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InterviewSolution
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.
| 51. |
In the circuit shown in figure, the capacitor is charged with a cell of `5V`.If the switch is closed at `t=0`, then at `t=12s`, charge on the capacitor is A. `(0.37)10muC`B. `(0.37)^210muC`C. `(0.63)10muC`D. `(0.63)^210muC` |
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Answer» Correct Answer - B `tau_c=CR=6s` `q_0=CV=10muC` `Now, q=q_0e^(-t/tauC)=(10muC)e^(-12/6)` `=(1/6)^2(10muC)` `=(0.37)^2(10muC)` |
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| 52. |
Find the charge stored in the capacitor. |
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Answer» Correct Answer - B::C::D In steady state, current flows in lower loop of the circuit. `i=30/(6+4)=3A` Now, potetial difference across capacitor `=` potential difference across `4Omega` resistance `=iR` `=(3)(4)12V` `:. Q=CV=(2muF)(12V)` `=24muF` |
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| 53. |
Two parallel plates have equal and opposite charges. When the space between the plates is evacuted, the electric field is `E_0=3.20xx10^5V//m`. When the space is filled with electric the electric field is `E-2.50xx10^5V//m` a. What is the dielectric constant? b. What is the charge density on each surface of the dielectric? |
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Answer» a. `K=E_0/E=(3.20xx10^5)/(2.50xx10^5)` `=1.28` b. `sigma_i=sigma_0(1-1/K)` `=(E_0epsilon_0)(1-1/K)` `=(93.20xx10^5)(8.86xx10^-12)(1-1/1.28)` `=6.2xx10^-7C//m^2` |
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| 54. |
Find the charge stored in the capacitor. |
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Answer» Correct Answer - 40uC Capacitors and resistor both are in paralel with the battery.`PD` across capacitor is `10 V`. Now, apply `q=CV`. |
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| 55. |
Find the charged stored in all the capacitors |
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Answer» Correct Answer - C All three capacitors are in parallel with the battery. `PD` across each of them is 10 `V`. So, apply `q=CV` for all of them. |
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| 56. |
Two capacitors of equal capacitance when connected in series hae net capacitance `C_(1)` and when connected in parallel have net capacitance `C_(2)` what is the value of `C_(1)//C_(2)`? |
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Answer» Correct Answer - `C_(1)//C_(2)=1//4` |
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| 57. |
Calculate the equivalent capacitance between the points A and B of the circuit shown in figuer. |
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Answer» Correct Answer - `C=(8)/(7)muF` |
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| 58. |
A parallel-plate capacitor has each plate of 6 cm diamter. If its two plates are separated by 0.05 cm of air, what should be the capacitance of the capacitor? What would be the radius of a spherical conductor having the same capacitance? |
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Answer» Correct Answer - 50pF,0.45m |
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| 59. |
A capacitor is given a charge `q`. The distance between the plates of the capacitor is `d`. One of the plates is fixed and the other plate is moved away from the other till the distance between them becomes `2d`. Find the work done by the external force. |
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Answer» Correct Answer - A::B::D When one plate is fixed, the other is attracted towards the first with a force `F=q^2/(2Aepsilon_0)=constant` Hence, an external force of same magnitude will have to be applied in oppositve direction to increase the separation between the plates. `:. W=F(2d-d)=(q^2d)/(2Aepsilon_0)` Alternate solution: `W=/_U=U_f-U_i=q^2/(2C_f)-q^2/(2C_i)`..........i Here, `C_f=(epsilon_A)/(2d)` and `C_i=(epsilonA)/d` Substituting in eqn i we have `W=q^2/(2((epsilonA)/(2d)))-q^2/(2((epsilon_0A)/d))=(q^2d)/(2epsilon_0A)` |
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| 60. |
In the circuit shown in figure find a. the equivalent capacitance and b. the charge stored in each capacitor. |
