) Presence of electric fieldA. randomB. unidirectionalC. bidirectionalD. circular |
| Answer» Correct Answer - A | |
| 2119. |
A large number of free electrons are present in metals. Why is there no current in the absence of electric field across it, but threre is a current in the presence of electric field?A. net flow of charge of one directionB. net flow of charge in two directionsC. no net flow of charge in any directionD. none of these |
| Answer» Correct Answer - C | |
| 2120. |
Threre is a current of 1.344 a in a copper wire whose area of cross-sectional normal to the length of wire is `1mm^(2)`. If the number of free electrons per `cm^(2)` is `8.4xx10^(28)m^(3)`, then the drift velocity would beA. `1mm//s`B. `1mm//s`C. `0.1mm//s`D. `0.01mm//s` |
| Answer» Correct Answer - C | |
| 2121. |
A large number of free electrons are present in metals. Why is there no current in the absence of electric field across ot, but threre is a current in the presence of electric field? |
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Answer» The current is due to flow of charge in a definite direction. In a metal, the free electrons move in all directions haphazardly, in the absence of electric field , the average thermal velocity of electrons is zero. Therefore, there is no current in metal in the absence of electric field. In the presence of electric field, each free electron experiences a force opposite to the direction of experiences a force opposite to the directgon of field. Therefore, the electrons acquire a drift velocity and there is a net flow of charge in a definite direction which constitutes the current. |
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| 2122. |
A non-conducting ring of radius r has charge q distributed unevenly over it. What will be the equivalent current if it rotates with an angular velocity `Omega`? |
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Answer» In one rotation, charge q crosses any fixed point near the ring. Time period of rotation `T=2 pi//omega` `:.` Current = charge crossing per second `q/T = q/( 2pi // omega) =q omega/2pi` |
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| 2123. |
A uniform wire of resistance `4Omega` is bent into circle of radius `r`. As specimen of the same wire is connected along the dimeter of the circle. What is the equivalent resistance across the ends of this wire?A. `(4)/((4+pi))Omega`B. `(3)/((3+pi))Omega`C. `(2)/((2+pi))Omega`D. `(1)/((1+pi))Omega` |
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Answer» Correct Answer - C |
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| 2124. |
A uniform wire of resistance `4Omega` is bent into circle of radius `r`. As specimen of the same wire is connected along the dimeter of the circle. What is the equivalent resistance across the ends of this wire? |
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Answer» Correct Answer - `(4)/(4 + pi) Omega` Circumference of the circuit `= 2 pi r` its resistance `R = 4 Omega` resistance per unit length of the wire `= (R )/(2 pi R)` Resistance of the specimen connected along the diameter is `R_(1) = (R )/(2 pi r) xx 2 r = (R )/(pi)` The resistance of each senicircular wire `R_(2) = (R )/(2)` Equivalent resistance `R_(3)` across the specimen is `(1)/(R_(3)) = (2)/(R ) + (2)/(R ) + (pi)/(R ) + (4 + pi)/(R )` or `R_(3) = (R )/(4+ pi) = (4)/(4 + pi) Omega` |
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| 2125. |
A microammeter has a resistance of 100W and a full scale range of 50 µA. It can be used as a voltmeter or a higher range ammeter provided a resistance is added to it. Pick the correct range and resistance combination(s). (1) 10V range with 200 kW resistance in series. (2) 50V range with 10 kW resistance in series. (3) 5 mA range with 1W resistance in parallel. (4) 10 mA range with 1 kW resistance in parallel.A. 10V range with `200KOmega` resistance in seriesB. 50V range with `10kOmega` resistance in seriesC. 5mA range with `1kOmega` resistance in parallelD. 10mA range with `50kOmega` resistance in parallel |
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Answer» Correct Answer - 1 Given G=`100Omega, lg=50muA` Consider the case (a) `V=lg(G+R)` when R is in series `10=(50xx10^(-6))(100 +R)` or `R+100=2xx10^(5)` or `R=200x x10^(3)-100` or `R=200` kilo ohm approximately This is the correct range and resistance combination. |
