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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.
| 1. |
The massses of the three wires of copper are in the ratio 1 : 3 : 5. And their lengths are in th ratio 5 : 3 : 1. the ratio of their electrical resistance isA. `1:03:05`B. `5:03:01`C. `1 : 15 : 125`D. `125 : 15 : 1` |
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Answer» Correct Answer - D `R = (rhol)/(A) = (rhol^(2))/(Al) = (rhol^(2))/(V) = (rhol^(2))/(m//d) = (rhodl^(2))/(m) or R prop (l^(2))/(m)` `R_(1) : R_(2) : R_(3) = (l_(0)^(2))/(m_(1)): (l_(1)^(2))/(m_(2)) : (l_(3)^(2))/(m_(3)) = (25)/(1) : (9)/(3) : (1)/(5) = 125 : 15 : 1` |
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| 2. |
In the circuit shown in figure find: a. the current in the `3.00 Omega` resistor,b. the unknown emfs`E_1` and `E_2` and c the resistance `R`. |
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Answer» a. The sum of the currents that enter the junction below the resistor equals `3A + 5A = 8A` b. Using the lower left loop, we get `epsilon _(1) - (4Omega)(3A)- (3Omega)(8A) = 0 or epsion_(1) = 36V` Using the lower right loop , we get `epsilon - (6Omega)(5A) - (3Omega)(8A) = 0` or `epsilon_2 =54V` c. Using the top loop, we get `54 V-R(2A) - 36 V = 0` or `R(18V)/(2A) = 9Omega` |
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| 3. |
A single battery is connected to three resistances as shown in fig. 5.316 A. The current through `7Omega` resistance is 4AB. The current through`3 Omega`resistance is 4A.C. The current through `6 Omega` resistance is 4A.D. The current through `7Omega` resistance is 0. |
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Answer» Correct Answer - B::C::D The `7Omega` resistor si short - circuited, so no current will flow through it. Potential difference across each of `3Omega and 6Omega` is 12 V , so we can find current in them. |
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| 4. |
For the batteries shown in fig. `R_1, R_2 and R_3` are the internal resistance of `E_1, E_2, and E_3,` respectively Then, which of the following is / are correct? A. The equivalent internal resistance R of the system is given by `(R_1R_2R_3)//(R_1R_2 +R_2R_3+R_3R_1).`B. If `E_3 = (E_1R_2 +E_2R_1)//(R_1+R_2),` the equivalent emf of the batteries will be equal to `E_3`C. The equivalent emf to the battery is equal to `E = (E_1 +E_2 +E_3)//3`D. The equivalent emf of the battery not only depends upon values of `E_1,E_2, and E_3` but also depends upon values of `R_1, R_2 and R_3.` |
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Answer» Correct Answer - A::B::D We know that the equivalent internal resistance is `R_(eq) = (1)/(sum(1)/(2)) = (1)/((1)/(R_(1)) +(1)/(R_(2)) +(1)/(R_(3)))` `=(R_(1)R_(2)R_(3))/(R_(1)R_(2)+R_(2)R+3+R_(3)+R_(1))` and the equivalent emf is `E_(eq)=(sum(E )/(R ))/(sum(1)/(R ))=((E_(1))/(R_(1))+(E_(2))/(R_(2))+(E_(3))/(R_(3)))/((1)/(R_(1))+(1)/(R_(2))+(1)/(R_(3)))` Use `E_(3) = (E_(1)R_(2) + E_(2)R_(1))/(R_(1) + R_(2))` to get `E_(eq) = (E_(1)R_(2) +E_(2)R_(1))/(R_(1)+R_(2)) = E_(3)` |
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| 5. |
In charging singe loop R - C circuit, after how many time constant, kthe potential energy in the capacitor will be 25% of its maximum value. |
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Answer» The potential energy of the capacitor at time t is `U = (q^(2))/(2C)` where `q = C epsilon(1-e^(t//RC))` Then `U = (C epsilon)/(2) (1 - e^(-t//RC))^2` where `(C epsilon^(2))/(2) = U_(0)` Putting `U = (U_(0))/(4) = (C epsilon^(2))/(8)`, we have `(1- e^(-t//RC))^(2) = (1)/(4)` or `1=e^(-t//RC) = (1)/(2)` or `e^(-t//RC) = (1)/(2)` or `(t)/(RC) In e = - In 2 or tau = (t)/(RC) = ln2` |
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| 6. |
How many time constant will elapse before the power delivrered by the battery drops to half of its maximum value. |
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Answer» Power delivered by a battery is given by `P = EI. P prop I`, so if power becomes half, current becomes half. `I = I_0e^(-t//tau)(I_(0))/(2)=I_(0)e^(-t//tau)`or `e^(t//tau)=2` or `t = tau In 2 = 0.69tau` |
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| 7. |
