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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.
| 451. |
Current provided by a battery is maximum whenA. Internal resistance equal to external resistanceB. Internal resistance is greater than external resistanceC. Internal resistance is less than external resistanceD. None of these |
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Answer» Correct Answer - A |
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| 452. |
The current in the following circuit is A. `(1)/(8)A`B. `(2)/(9)A`C. `(2)/(3)A`D. 1A |
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Answer» Correct Answer - D |
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| 453. |
What is the equivalent resistance of the circuitA. `6 Omega`B. `7 Omega`C. `8 Omega`D. `9 Omega` |
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Answer» Correct Answer - C |
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| 454. |
The equivalent resistance of the circuit shown in the figure is A. `8 Omega`B. `6 Omega`C. `5 Omega`D. `4 Omega` |
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Answer» Correct Answer - C |
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| 455. |
Eels are able to generate current with biological cells called electroplaques. The electroplaques in an eel are arranged in 100 rows, each row stretching horizontally along the body of the fish containing 5000 electroplaques. The arrangment is suggestively shown below. Each electroplaques has an emf of `0.15 V` and internal resistance of `0.25 Omega` The water surrounding theeel completes a cricuit be ween the head and its tail. If the water surrounding it has a resistance of `500 Omega`, the current an eel can produce in water is about A. 1.5AB. 3.0AC. 15AD. 30A |
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Answer» Correct Answer - A |
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| 456. |
In the above question, if the internal resistance of the battery is 1 ohm , then what is the reading of ammeterA. `5//3A`B. `40//29A`C. `10//9 A`D. `1A` |
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Answer» Correct Answer - B |
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| 457. |
10 wires (same length, same area, same material) are connected in parallel and each has `1Omega` resistance, then the equivalent resistance will beA. `10 Omega`B. `1 Omega`C. `0.1 Omega`D. `0.001 Omega` |
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Answer» Correct Answer - C |
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| 458. |
In the circuit shows in Fig. `6.41`, the reading of the ammeter is (assume internal resistance of the battery be to zero) A. `(40)/(29)A`B. `(10)/(9)A`C. `(5)/(3)A`D. 2A |
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Answer» Correct Answer - D |
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| 459. |
In the arrangement shown, the magnitude of each resistance is `1Omega`. The equivalent resistance between O and A is given by A. `(14)/(13)Omega`B. `(3)/(4)Omega`C. `(2)/(3)Omega`D. `(5)/(6)Omega` |
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Answer» Correct Answer - B |
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| 460. |
Fing the reading of the ideal ammeter connected in the given circuit. Assume that the cells have negligible internal resistance. A. `0.8A`B. `0.25A`C. `1.95A`D. `1.0A` |
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Answer» Correct Answer - C |
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| 461. |
In the circuit all the measuring instruments are ideal the reading of ammeter`A_(2)` is …….. |
| Answer» Correct Answer - zero | |
| 462. |
The thermo e.m.f. E in volts of a certain thermocouple is found to vary with temperature difference q in `.^(@)C` between the two junctions according to the relation `E=30 theta-(theta^(2))/(15)` The neutral temperature for the thermo-couple will be:-A. `400^(@)C`B. `225^ (@)C`C. `30^(@)C`D. `450^(@)C` |
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Answer» At neutral temperature `(dE)/(dtheta)=0` so `30-(2thetan)/(15) =0rArrtheta_(n)=225^(@)C` |
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| 463. |
With the cold junction at `0^(@)C` the neutral temperature for a thermo-couple is obtained at `270^(@)C`. The cold junction temperature is now lowered to `-10^(@)C` Obta in the (a) neutral temperature (b) the temperature of inversion in this case. |
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Answer» (a) Neutral temperature remains same (i.e. `270^(@)C`) (b) Temperature of inversion becomes `T_(n)=(T_(0)+T_(1))/( 2)rArr 270=(-10+T_(1))/(2)rArrT_(1)=540+10=550^(@)C` |
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| 464. |
In the circuit shown in the figure, if potentail at point `A` is taken to be zero, the potential at point `B` is A. `+ 1V`B. `-1V`C. ` +2V`D. ` -2V` |
| Answer» Correct Answer - 1 | |
| 465. |