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| 2126. |
A primary and a secondary cell have the same EMF which of these will provided higher value of the maximum currentthat can be drawn? Explain briefly. |
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Answer» Correct Answer - D |
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| 2127. |
A nonideal battery is connected to a resistor. Is work done by thr battery equal to the hermal energy developed in the resistor?Does your answer change If the battery is ideal? |
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Answer» Correct Answer - A |
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| 2128. |
For a cell, the terminal potential difference is `2.2 V`, when circuit is open and reduces to `1.8 V `. When cell is connected to a resistance `R=5Omega`, the internal resistance of cell `(R)` isA. `10/9Omega`B. `9/10 Omega`C. `11/9 Omega`D. `5/9 Omega` |
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Answer» Correct Answer - A `r=R(E/V-1)=5(2.2/1.8-1)=10/9Omega` |
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| 2129. |
The potential difference between points `A` and `B` in the circuit shown in figure will be A. `1V`B. `2V`C. `-3V`D. none of these |
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Answer» Correct Answer - D `i=(10-5)/(2.5+2.5+40)=1/9A` (clockwise) `V_B-15i-25i=V_A` `:. V_A-V_BH=40 i = - 40/9V` |
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| 2130. |
For the post office arrangement to determine the value of unknown resistance, the unknown resistance should be connected between. A. B and CB. C and DC. A and DD. B and C |
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Answer» Correct Answer - C |
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| 2131. |
For the post office arrangement to determine the value of unknown resistance, the unknown resistance should be connected between. A. `B` and `C`B. `C` and `D`C. `A` and` D`D. `B_1` and `C_1` |
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Answer» Correct Answer - C `BC, CD` and `BA` are known resistance. The unknown resistance is connected between `A` and `D`. |
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| 2132. |
For the post office arrangement to determine the value of unknown resistance, the unknown resistance should be connected between. A. `B^(1)` and `C^(1)`B. `A` and `D`C. `C` and DD. B and D |
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Answer» Correct Answer - B Total external resistance will be total resistance of the whole length of box. It should be connected between A and D, hence correct option is (b) |
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| 2133. |
A cell has an emf 1.5V. When connected across an external resistance of `2Omega`, the terminal potential difference falls to 1.0V. The internal resistance of the cell is:A. `2Omega`B. `1.5Omega`C. `1.0Omega`D. `0.5Omega` |
| Answer» Correct Answer - C | |
| 2134. |
Six identical resistors are connected as shown in the figure. The equivalent resistance will be A. `P` and `Q`B. `Q` and `R`C. `P` and `R`D. any tow points |
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Answer» Correct Answer - A `R_(PQ)=R||(R/3+R/2)=R||(5R)/6=(11R)/6` `R_(PR)=(R/3)||(R+R/2)=R/3||(3R)/2=(3R)/11` `P_(QR)=R/2||(R+R/3)=R/2||(4R)/3=(4R)/11` so, `R_(PQ)gtR_(QR)gtR_(PQ)` |
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| 2135. |
Six identical resistors are connected as shown in the figure. The equivalent resistance will be A. Maximum between P and RB. Maximum between Q and RC. Maximum between P and QD. All are equal |
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Answer» Correct Answer - C (c) For various combinations equivalent resistance is maximum between P and Q. |
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| 2136. |
A capacitor is charged using an external battery with a resistance x in series. The dashed line showns the variation of In I with respect to time. If the resistance is changed to 2x, the new graph will be A. PB. QC. RD. S |
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Answer» Correct Answer - B (b) KEY CONCEPT : The current in RC circuit is given by `I = I_0e^(-t//RC)` or `lnI = lnI_0 - t/(RC) or lnI = (-t/(RC)) + lnI_0` `lnI = (-t/(RC)) + ln (E_0/R)` On comparing with `y = mx +C` Intercept = `ln (E_0/R) and slope = -1/(RC)` When R is changedto 2R then slope increases and current becomes less. New graph is Q. |