If 0.6 mol of electorns flows throutgh a wire in 45 min, what are (a) the total charges that passes through the wire and (b) the magnitude of the current? |
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Answer» a. The number N of electrons in 0.6 mol is `N =(0.6mol)(6.02xx10^(23)"electrons"//"mol"^(-1))` `=3.6xx10^(23)"electrons"` `q = Ne = (3.6xx10^(23))(1.6xx10^(-19)) = 5.78xx10^4` b. `t = (45min)(60smin^(-1)) = 2.7xx10^3S` `I = (q)/(t) = (5.78x10^4)/(2.7xx10^3Cs)=21.4A` |
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| 8. |
The current through the `8Omega` resistor (shown in fig.) is A. 4AB. 2AC. zeroD. 2.5A |
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Answer» Correct Answer - C The lower limit is zero volt (0 V) when X is at the lower end of the `4 k Omega` resistor. The upper limit is the potential differnece across the `4 k Omega` resistor when X is at the upper end of the `4 k Omega` resistor. That is `V= ((25)/(1K + 4K))4K = ((25)/(5)) 4 = 20V` Thus, the limits are 0 and 20 V. |
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| 9. |
In the network shown in fig. , the potential differene across A and B is A. 6VB. zeroC. 2VD. 4V |
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Answer» Correct Answer - B Notice the polarities of the batteries. The batteries will cancel each other and finally there will be no current anywhare in the circuit. |
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| 10. |
The voltage-current variations of two metallic wire X and Y at constant temperature is shown in fig Assumign that the wires have the same length and the same diameter, explain which of the two wires will have larger resisitivity. |
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Answer» `R = (rhol)/(A), "so " rho prop R. "But" R_Y gt R_X.` Because for same V, current in Y is less. So `rho_y gt rho_x.` |
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| 11. |
Relation between current in a conductor and time is shown in fig. Write the expression of current in terms of time.A. `i = i_0(t)/(t_0)`B. `i= i_0(1+(t)/(t_0))`C. `i=i_0((t)/(t_0) - 1)`D. `I = i_0(1-(t)/(t_0))` |
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Answer» Correct Answer - D `q = int idt =` area of given curve `= (1)/(2)i_0t_0` `(i)/(i_0) +(t)/(t_0) 1` or `I = i_0(1-(t)/(t_0))` Heat `= int i^2 Rdt = int_0^(t_0)i_0^2R(1-(t)/(t_0))^2 dt` `H = (i_0^2R(1-(t)/(t_0))^3)/(3(-(1)/(t_0)))]_0^(t0) or H = (Rt_0i_0^2)/(3)` |
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| 12. |
Relation between current in a conductor and time is shown in fig. Total charge flowing through the conductor isA. `i_(0)t_(0)//2`B. `i_(0)t_(0)`C. `i_(0)t_(0)//4`D. `2i_(0)t_(0)` |
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Answer» Correct Answer - A `q = int idt =` area of given curve `= (1)/(2)i_0t_0` `(i)/(i_0) +(t)/(t_0) 1` or `I = i_0(1-(t)/(t_0))` Heat `= int i^2 Rdt = int_0^(t_0)i_0^2R(1-(t)/(t_0))^2 dt` `H = (i_0^2R(1-(t)/(t_0))^3)/(3(-(1)/(t_0)))]_0^(t0) or H = (Rt_0i_0^2)/(3)` |
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| 13. |
The plates of a `50(mu)F` capacitor charged to `400(mu)C` are connected through a resistance of `1.0k(Omega)`. Find the charge remaining on the capacitor 1s after the connection is made. |
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Answer» `q_(1) = 400 muC, C = 50muF` `R = 1000Omega, tau = RC = 50xx10^(10)xx1000= 5xx10^(-2)s` `q = q_(0)e^(-t//tau) = 400e^((1)/(5xx10^(-2))) = (400)/(e^(+20))muC` |
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| 14. |
The space between the plates orf a parallel plate capacitor is completely filled with a meterail of resistaivity `2xx10^(11) Omegam` and dielectric constant 6. Capacity of the capacitor with the given dielectric medium between the paltes is 20uF. Find the leakage current if a potential difference 2500 V is applied across the capacitor. |
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Answer» Charge on the capacitor, `Q = CV = 20xx10^(-19) xx2500 = 5xx10^(-5)C` Surface charge density is `sigma_s = (A)/(Q)` [A is the plate area.] Electric field strength between the plates is `E = (sigma_(s))/(K epsilon_0) = (Q)/(Kaepsilon_0)` Using `J = simgaE = (E )/(rho)` [Here `simga` is the conducitivity and `rho` is the resistivity.] Current density is `J = Q//KA epsion_0rho` and current is `JA = (Q)/(Kepsilonrho) =(5xx10^(-5))/(6xx(8.85xx10^(-22))xx(2xx10^(11))` `=4.7xx10^(-6) A = 4.7muA` |