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Answer» a. the capacitors are in parallel. Hence, the equivalent capacitance is `C=C_1+C_2+C_3` or `C=(1+2+3)=6muF` b. total charge drawn from the battery `K` `q=CV-6xx100muC=600muC` This charge will be distributed in the ratio of their capacities. Hence, `q_1:q_2:q_3=C_1:C_2:C_3=1:2:3` `:. q_1=(1/(1+2+3))xx600=100muC` `q_2=(2/(1+2+3))xx600=200muC` and `q_3=(3/(1+2+3))xx600=300muC` Alternate solution: Since the capacitors are in parallel the `PD` across each of them in `100V`. Therefore from `q=CV`, the charge stored in `1muF` capacitor is `100muC`, in `2muF` capacitor is `200muC` and that in `3muF` capacitor is `300muC` |
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| 61. |
An insulated conductor initiallly free from charge is charged by repeated contacts with a plate which after each contact is replenished to a charge `Q`. If `q` is the charge on the conductor after first operation prove that the maximum charge which can be given to the conductor in this way is `(Qq)/(Q-q)` |
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Answer» Let `C_1` be the capacity of plate and `C_2` that of the conductor. After first contact charge on conductor is `q`. Therefore, charge on plate will remain `Q-q`. As the charge redistributes in the ratio of capacities. `(Q-q)/q=C_1/C_2`………i Let `q_n` be the maximum charge which can be given to the conductor. The flow of charge from the plate to the conductor will stop when `V_("conductor")=V_(plate)` `q_m/C_2=Q/C_1implies q_m=(C_2/C_1)Q` Substituting `C_2/C_1` from eqn i we get `q_m=(Qq)/(Q-q)` |
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| 62. |
Two parallel plate vacuum capacitors have areas `A_1` and `A_2` and equal plate spacing `d`. Show that when the capacitors are connected in parallel, the equivalent capacitance is the same as for a single capacitor with plate area `A_1+A_2` and spacing `d`. |
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Answer» `C=C_1+C_2`(in parallel) `:.(epsilon_A)/d=(epsilonA_1)/d=(epsilon_0A_2)/d` or `A=A_1+A_2` |
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| 63. |
Three capacitors each of capacitor `2muF` are connected in series. Find resultance capacity in farad. |
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Answer» Correct Answer - `0.67xx10^(-6)F` |
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| 64. |
A capacitor of capacity `2muF` is charged to `100 V`. What is the heat generated when this capacitor is connected in parallel to n another capacitor of same capacity?A. `2.5mJ`B. `5.0mJ`C. `10mJ`D. `4mJ` |
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Answer» Correct Answer - B `q=CV=200muC` In parallel the common potential is given by `V=("Total charge")/("Total capacity")` `=(200muC)/((2+2)muF)=50V` Heat loss=`U_i-U_f` `=1/2(2xx10^-6)(100)^2-1/2(4xx10^-6)(50)^2` `=5xx10^-3J` `=5mJ` |
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| 65. |
A charged capacitor is discharged through a resistance. The time constant of the circuit is `eta`. Then the value of time constant for the power dissipated through the resistance will beA. `eta`B. `2eta`C. `eta/2`D. zero |
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Answer» Correct Answer - C `P=i^2R=(i_0e^(-t//eta))^2R` `=(i_0^2R)e^(-2t//eta)` `=P_0e^(-t//(n//2))` Hence, the time constant is `eta/2`. |
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| 66. |
A capacitor is charged by a cell of emf `E` the chaging battery is then removed. If an identical capacitor is now inserted in te circuit in parallel with the prevous capacitor, the potential difference across the new capacitor isA. `2E`B. `E`C. `E/2`D. zero |
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Answer» Correct Answer - C common potential in parallel grouping `=("Total charge")/("Total capacity")` `(EC)/(2C)=E/2` |
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| 67. |
Figure shows a parallel plate capacitor with plate area `A` and plate separation `d`. A potential difference is being applied between the plates. The battery is then disconnected and a dielectric slab of dieletric constant `K` is placed in between the plates of the capacitor as shown. The electric field in the gaps between the plates and the electric slab will beA. `(epsilon_0AV)/d`B. `V/d`C. `(KV)/d`D. `V/(d-t)` |
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Answer» Correct Answer - B `E_(air)=E_0=V/d` |
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| 68. |
A capacitor of capacitance `5muF` is connected to a source of constant emf of `200 V`,Then switch was shifted to contact 2 from contact 1. Find the amount of heat generated in the `400Omega` resistance. |