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| 2137. |
An electric kettle used to prepare tea, takes 2 minutes to boil 4 cups of water (1 cup contains 200 cc of water) if the room temperature is `25^@ C`, (a) If the cost of power consumption is `Re 1.00` per unit `(1 unit = 1000 watt-hour)`, calculate the cost of boiling 4 cups of water. (b) What will be the corresponding cost if the room temperature drops to `5^@ C`? |
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Answer» Correct Answer - (i)`21 palse`(ii)27 palse` Volume of water in `4` cups `upsilon = 4 xx 200 = 800 c.c` mass of wire m = volume `xx` desity `= 800 xx 1 = 800` gram (a) Energy spent is boding the water `= ms d theta ` `= 800 xx 1 xx (100 - 25)` cal `= 800 xx 75 xx 4.2J` `= (800 xx 75 xx 4.2)/(3.6) xx 10^(6)) kWh =0.07 kWb` cost of bospend in boliing the water `= 0.07 xx 3 = 0.21 Rs = 21` pulse (ii) Energy spent in boding the water `= 800 xx 1 xx (100 - 5) cal` `= 800 xx 95 xx 4.2J` `= (800 xx 95 xx 4.2)/(3.6) xx 10^(6)) kWh =0.09 kWb` cost of boiling the water `= 0.09 xx 3 = 0.27 Rs = 21` pulse |
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| 2138. |
A fixed chamber isolated from surrounding is divided into equal halves A and B as shown in figure. Part A contains one mole of oxygen and part B contains one mole of helium. The separator C is thermally conducting and kept fixed. Initial temperature of oxygen chamber is 600 K and that of helium 300 K. Specific heat capacity of separator and chamber is negligible. Choose the correct statement. A. change in temperature of helium gas is equal to change in temperature of oxygen gasB. change in internal energy of helium gas is equal to 200 RC. temperature of oxygen gas in steady state condition is 487 K (approximately)D. All of the above |
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Answer» Correct Answer - C |
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| 2139. |
In the shown arrangement of the experiment of the meter bridge if AC corroesponding to null deflection of galvanometer is x, what would be its value if the radius of the wire AB is doubled? . |
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Answer» For null deflection of galvanometer in a metrebridge experiment, `(R_(1))/(R_(2))=(R_(AC))/(R_(CB)) or (R_(1))/(R_(2))=(x)/((100-x))` Since `(R_(1))/(R_(2))` remains constant `(x)/((100-x))` also remains constant, the value of `x` remains as such. `therefore` Length of `AC=x` |
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| 2140. |
In the meterbridge experimental set up, shown the null point `D` is obtained at a distance of `40 cm` from end A of the meter bridge wire if a resistance of `10 Omega` is connected in series with `R_(1)`, null point is obtained at `AD = 30 cm` |
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Answer» Correct Answer - `8 Omega` Case (i) `(R_(1))/(R_(2)) = (40)/(60) = (2)/(3) or R_(1) = (2)/(3) R_(2)`….(i) Case (ii) , `(R_(1) + 10)/(R_(2)) = (60)/(40) = (2)/(3)` or `2R_(1) + 20= 3R_(2)` or `2 xx (2)/(3) R_(2) + 20 = 3R_(2) or 4R_(2) + 60 = 9 R_(2)` or `R_(2) = 12 Omega` from (i) `R_(1) = (2)/(3) xx 12 = 8 Omega` |
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| 2141. |
The figure shows a network of currents. The magnitude of currents is shwon here. The magnitude of current l will be A. `3 A`B. `13 A`C. `23 A`D. `-3 A` |
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Answer» Correct Answer - C `sum l=0` `(7+3)+(8+5)+l=0` `l=-23A implies |l|=23A` |
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| 2142. |
Shown is a network of currents. The magnitude of the current is also shown there. Find the current `i`. A. `3A`B. `13 A`C. `23 A`D. `-3 A` |
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Answer» Correct Answer - C |
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| 2143. |
Shown is a network of currents. The magnitude of the current is also shown there. Find the current `i`. A. `3.4`B. `13A`C. `23A`D. `-3A` |
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Answer» Correct Answer - C |
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| 2144. |
The magnitude in `i` in ampere unit is A. 0.1B. 0.3C. 0.6D. None of these |
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Answer» Correct Answer - A |