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| 15. |
Determine the voltage drop across the resistor `R_1` in the circuit given below with `E = 65V, R_1 = 50 OmegaR_2 = 100 Omega, R_3 = 100 Omega, and R_4 300 Omega. |
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Answer» `R_3 and R_4` are in series. The combined resistance is `400 Omega.` Now `R_2 (-100Omega)and 400Omega` resistance are in parallel. The combined resistance is `(100xx400)//500 = 80Omega.` Total resistance is `R = 80 +50 = 130 Omega` `I = 65//130 = 1//2A.` So, `V = IR_1 = 1.//2 xx50 = 25V` |
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| 16. |
Consider the circuit shown in fig. If the switch is closed at t =0, then calculate the values of `I, I_(1)` and `I_(2)` at a. t =0 , b. `= oo` |
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Answer» At t = 0, C wil behave as a short circuit, so no current passes through `R_(2)` and `I = I_(2) = epsilon //R_(1).` b. `At t = oo, C` will block thr current. So `I_(2) = 0.` and `I = I_(1) = (epsilon)/(R_(1) + R_(2))` |
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| 17. |
Current through the battery at the instance when the switch S is closed is A. zeroB. 2AC. 4AD. 5A |
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Answer» Correct Answer - D During charging, `tau_(1) =RC,` during discharging, `tau_(2) = 2RC.` Therefore, ratio is `(tau_(1))/(tau_(2)) = (1)/(2) =1 : 2` |
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| 18. |
Determine the current through the battery in the circuit shown in figure. (a) immediately after the switch S is closed (b) after a long time. |
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Answer» a. Immediately after the swithc is closed,`C_(1)` wil act as a simle wire due to which `R_(2)` and `R_(3)` will be short - circuited. So, `I = E//R_(1)` b. After a long time, `C_(1)` and `C_(2)` will block the current. Current will pass through onlu `R_(1)` and `R_(2) I =(E )/(R_(1) + R_(3))` |
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| 19. |
In fig , after closing switch S, what is the change in curent flowing through A? the battery is ideal. A. no changeB. decreasesC. increasesD. cannot say |
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Answer» Correct Answer - A All the four resistacne are in parrallel to E. So current in them flows independiently. Hence, no change in current flowing in A after closing the swithc. |
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| 20. |
In the above question, what would have been the change in current in A if battery were some internal resistance.A. no changeB. decreasesC. increasesD. cannot say |
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Answer» Correct Answer - B On connecting the switch, the current drawn by the resistance through the battery will increase. This will decrease the terminal potential difference (V = E - ir) across the cell and hence the potential differnce across A will also decrease. So the current through A will decrease. |
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| 21. |
N identical calls are cannected to from a battery. When the terminals of the battery are joined directly. When the terminals of the battery are joined directly (short - ciruited), current I flows in the circuit. To obtain the maximum value of I,A. all the cells should be joined in seriesB. all the cells should be joined in parallelC. two rows fo N //2 cells each should be joined in parallelD. `sqrtN` rows of `sqrtN` cells each should be joined in parlallel, given that `sqrtN` is an interger |
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Answer» Correct Answer - B For series connection , `I_(min) = (N epsilon)/(Nr) = (epsilon)/(r )` For parallel connection, `i_(max) = (epsilon)/(r//N)= (N epsilon)/(r )` |
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| 22. |