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Answer» Correct Answer - D Potential energy stored in the capacitor `U=1/2 CV^2=1/2xx5xx10^-6xx(200)^2=0.1J` During discharging this `0.1J` will distribute in direct ratio of resistance `:. H_400=400/(400+500)xx0.1` `=44.4xx10^-3J` `=44.4mJ` |
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| 69. |
In the circuit shown in figure. If point B is earthed and A is kept at 1500 V, then calculate the potential at the point P. |
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Answer» Correct Answer - 750 V |
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| 70. |
A parallel-plate capacitor consists of two square plates of effective area A, length L and plate separation d. If one of the plates is turned through a small angle `theta`, what is the change in the capacitance of the capacitor? |
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Answer» Correct Answer - `(epsilon_(0)ALtheta)/(2d^(2))` |
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| 71. |
A parallel plate air capacitor consists of two circular plates of diameter `8 cm` . At what distance should the plates be held so as to have the same capacitance as that of a sphere of a diameter `20 cm` ? |
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Answer» Correct Answer - 4mm |
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| 72. |
find the capacitance between the points A and B of the assembly shown in figure. |
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Answer» Correct Answer - `2.25muF` |
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| 73. |
The capacitor `C_1` in the figure shown initially carries a charge `q_0`. When the switches `S_1` and `S_2` are closed,capacitor `C_1` is connected in series to a resistor `R` and a second capacitor `C_2` which is initially uncharged. The flown through wires as a function of time is where `C=(C_1C_2)/(C_1+C_2)`A. `q_0e^(-t/(RC))+C/C_2q_0`B. `(q_0C)/C_1x[1-e^(-t/(RC))]`C. `q_0C/C_1e^(-t/(CR))`D. `q_0e^(-t/(RC))` |
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Answer» Correct Answer - B Finally the capacitors are in parallel and total charge `(=q_0)` distributes between them in direct ratio of capacity. `:. q_(C_2)=(C_2/(C_1+C_2))q_0rarr` in steady state . but this chasrge increases exponentially. Hence, charge on `C_2` at any time t is `q_(C_2)=((C_2q_0)/(C_1+C_2))(1-e^(-t/tau_C))` Initially `C_2` is uncharged so, what ever is the charge on `C_2` it is charge flown through switches. |
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| 74. |
The capactor `C_1` in the figure initially carries a charge `q_0`. When the i switch `S_1` and `S_2` are closed, capacitor `C_1` is connected to a resistor `R` and a second capacitor `C_2`, which initially does not carry any charge. (a) Find the charges deposited on the capacitors in steady state and the current through R as a function of time. (b) What is heat lost in the resistor after a long time of closing the switch? |
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Answer» Correct Answer - A::B::C In steady state, capacitors will be in parallel. Charge will distribute in direct ratio of their capacity. `:. q_1=(C_1/(C_1+C_2))q_0` and `q_2=(C_2/(C_1+C_2))q_0` Initial emf in the circuit is potential difference across capacitor `C_1` or `q_0/C_1`. Therefore, initial current would be `i_0=(q_0//C_1)/(R)=(q_0)/(C_1R)` Current as function of time will be `i=i_0e^(-t//tau_C)` Here, `tau_C=((C_1C_2)/(C_1+C_2))R` b. `U_i=1/2q_0^2/C_1` and `U_f=1/2q_0^2/(C_1+C_2)` Heat lost in the resistor `=U_i-U_f=q_0^2/2[C_2/(C_1(C_1+C_3))]` |
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| 75. |
For the arrangement shown in the figure, the switch is closed at `t = 0`. `C_2` is initially uncharged while `C_1` has a charge of `2muC` (a) Find the current comingn out of the battery just after the switch is closed. (b) Find the charge on the capacitors in the steady state condition. |
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Answer» Correct Answer - B At `t=0`, equivalent resistance of an uncharged capacitor is zero and a charged capacitor is like a bttery of emf = potential difference across the capacitor. a. `V_(C_1)=q_1/C_1=2V` `:.` Net emf of the circuit `=9-2=7V` or `92=11V` Net resistance `=30+(30xx60)/(30+60)=50Omega` `:.` Current at `t=0` would be `i_0=7/50A` or `11/50A` b. In steady state no current will flow through the circuit `C_2` will therefore be short circuited, while `PD` across `C_1` will be 9 V `:. Q_2=0 and Q_1=9muC` |