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| 2145. |
The figure shows a network of currents. The magnitude of currents is shown here. The current I will be A. 3AB. 9AC. 13AD. 19A |
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Answer» Correct Answer - C |
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| 2146. |
A circuit shown in the figure has resistances `20 Omega` and `30 Omega`. At what value of resistance `R_x` will the thermal power generated in it be practically independent of small variations of that resistance? The voltage between points `A` and `B` is supposed to be constant in this case. |
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Answer» Let V be the potential difference applied across across A and B. Let `I,I_(1),I_(2)` be the currents through `R_(1),R_(2)` and `R_(x)` respectively. Here `R_(2)` and `R_(x)` are connected in parallel, their effective resistance is `R_(P) = (R_(2) R_(x))/(R_(2) + R_(x))` Total resistance between A and B is `R_(1) = (R_(2) R_(x))/(R_(2) + R_(x))` Current through the circuit, `I = V/(R_(1) + (R_(2) R_(x))/(R_(2) + R_(x)))` Potential difference across C and D `V_(G//D) I R_(P) = I = V/(R_(1) + (R_(2) R_(x))/(R_(2) + R_(x))) xx (R_(2) R_(x))/(R_(2) + R_(x))` `(VR_(2)R_(x))/(R_(1) (R_(2) + R_(x)) + R_(2)R_(x))` Current through resistance `R_(x)` will be `I_(2) = V_(CD)/R_(x) = (VR_(2))/(R_(1) (R_(2) + R_(x)) + R_(2)R_(x))` `= (VR_(2))/(R_(1) (R_(2) + R_(x)) + R_(2)R_(x))` Thermal power generated in resistance `R_(x)` will be `P = I_(2)^(2) R_(x) = (V^(2)R_(2)^(2)R_(x))/[R_(1)R_(2)+(R_(1) + R_(2))+R_(x)]^(2)` For P to be independent of `R_(x),dP/dR_(x) = 0` So, `d/dR_(x) [ (V^(2)R_(2)^(2)R_(x))/[R_(1)R_(2)+(R_(1) + R_(2))+R_(x)]^(2) ] = 0` On simplifying, we get `R_(1)R_(2) - (R_(1) + R_(2)) R_(x) = 0 or R_(x) = R_(1)R_(2)/(R_(1) + R_(2))` `:. R_(x) = (20 xx 30)/(20 + 30) = 12 Omega` |
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| 2147. |
To get the maximum current from a parallel combination of `n` identical cells each of internal resistance `r` in an external resistance `R`, whenA. SeriesB. ParallelC. MixedD. Depends upon the relative values of exterbal and internal resistane |
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Answer» Correct Answer - D |
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| 2148. |
A series-parallel combination battery consists of 300 identical cells, each with an internal resistance `0.3 Omega`. It is connected to the external resistance `10 Omega`. Find the number of parallel goups cosisting of equal number of cells connected in series, at which the external resistance generated the higher thermal power. |
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Answer» Let there be m rows of cells and n cells in each row of a battery. Total number of cells, `N = nm or n = N//m` Let `epslion`,r be the mef and internal resistance of each cell and R be the external resistance. Then total internal resistance of all the m rows of cells ` = nr//m`. Total resistance of the whole circuit `= (R + nr//m)` Total emf of all the cells `=n epsilon` Current in the external resistance R will be `I = (n epsilon)/(R + nr//m) = ((N//m)epsilon)/([R + (N//m) r//m]) = (m N epsilon)/(m^(2)R + N r)` Heat generated in resistance R is `H = I^(2)R= ((m N epsilon)/(m^(2)r + Nr))^(2) R` For H to be maximum, `(dH)/(dm) = 0` Differentiating (ii), w.r.t. m and equation to zero, we get `m =sqrt(Nr//R) = sqrt(300 xx 0.3//10) = 3` |
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| 2149. |
To get the maximum current from a parallel combination of `n` identical cells each of internal resistance `r` in an external resistance `R`, whenA. `R gt gt r`B. `R lt lt r`C. `R = r`D. None of these |
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Answer» Correct Answer - B (b) Cells are joined in parallel when internal resistance is higher then a external resistance . `(R lt lt r)` `i= (E)/(R + (r )/(n))` |
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| 2150. |
Two identical calls send the same current in `2 Omega` resistance, whether connected in series or in parallel. The internal resistance of the cell should beA. `1 Omega`B. `2Omega`C. `(1)/(2)Omega`D. `2.5 Omega` |
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Answer» Correct Answer - B |
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