The emf of a cell is `epsilon` and its internal resistance is r. its termnals are cannected to a resistance R. The potential difference between the terminals is `1.6V for R = 4 Omega, and 1.8 V for R = 9 Omega. Then,A. `epsilon =1 V, r = 1 Omega`B. `epsilon =2V, r = 1 Omega`C. `epsilon =2 V, r =2Omega`D. `epsilon =2.5 V, r = 0.5 Omega` |
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Answer» Correct Answer - B Current in the cirut is `I = (epsilon)/(R +r)` Potential difference across cell - potential differnce across R ` = iR = (epsilonR)/(R +r)` Set up two equations with the given data and sovle for `epsilon, r.` |
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| 23. |
The time constant of and R- C cirucit during charging is them time in which the charge on the condenser plates, as compared to maximum charge `(q_0)` becomes `(q//q_0)xx `100,` which is equal to …….. . |
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Answer» `q = q_(0)(1 - e^(-t/tau))`. Putting `t = tau,` we get `(q)/(q_(0)) = 1-e^(-1)` or `(q)/(q_(0))xx100 = (e-1)/(e ) xx100 = 63.2%` |
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| 24. |
In the circuit shwon in fig. the current I has a value equal to A. 1AB. 2AC. 4AD. `3.5A` |
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Answer» Correct Answer - A The equivalent resistance of resistor is `R =2 +(4)/(2) +(15)/(3) = 9Omega, I = (E )/(r+r) = (10)/(1+9) =1A` |
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| 25. |
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 timeC. decreases with timeD. first increases, then decreases |
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Answer» Correct Answer - B From the given situation, we have `i_(1) = (V)/(R ) (e^(-t//RC))` and `i_(2) = (V)/(R )(e^(-t//3RC))` `(i_(1))/(I_(2))=(e )^((2t)/(3RC))` which increases with time |
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| 26. |
In the circuit shwon in fig. switch S is closed at time t = 0. Let `I_1 and I_2` be the currents at any finite time t, then the ratio `I_1// I_2` A. is constantB. increases with timeC. decreases with timeD. first increases, then decreases |
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Answer» Correct Answer - C `I = I_(0)e^(-t//RC) or (I_(0))/(2) = I_(0)e^(-t//RC) or t = RC in 2` or `10^(-6) xx In4 =(2+r)xx0.5xx10^(-6)In 2` Solving to get `r = 2 Omega` |
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| 27. |
In the circuit given in fig. swithc S is at position 1 for long time. Find the total heat generated in resistor of resistance `(2r_0),` when the switch S is shifted from position 1 to position 2. A. `(C_0E_0^2)/(2)`B. `C_0E_0^2`C. `(C_0E_0^2)/(3)`D. none |
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Answer» Correct Answer - C Uncharged capacitor behaves as a short circuit just after closing the switch. But charged capacitor behaves as the battery of emf V just after closing the switch. Therfore, `I = (q_0)/(C_1(2R)) = (q_0)/(2RC_1) = (V)/(2R)` |
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| 28. |
For the resistor network shown in the potential drop between a and b s 12V. a. Current through resistance of `6Omega is ………..` b. Current through resistance `2 Omega is ………..` c. Current through resistance of 8 Omega is ...........` |
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Answer» `V_(ab) = I_1(6+(4xx6)/(4+6)) = I_1(8.4) (i)` `V_(ab) I_2(4 +(8xx8)/(8+8)) = I_2 (8) (ii)` a. From Eq. (i) we get `I_= (12)/(8.4) (10)/(7)A` `I_(1)` is the current in the `6Omega` resistor. b. `I_(2,4) = (4)/(6+4)xx(10)/(7) =(4)/(7)A` c. From eq (ii) we get `I_(2) = 3//2 A,` so current in the `4Omega` resistacne is 3//2A. It will be equally dicided inot `8Omega` resistance which are in parallel. Therefore, `I_(8) = (3)/(2xx2) = (3)/(4)A` |
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| 29. |
In an experiment a graph was plotted of the potential difference V between the terminals of a cell against circuit current I by varying load rheostat. Find the internal conductance fo the cell. |
| Answer» Here internal resistacne is given by the slopel of graph, i.e., x//y but conductance `=1/ resistance = x/y` | |
| 30. |
An uncharged capacitor is connected to a 15 V battery through a resistance of `10Omega.` It is found that in a time fo `2mus,` potential difference across the capacitor. Take in (1.5) = 0.4. |