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| 76. |
In the circuit shown switch `S` is closed at `t=0`. Let `i_1` and `i_2` be the current at any finite time t then the ratio `i_1/i_2` A. is constantB. increases with lineC. decrease with timeD. first increases then decreases |
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Answer» Correct Answer - B `i_1=(V/(2R))e^(-t/(6CR))` `:. i_2=(V/R)e^(-t/(CR))` `:. i_1/i_2=e^((5t)/(6CR))/2` we can that this ratio is increasing with time. |
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| 77. |
In the arrangement shown in figure dielectric constant `K_1=2` and `K_2=3`. If the capacitance across `P` and `Q` are `C_1` and `C_2` respectively, then `C_1//C_2` wil be (the gaps shown are negligible) `A. `1:1`B. `2:3`C. `9:5`D. `25:24` |
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Answer» Correct Answer - D `C_1=C_(RHS)+ C_(LHS)` `=(K_2epsilon_0(A/2))/d=(K_1epsilon_0(A/2))/d` `(epsilon_0A)/(2d)(K_1+K_2)=(5epsilon_0A)/(2d)` `C_2=(epsilon_0A)/(d-d/2-d/2+(d/2)/K_1+(d/2)/K_2` `=(2epsilon_0A)/d((K_1K_2)/(K_1+K_2))` `=(12epsilon_0A)/(5d)` `:. C_1/C_2=25/24` |
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| 78. |
Three capacitors of capacitance `2 pF, 3 pF and 4 pF` are connected in parallel. (a) what is the total capacitance of the combination ? (b) Determine the charge on each capacitor, If the combination is connected to 100V supply. |
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Answer» Correct Answer - `9pF,2xx10^(-10)C,3xx10^(-10),4xx10^(-10)C` |
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| 79. |
Three capacitors each of capacitance 9 pF are connected in series. (a) What is the total capacitance of the combination ? (b) What is the potential difference across each capacitor if the combination is connected to a 120V supply. |
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Answer» Correct Answer - 3pF, 40V |
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| 80. |
A combination of four identical capacitors is shown in Fig. IF resultant capacitance of the combination between the points `P` and `Q` is `1 mu F`, calculate capacitance of each capacitor. |
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Answer» Correct Answer - `4muF` |
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| 81. |
A spherical shell of radius `R_(1)` with uniformly distributed charge q has a point charge `q_(0)` at its centre. Find the work performed by the electric forces during the expansion of the shell from `R_(1)` to `R_(2)`. [Hint: W=`W_(1)`(decrease in self potential energy)`+W_(2)`(work done by electrical force of `q_(0)` on q)] |
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Answer» Correct Answer - `(q(q_(0)+q//2))/(4piepsilon_(0))((1)/(R_(1))-(1)/(R_(2)))` |
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| 82. |
The equivalent capacitance of the combination shown in figure (31 -Q1) is(a) C (b) 2C (c) C/2 (d) none of these. |
| Answer» The correct answer is (b) 2C. | |
| 83. |
Assertion: In the circuit shown in figure no charge will stored in the capacitor. Reason: Current through `R_2` will be zero.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - B Capacitor and `R_2` are short circuited. Hence, current through `R_2` is zero and capacitor is not charged. |
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| 84. |
Assertion: Two capacitors are connected in series with a battery. Energy stored across then is in inverse ratio of their capacity. Reason: `U=1/2qV` or `UpropqV`A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - B `U=1/2 q^2/C or Uprop1/C ` as `q` is same in capacitors (if initially they are uncharged)` |
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| 85. |
An electrical circuit is shown in the given figure. The resistance of each voltmeter is infinite and ech ammeter is `100Omega`. The charge on the cpacitor of `100muF` in steady staste is `4mC`. Chose correct statement(s) regarding the given circuit. A. Reading of voltmeter `V_2 is 16`B. Reading of ammeter `A_1` is zero and `A_2` is `1/25A`C. Reading of voltmeter `V_1` is `40 V`D. Emf of the ideal cell is `66 V` |