A thermocouple of negligible resistance produces an emf fo `40 mu V//^(@)C` in the linear range of temperature. A galvanometer of resistance `10 Omega` whose sensitivity is `1 mu A//div` is employed with the thermocouple. The smallest value of temperature difference that can be detected by the system willA. `0.5^(@)C`B. `1^(@)C`C. `0.1^(@)C`D. `0.25^(@)C` |
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Answer» Correct Answer - D (d) For minimum deflection of 1 divison require current `= 1 mu A` Voltage required `= iR = 1 xx 10 = 10 mu V` `:. 40 mu V~= 1 ^(@)C` `10 mu V ~= (1)/(4)^(@)C` `= 2.50^(@)C` |
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| 466. |
The thermo emf of a thermocouple is given by `E=alpha(theta-theta_(g))+beta(theta^(2)-theta_(0)^(2))` Determine the neutral temperature and inversion temperature. |
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Answer» At neutral temperature `(dE)/(dtheta)=0rArr alpha+beta(2theta_(n))=0=-(alpha)/(2beta)` At inversion temperature `E=0rArr alpha(theta_(1)-theta_(0))+beta(theta_(1)^(2)-theta_(0)^(2))=0 rArr theta_(1)=-theta_(0)-(alpha)/(beta)` |
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| 467. |
A thermocouple produces a thermo emf of `40muV//.^(@)C`. A thermopile is made by using 150 such thermocouples. The temperature of the two junctions is `20^(@)C` and `40^(@)C` How much thermo emf generates in the thermopile |
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Answer» EMF for 1 couple `=40muV//.^(@)C` now in thermopile, thermo, couples are joined in series so emf For 150 couples is equal to `150xx40muVV//.^(@)C=6mV//.^(@)C` Now temperature difference is `=40-20=20^(@)C` Now total emf for `20^(@)` temp. diff, `E=(6mV//.^(@)C)xx20^(@)C=120mV` |
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| 468. |
In the circuit shown in the figure, if potentail at point `A` is taken to be zero, the potential at point `B` is A. `- 1 V`B. `+ 2 V`C. `- 2 V`D. `+ 1 V` |
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Answer» Correct Answer - D (d) By `KVL` along path `ACDB` `V_(A) + 1 (1) (2) - 2 = V_(B)` `0 + 1 = V_(B) implies V_(B) = 1V` |
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| 469. |
The rate of increase of thermo-emf with temperature at the neutral temperature of a thermocoupleA. is zeroB. depends upon the choice of the two material the thermocoupleC. is negativeD. is positive |
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Answer» Correct Answer - A (a) We have `e = at + bt^(2)` `(de)/(dt) = a + 2 bt` Natural temperature `R_(n) = - (a)/(2 b)` At neutral temperature `(de)/(dt) = 0` |
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| 470. |
A wire is stretched so as to change its diameter by `0.25%` . The percentage change in resistance isA. `4.0%`B. `2.0%`C. `1.0%`D. `0.5%` |
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Answer» Correct Answer - C On stretching, volume (V) remains constant. So, V = Al or l = V/A Now, `R=(rho l)/(A)=(rho V)/(A^(2))=(rho V)/(pi^(2)D^(4)//16)=(16rho V)/(pi^(2)D^(4))` Taking logarithm of both the side and differentiating it, we get `(Delta R)/(R )=-4 (Delta D)/(D)` or `(Delta R)/(R )=-4xx(-0.25)=1.0%` |
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| 471. |
If power dissipated in the `9 Omega` resistor in the resistor shown is `36 W`, the potential difference across the `2 Omega` resistor is A. `8 V`B. `10 V`C. `2 V`D. `4 V` |
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Answer» Correct Answer - B (b) Electric power, `p = i^(2) R` `:.` Current, `I = sqrt((p)/(R ))` Fro resistance of `9 Omega` `i_(1) = sqrt((36)/(9)) = sqrt4 = 2 A` `i_(2) = (i_(1) xx R)/(6) = (2 xx 9)/(6) = 3 A` `i- i_(1) + i_(2) = 2 + 3 = 5A` `V_(2) = IR_(2) = 5 xx 2 = 10 V` |
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| 472. |
A potential difference of 100 V is applied to the ends of a copper wire one metre long . What is the average drift velocity of electrons ? (Given, `sigma=5.81xx10^(7)Omega` or `n_(Cu)=8.5xx10^(28)m^(-3)`)A. `0.43 ms^(-1)`B. `0.83 ms^(-1)`C. `0.52 ms^(-1)`D. `0.95 ms^(-1)` |
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Answer» Correct Answer - A Since, `Delta V =100 V, l = 1m` `therefore` Electric field `=(Delta V)/(l)=(100)/(1)=100 Vm^(-1)` Also, conductivity `sigma = 5.81xx10^(7)Sigma^(-1)m^(-1)` `n=8.5xx10^(28)m^(-3)` The average drift velocity of the electrons `v_(d)=(l)/("neA")` `because " " J = sigma E, J=(l)/(A) " (given)"` `therefore " " v_(d)=(sigma)/("en")E` `= (5.81xx10^(7)xx100)/(1.6xx10^(-19)xx8.5xx10^(28))` `=0.43 ms^(-1)` |
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| 473. |
If power dissipated in the `9 Omega` resistor in the resistor shown is `36 W`, the potential difference across the `2 Omega` resistor is A. 8VB. 10VC. 2VD. 4V |