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Answer» We know that charge on the capacitor at any tieme is given by ` q= Q(1 -e^(-t//pi))` where `Q = EC -15C`. Here cahrge q at any time is given by `q =VC` where V is potential difference across capacitor at that time. Here `V = -5 V`, so `q= 5C`, putting the values, we get `5C =15C(1-e^(-t//pi)) or e^(-t//pi) = 2//3` or` (t)/(pi) = In((3)/(2)) or (t)/(Rc) =In ((3)/(2))` or `C =(t)/(R In (3//2)) = 0.5 muF` |
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| 31. |
In the circuit shown in fig. `C_1 = 2C_2` Initially, capacitor `C_1` is charged to a potential of V. The current in the circuit just after the switch S is closed is |
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Answer» Correct Answer - D Here, `I_(1) = (V)/(R )e^(-t//RC), I_(2) = (V)/(R )e^(-t//2RC)` `:. (I_(1))/(I_(2))=e^(-t//RC +t//2RC) = e^(-t//2RC) = (1)/(e^(t//2RC)` From this expression, it is clear that when t increases, ratio decrease. |
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| 32. |
In the circuit shown in fig. find the maximum energy stored on the capacitor. Initially, the capacitor was uncharged. A. `150muC`B. `100muC`C. `50muC`D. zero |
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Answer» Correct Answer - D The current in the circuit is `I = (12//12) = 1A.` Potential across d and C is `12 V -3 xx1V = 9V` Capacitance across d and e is `C = (1xx2)/(1+2) = (2)/(3)muF` Therefore, charge on either capacitor is `(2)/(3)xx9 = 6muC.` |
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| 33. |
n identical cells, each of emf `epsilon` and internal resistance r, are joined in series to from a closed circuit. One cell a is joined with reversed polarity. The potentia difference across each cell, except A, isA. `(2epsilon)/(n)`B. `(n-1)/(n)epsilon`C. `(n-2)/(n)epsilon`D. `(2n)/(n-2)epsilon` |
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Answer» Correct Answer - A `I = ((n - 2)epsilon)/(nr)` `V_(B) - V_(A) = -ir + epsilon =epsilon - ((n -2)epsilon)/(nr)r= epsilon[1 - (n-2)/(n)] = (2 epsilon)/(n)` |
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| 34. |
Five resistors are connected between points A and B as shown in fig. A current of 10A flows from A to B. which of the following is correct? A. `V_(AC) = V_(CB)`B. `V_(AC) gt V_(CB)`C. `(V_(AC) lt V_(CB)`D. none of these |
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Answer» Correct Answer - B For resistance in sereis connection, current (I) is the same through the resistors. `I = (5-3)/(R_(1)) = (3-2)/(R_(2)) = (2)/(R_(3)), i.e., R_(1) : R_(2) : R_(3) = 2 : 1 :2` |
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| 35. |
Calculate the current in bracnhes containing `R_1, R_2 and R_3` in the circuit shown in |
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Answer» Potential difference across `R_1and R_3` is independent of time and remains constant. So current in them remains constant. Veriation in current will occur only in `R_2`. So `i_1 =(V)/(R_1) = constant, i_2 = i_0e^(-t//R_2C)` where `i_0 = (V)/(R_2) and i_3 = (v)/(R_3)` |
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| 36. |
A potential divider is used to given outputs of 2V and 3V form a 5V source, as shown in fig. Which combination of resistance `R_1, R_2, and R_3` gives the correct voltages?A. 2B. 2C. 2D. 3 |
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Answer» Correct Answer - B For cell A, the current I flows opposite to the direction of its emf. Potential difference is `epsilon + ir = epsilon +((n -2)epsilon)/(nr)r` `=epsilon[1+(n - 2)/(n)] =epsilon((2n-2)/(n))` |
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| 37. |
Find the minimum number of k cells required to produce a current of 1.5A through a resistance of `30Omega.` Given that the emf of each cell is 1.5V and the internal resistance is `1Omega` |
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Answer» `nm = ? , 30 = (nxx1)/(m) o rln = 30m`, `1.5 = (nxx1.5)/(30+ (n)/(m))` On solving we get n 60, m=2, total number nof batteries = mm = 120. |
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| 38. |