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Answer» Correct Answer - B::C::D In steady state current through capacitor wire is zero. Current flows `200Omega, 900Omega` and `A_2`. `V_C=q/C=(4xx10^-3)/(100xx10^-6)` `=40V` This is also potential drop across `900Omega` resistance and `100Omega` ammeter `A_2` (Total resistance `=1000Omega`). Now, this `1000Omega` and `200Omega` are in series. Therfore, `V_2=V_(200Omega)=V_(1000Omega)/5` `40/5=8V` `Emf=V_(1000Omega)+V_(200Omega)=48V` `i=("Emf")/("Net resistance")` `=48/1200=1/25A` |
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| 86. |
Assertion: In the circuit in figure, time constant of charging of capacitor is s`(CR)/2` Reason: In the absence of capcitor in the circuit, two resistors are in parallel with the battery.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - D Capacitor and resistance in its own wire are directly connected with the battery. Hence, time constant during charging is `CR`. |
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| 87. |
In the circuit shown, `A` and `B` are equal resistances. When `S` is closed, the capcitor Ccharges from the cell of emf epsilon and reaches a steady state. A. During charging, more heat is produced in A than in BB. In stedy state heat is produced at the same rate in A and BC. In the steady state, energy stored in C is `1/4Cepsilon^2`D. In the steady state energy stored in C is `1/8Cepsilon^2` |
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Answer» Correct Answer - A::B::D Current through `A` is the main current passing through the battery. So, the current is more than the current passing through `B`. Hence, during charging more heat is produced is `A`. In steady state `i_C=0` `and i_A=i_B` Hence, heat is produced at the same rate in `A` and `B`. further, in steady state `V_C=V_B=epsilon/2` `:. U=1/2CV_C^2=1/8Cepsilon^2` |
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| 88. |
A `5.80 muF` parallel-plate air capacitor has a plate separation of `5.00 mm` and is charged to a potential difference of `400 V`. Calculate the energy density in the region between the plates, in `J/m^3` |
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Answer» `u=1/2epsilon_0E^2` `=1/2epsilon_0(V/d)^2` |
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| 89. |
Two condensers are in parallel and the energy of the combination is `0.1 J`, when the difference. of potential between terminals is `2 V`. With the same two condensers in series, the energy `1.6 xx 10^-2` J for the same difference of potential across the series combination. What are the capacities? |
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Answer» Correct Answer - A `0.1=1/2(C_1+C_2)(2)^2`………….i `1.6xx10^-2=1/2((C_1C_2)/(C_1+C_2))(2)^2`…………ii Solving these two equations, we can find `C_1` and `C_2` |
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| 90. |
The plates of a parallel plate capacitor have an area of `100cm^(2)` each and area separated by 2.5 mm. the capacitor is charged to 200V. Calculate the energy stored in the capacitor. |
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Answer» Correct Answer - `7.08xx10^(-7)J` |
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| 91. |
An electron enters the region between the plates of a parallel plate capacitor at an angle `theta` to the plates. The plate width is `l`, the plate separation is `d`. The electron follows the path shown just missing the upper plate. Neglect gravity. Then, `A. `tantheta=2d/l`B. `tantheta=4d/l`C. `tantheta=8d/l`D. The data given is insufficient to find a relation between d, l and `theta` |
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Answer» Correct Answer - B Horizontal range, `R=(2u_x xxu_y)/g=l`…….i Maximum height `H=u_y^2/(2g)=d`……..ii Dividing eqn ii by eqn i we have `1/4(u_y/u_x)=d/l` or `u_y/u_x=(usintheta)/(ucostheta)` `=tantheta=(4d)/l` |
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| 92. |
The separation between the plates of a charged parallel-plate capacitor is increased. Which of the following quantities will change ? (a) charge on the capacitor (b) potential difference across the capacitor (c) energy of the capacitor (d) energy density between the plates. |
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Answer» The correct answer is (b) (c) (b) potential difference across the capacitor |
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| 93. |
When the switch is closed, the initial current through the `1Omega` resistor is `A. `2A`B. `4A`C. `3A`D. `6` |
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Answer» Correct Answer - B At `t=0` , when capacitor is under charged, equivalenresistance of capacitor `=0` in this case `6Omega` and `3Omega` are parallel (equivalent `=2Omega)` `:. R_(n et)(1+2)Omega=3Omega` `:. "Current from battery" =12/3 =4A` `= "Current through" `1Omega` resistor |