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Answer» Correct Answer - B Electric power, `P = i^(2)R` `therefore " "` Current, `I = sqrt((P)/(R ))` For resistance of 9 `Sigma`, `i_(1)=sqrt((36)/(9))=sqrt(4)=2A` `i_(2)=(i_(1)xxR)/(6)=(2xx9)/(6)=3A` `i=i_(1)+i_(2)=2+3=5A` `V_(2)=lR_(2)=5xx2=10V` |
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| 474. |
The effective resistance between points A and C for the network shown figure is A. `(2)/(3)R`B. `(3)/(2)R`C. 2RD. `(1)/(2R)` |
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Answer» Correct Answer - A As `R_(AB)=R_(AO)` and `R_(BC)=R_(OC)`, so points O and B will be at same potential and hence, resistance `R_(OB)` becomes ineffective. Similarly, as `R_(AO)=R_(AD)` and `R_(OC)=R_(DC)`, resistance `R_(OD)` bcomes ineffective. So, excluding `R_(BO)` and `R_(OD)`, equivalent resistance `R_(eq)` of the given network between points A and C will be given by, `(1)/(R_(eq))=(1)/(2R)+(1)/(2R)+(1)/(2R)` i.e., `R_(eq)=(2)/(3)R` |
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| 475. |
A resistor of `6kOmega`with tolerance 10% and another resistance of `4kOmega` with tolerance `10%` are connected in series. The tolerance of the combination is aboutA. 0.05B. 0.1C. 0.12D. 0.15 |
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Answer» Correct Answer - B In series combination, `R = R_(1)+R_(2)=6+4=10 k Omega` Error in combination, `Delta R = Delta R_(1)+Delta R_(2)` `= (10)/(100)xx6+(10)/(100)xx4` `= 0.6+0.4=1` `(Delta R)/(R )=(1)/(10)=10%` |
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| 476. |
All resistance shown in the circuit are `2Omega` each. The current in the resistance between `D` and `E` is A. `5A`B. `2.5A`C. `1A`D. `7.5A` |
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Answer» Correct Answer - B Resistance between `A` and `B` can be removed due to balanced wheastone bridfge concept. Now, `R_(DE)` and `R_(GH)` are in series and they are connected in parallel with `10 V` battery. `:. I_(DE)=10/(R_(DE)+R_(HG))=10/(2+2)` `=2.5A` |
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| 477. |
Iri the circuit shown in figure-3.334, the resistance of voltmeter is` 6kOmega`. The voltmeter reading will be : A. `6V`B. `5V`C. `4V`D. `3V` |
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Answer» Correct Answer - B Net resistance of `3kOmega` and voltmeter is also `2kOmega`. Now, the applied `10 V` is equally distributed between `2Omega and 2kOmega` Hence, reading of voltmeter `=10/2=5V` |
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| 478. |
There are two metallic wires of same material, same length but of different radii. When these are connected to an ideal bettery in series, heat produced is `H_(1)` but when connected in parallel, geat produced is `H_(2)` for the wsame time.Then the correct statement isA. `H_(1)=H_(2)`B. `H_(1)ltH_(2)`C. `H_(1)gtH_(2)`D. No relation |
| Answer» Correct Answer - B | |
| 479. |
10 g of ice at `0^(@)C` is mixed with m mass of water at `50^(@)C`. What is minimum value of m so that ice melts completely. (L = 80 cal/g and s = 1 cal/`g-.^(@)C`)A. 32 gB. 20 gC. 40 gD. 16 g |
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Answer» Correct Answer - D |
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| 480. |
Using meter bridge, it is advised to obtain the null point in the middle of bridge wire. Why? |
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Answer» The error in the measured value of unknown resistance S using meterbridge wil be minimum, when the null point is obtained at the middle of meterbridge wire. In this situation, the end error of the bridge will become ineffective. To understand it, let `alpha` be the end resistance of left and right copper strips in meterbridge. Let unknown resistance R be in the left gap of meter bridge. If null point is obtained at the middle of bridge wire (i.e., `l = 50cm`) then `(S + alpha)/(R + alpha) = (100 -50)/50 = 1 or S = R` It means the effect of `alpha` gets cancelled. In this situation the bridge is most accurate ans sensitive. If the null point is not at the centre, there will be some error becasue S will then not be equal to R. |
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| 481. |
A resistance R is to be measured using a meter bridge. Student chooses the standared resistance S to be `100 Omega`. He finds the null point at `l_(1)=2.9cm`. He is told to attempt to improve the accuracy. Which of the folllowing is a useful way?A. He should measure `I_(1)` more accuratelyB. He hould change S to `1000Omega` and repeat the experimentC. He should change S to `3Omega` and repeat the experiment.D. He should given up hope of a more accurate measurement with a meter bridge |
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Answer» Correct Answer - C The percentage error in R can be minimised by adjusting the balance point near the middle of the bridge, i.e., when `I_(1)` in close to 50 cm. this requires a suitable choice of S. Since, `(R)/(S)=(Rl_(1))/(R(100-l_(1)))=(l_(1))/(100-l_(1))` Since here `R:S::2.9:97.1` imply that the S is nearly 33 times to that of R. in orded to make this ratio 1:1, it is neccesary to reduce the value of S nearly `(1)/(33)` times i.e., nearly `3Omega`. |