100 cells each of emf 5V and internal resistance `1 Omega` are to be arranged to produce maximum current in a `25 Omega` resistance. Each row contains equal number of cells. Find the number of rows. |
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Answer» Total number of cells is mn = 100. (i) Current will be maximum when `R =(nr)/(m) or 25 = (nxx1)/(m)` n =25m (ii) From Eqs. (i) and (ii) we get n = 50 and m =2. |
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| 39. |
Find out the potential difference between the points x and y in fing 5.169. |
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Answer» Given `E = 24V, R = 2Omega, R_1 = 4Omega , R_2= 6Omega, C_1 = (4//3)muF,` `C_2 =4muF.` As the capacitor offers a very high resistance to the current in the steady state, so the current is prevented to pass through the capacitors. Now, the total resistance in the circuit is `R_(eq) = R + R_1+R_2 = 2 +4+6 = 12 Omega` Hence, the net current in the circuit is `(E)/(R_(eq)) = (24)/(12) =2A` Therefore, the terminal potential differecne across is `AB =24 - 2xx2 = 20V.` |
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| 40. |
A capacitor of capacitance `10muF` is connected to a resistance `2Omega` and a battery of emf 5V of negligible interanl resistance. After `20mus` of comleting the circuit find. a. power delivered by the battery b. power dissipated as heat c. rate of energy stored in the capacitor |
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Answer» Here `C = 10muF`, `R = 2Omega, E = 5v, tau =RC = 2xx10xx10^(-6)` `s =20 mus` a. Current at` t = 20mu` s is `I = (E )/(e )e^(9-t//tau) = (5)/(2) e^(-20//20) = (5)/(2e)` Power delivered by battery is `P_1 = EI =5xxx(5)/(2e) = (25)/(2e) =4.6W` b. Power dissipated as heat is `P_2 = I^2R = ((5)/(2e))^2 2 = (25)/(2e^2) = 1.7 W` c. Energy on the capacitor is `U = (q^(2))/(2C)` Rate of energy stored in the capacitor is `P_3 = (dU)/(dt) = (2q)/(2C)(dq)/(dt) = (q)/(C) (dp)/(dt) = (Q(1-e^(-t//tau)))/(C) I` `= (EC)/(C)(1-e^(-t//tau)) I = EI(1 -e^(-t//tau))` `=(5xx5)/(2e)(1-e^(-20//20))W =(25)/(2e) -(25)/(2e^2) = 2.9` |
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| 41. |
Consider the circuit shown in fig 5.174 where a battery of emf 4V and a capacitor of capacitance 1 `muF` is connected to a combination of resistacnes. Find out the stedy-state current in the battery. |
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Answer» Let us consider I to be the steady-state current through the cirucuit. In the steady-state condition , current I is constant in the circuit and the capacitor offers infinite resistance. So the resistance `10Omega` becomues ineffective in the circuit. So in this case, the euqivalent resistance of resistors `2Omega and 4Omega` that are connected in parallel is `(1)/(R_P) = (1)/(2) +(1)/(4) = (3)/(4) or R_P = (4)/(3)Omega` Therefore, the total resistance of the circuit is `(4)/(3) +(2)/(3) +(2)/(3) = (6)/(3) = 2Omega` Hence, the steady state current in the circuit is `4/2 = 2A.` |
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| 42. |
Find out the value of resistance R in fig. A. `100 Omega`B. `200 Omega`C. ` 50 Omega`D. `150 Omega` |
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Answer» Correct Answer - A `((20R)//(20 +R))/(R ) = (20)/(10) or R = 20 Omega` |
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| 43. |
Shows a network of currents. The magnitude of the current is shown here. Find the current. i. |
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Answer» The net incoming current in circuit is 15+3+5 =23A.` As incoming current in circuit = Outgoing current from circuit hence I = 23A. |
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| 44. |
The capacitor C is initially without charge. X is now joined to Y for a long time, during which `H_1` heat is produced in the resistance R. X is now joined to Z for a long time, during which `H_2` heat is produced in R A. `H_1 = H_2`B. `H_1 = (1)/(2)H_2`C. `H_1 = 2H_2`D. The maximum energy stored in C at any time is `H_1.` |
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Answer» Correct Answer - A::D When X is joined to Y for a long time (charging), energy stored in the capacitor is equal to heat produced in `R =H_(1)`. When X is joined to Z (discharging), the energy stored in `C = (H_(1))` reappears as heat `(H_(2))` in R. Thus, `H_(1) = H_(2)`. |