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| 94. |
For the circuit arrangement shown in figure, in the steady state condition charge on the capacitor is `A. `12muC`B. `14muC`C. `2muC`D. `18muC` |
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Answer» Correct Answer - D In steady state condition, current flows from outermost loop. `i=12/(6+2)=1.5A` Now, `V_C=V_(6Omega)=iR` `=1.5xx6=9V` `:. q=CV_C=18muC` |
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| 95. |
A dielectric slab is inserted between the plates of an isolated charged capacitor. Which of the following quantities will remain the same ?(a) the electric field in the capacitor(b) the charge on the capacitor (c) the potential difference between the plates(d) the stored energy in the capacitor. |
| Answer» (d) the stored energy in the capacitor. | |
| 96. |
The separation between the plates of a charged parallel-plate capacitor is increased. The force between the platesA. increasesB. decreasesC. remains sameD. first increases then decreases |
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Answer» Correct Answer - C `F=qE=q(sigma/(2epsilon_0))=q(q/(2Aepsilon_0))q` will not change `:.` F=constant |
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| 97. |
The current in `1Omega` resistance and charge stored in the capacitor are A. `4A,6muC`B. `7A,12muC`C. `4,12muC`D. `7A,6muC` |
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Answer» Correct Answer - B `V_(1Omega)=5+2=7V` `:. i_(1Omega)=V/R=7A` `V_(2muF)=6V` `:. q_(2muF)=CV=12muC` |
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| 98. |
An isolated parallel plate capacitor has circular plates of radius `4.0cm`. If the gap is filled with a partially conducting material of dielectric constant `K` and conductivity `5.0xx10^-14Omega^-1m^-1`. When the capacitor is charged to a surface charge density of `15muC//cm^2`, the initial current between the plates is `1.0muA`? a. Determine the value of dielectric constant `K`. b. If the total joule heating produced is `7500J`, determine the separation of the capacitor plates. |
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Answer» a. This is basically a problem of discharging of a capacitor from inside the capacitor. Charge at any time `t` is Here, `q_0=` (area of plates )(surface charge density) and discharging current, `i=((-dq)/(dt))=q_0/tau_C.e^(-t//tau_C)=i_0e^(-t//tau_C)` Here `i_0=q_0/tau_C=q_0/(CR)` `C=(Kepsilon_0A)/d` and `R=d/(sigmaA)` `:. CR=(Kepsilon_0)/sigma` Therfore, `i_0=q_0/((Kepsilon_0)/sigma)=(sigmaq_0)/(Kepsilon_0)impliesK.(sigmaq_0)/(i_0epsilon_0)` substituting the values, we have `K=((5.0xx10^-14)(pi)(4.0)^2(15xx10^-6))/((1.0xx10^-6)(8.86xx10^-12))=4.25` b. `U=1/2q_0^2/C=1/2q_0^2/((Kepsilon_0A)/d)` `:. d=(2Kepsilon_0AU)/q_0^2` `=(2xx4.25xx8.86xx10^-12xxpixx(4.0xx10^-2)^2xx7500)/((15xx10^-6xxpixx4.0xx4.0)^2)` `=5.0xx10^-3m=5.0mm` |
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| 99. |
Consider a capacitor charging circuit. Let `Q_1` be the charge given to the capacitor in time interval of `20 ms` and `Q_2` be the charge given in the next time interval of `20ms`. Let 10`muC` charge be deposited in a time interval `t_1` and the next `10muC` charges is deposited in the next time interval `t_2`. ThenA. `Q_1gtQ_2,t_1gtt_2`B. `Q_1gtQ_2,t_1ltt_2`C. `Q_1ltQ_2,t_1gtt_2`D. `Q_1ltQ_2,t_1ltt_2` |
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Answer» Correct Answer - B Initially the rate of charging fast |
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| 100. |
A parallel-plate capacitor has plates of area `A` and separation `d` and is charged to a potentialdifference `V`. The charging battery is then disconnected, and the plates are pulled apart until their separation is `2d`. Derive expression in terms of `A, d` and `V` for (a) the new potential difference (b) the initial and final stored energies, `U_i` and `U_f` and (c) the work required to increase the separation of plates from d to `2d`. |
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Answer» Correct Answer - A::B::D a. `q=C_iV=((epsilon_0A)/d)V` `V_f=q/C_f=((epsilon_0AV//d))/((epsilon_0A//2d))=2V` b. `U_i=1/2C_iV^2=1/2((epsilon_0A)/d)V^2` `U_f=1/2C_fV_f^2=1/2((epsilon_0A)/(2d))(2V)^2` `=((epsilon_0A)/d)V^2` c. `W=U_f-U-i=1/2((epsilon_A)/d)V^2` |
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