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| 45. |
Switch S of circuit shown in Fig 5.200 is in position 1 for a long time. At instant t = 0, it is thrown from position 1 ot 2 Calcuate thermal power `P_1 (t) P_2 (t)` generated across resistanace `R_1 and R_2,` respectively. |
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Answer» Initally, the switch was in position 1 for a long time, therefore, initially the capacitor was fully charged and potential difference across capacitor at t = 0 was equal to emf E of the battery. Initial charge on capacitor is `q_0 = CE` When the switch is thrown to position 2, the capacitor starts to discharge through resistance `R_1 and R_2` To calculate thermal power `P_1(t) and P_2(t)` generated across `R_1 and R_2` respectively, current I at time t through the circuit must be known. we know that I is given by `I = (E )/((R_1+R_2))e^(-t//(R_1+R_2)C` Hence, thermal power across `R_1 is ` `P_1 = I^2R_1 = (E^2R_1)/((R_1+R_2)^2) e^(-2t(R_1+R_2)C)` Similarly, thermal power across `R_2 is ` `P_2 = I^2R_2 = (E^2R_2)/((R_1_R_2)^2) e^(-2t//R_1+R_2)C` |
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| 46. |
In the given circuit, with steady current, the potential drop across the capacitor must be A. VB. `V//2`C. `V//3`D. `2V//3` |
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Answer» Correct Answer - C V = E -ir. When I = 0, the potential reading is 2V. Hence emf is 2 V. When V = 0, I = 5A. Thisk gives `r = 0.4 Omega.` |
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| 47. |
A resistance R of therimal coefficient of resistivity `alpha` is connected in parallel with a resistance 3R. Having thermal coeffiecinet of resistivity `2alpha` Find the value of `alpha_(eff)` |
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Answer» The equivalent resistance at `0^@C` is `R_0 = (R_(10)R_(20))/(R_(10) + R_(20))` …(i) The equivalent resistance at `t^@`C is `R = (R_1R_2)/(R_1+R_2)` .....(ii) But `R_1 = R_(10)(1 + alphat)` .....(iii) `R_2 = R_(20) (1+2 alphat)` ....(iv) and `R = R_0 (1+alpha_(eft) t)` ...(v) Putteing the value of (i), (iii), (iv) and (v) in Eq. (ii), we have `alpha_(eff) = (5)/(4)alpha` |
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| 48. |
In the circuit shown in fig. the cell is ideal with emf 15V. Each resistance is of `3Omega.` The potential difference across the capacitoro in steady state is |
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Answer» Correct Answer - C As the capacitors are identical, each of them finally have chargel Q//2. Initial energy of the system in `E_(1) = (Q^(2))/(2C)` Final energy of the system is `E_(f)=2[((Q//2)^(2))/(2C)]=(Q^(2))/(4C)` Heat produced = loss in energy `=E_(i) - E_(f) = (Q^(2))/(4C)` |
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| 49. |
Consider a wire of length l, area of cross section A, and resistivity `rho` with resistance `10 Omega`. Its length is increased by applying a force, and it becomes four times of its original value. Find the changed resistance of the wire. |
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Answer» Here `I, A_1 = A, and R = 10 Omega` Similarly, `I_2 = 4I and R_2 = ?` Resistivity is same in each case as the meterail is same. The volume of the wire will remain the same even after the increase in the length. `A_1I_1 = A_2I_2 or A_2 = (A_1I_1)/(I_2) = (AI)/(4I) = (A)/(4)` The formula used for measuring resistacne of wire is `R = rho (I/A)` Using this formula in both cases, we have `R_1 = rho(I_1)/(A_1) =(rhoI)/(A) (i)` and `R_2 = rho(I_2)/(A_2) = rho(4I)/(A//4) = 16 rho(1)/(A)` (ii) Dividing Eq. (ii) by Eq. (i), `(R_2)/(R_1) = 16 or R, 160 Omega` |
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| 50. |
A coil of wire has a resistance of `25.00Omega at 20^@ C` and a resistance of `25.17Omega at 150^@C`. What is its temperauture cofficient of resistance? |
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Answer» `R = R_0[1+alpha(T - T_0)] or alpha = Delta R |(R_0 DeltaT),` with `Delta R =R -R_0 = 0.17Omega and Delta T =T - T_0 = 15^@C.` then `alpha = (0.17)/(25.00xxx15) = 4.5xx10^(-4)C^(-1)` |
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