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| 482. |
The equivalent resistance between the points P and Q of the circuit given is A. `(R)/(4)`B. `(R)/(3)`C. 4RD. 2R |
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Answer» Correct Answer - B |
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| 483. |
Consider the double cube resistor network shown in fig. Each side of both cubes has resistance R and each of the wires joining the vertices of the two cubes also have same resistance R. find the equivalent resistance between points A & B. |
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Answer» Correct Answer - `(2R)/(3)` |
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| 484. |
Three identical wire rings have been placed symmetrically as shown in the figure A, B and C are centres of the three rings. Resistance of each ring is 3R. find the equivalent resistance of this wire mesh across points C and D. |
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Answer» Correct Answer - `(63 R)/(164)` |
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| 485. |
Which of the following is vector quantityA. Current densityB. CurrentC. Wattless currentD. Power |
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Answer» Correct Answer - A |
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| 486. |
The fig shows a network consisting of an infinite number of pairs of resistors `R_(1) = 2 Omega and R_(2) = 1 Omega`. Since the network is infinite, removing a pair of `R_(1) and R_(2)` from either end of the network will not make any difference. Using this calculate the equivalent `(R)` across points A and B. (b) Prove that `I_(n)=(I_(n-1)R_(2))/(R_(2)+R)=(I_(n-1))/(sqrt(3)+2)` ltbr Where `I_(n)` and `I_(n-1)` represent the current through `R_(1)` in `n^(th)` and `(n-1)^(th)` segment respectively (see Fig) (c) If a 20 V battery is connected across A and B find `I_(10)` |
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Answer» Correct Answer - (a) `R=(1+sqrt(3))Omega` (c) `(20)/((sqrt(3)-1)(sqrt(3)+2)^(9))` |
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| 487. |
By increasing the temperature , the specific resistance of a conductor and a semiconductorA. Increases for bothB. Decreases for bothC. Increases, decreasesD. Decreases, increases |
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Answer» Correct Answer - C |
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| 488. |
In the network shown in fig. each resistance is R. The equivalent resistance between A and G is |
| Answer» Correct Answer - B | |
| 489. |
The figure here shows a portion of a circuit. What are the magnitude and direction of the current i in the lower right-hand wire A. 7AB. 8AC. 6AD. 2A |
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Answer» Correct Answer - B |
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| 490. |
The equivalent resistance across the terminals of source of e.m.f. `24 V` for the circuit shown in the figure is A. `15 Omega`B. `10 Omega`C. `5 Omega`D. `4 Omega` |
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Answer» Correct Answer - C |
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| 491. |
In the circuit shown in figure, switch S is initially closed and S is open. Find V-V A. `4V`B. `8V`C. `12V`D. `16V` |
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Answer» Correct Answer - B |
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| 492. |
Most of the times we connected remote speaker to play music another room along with the buit-in speakers. These speakers are connected in parallel with the music system. At the instant represented in the picture, the a.c. voltage across the speakers is `6.00 V`. the main speaker resistance is `8.00 Omega` and the remote speaker has `4.00 Omega` resistance.The total current supplied by the music system is: |
| Answer» Correct Answer - A | |
| 493. |
Most of the times we connected remote speaker to play music another room along with the buit-in speakers. These speakers are connected in parallel with the music system. At the instant represented in the picture, the a.c. voltage across the speakers is `6.00 V`. the main speaker resistance is `8.00 Omega` and the remote speaker has `4.00 Omega` resistance. The power dissipated in the speaker of `4.00 Omega` resistance is : |
| Answer» Correct Answer - A | |
| 494. |
A resistor circuit is constructed such that 12 resistor are arranged to form a cube as shown in fig. Each resistor has a resistance of `2Omega.` The potential difference of 30V is applied across two of the opposing points as shown If we replace resistor between A and B and resistors between G and H by wires of zero resistance, then the points having the same potential are i. D,E,C,F ,ii. A,B ,iii. G,H |
| Answer» Correct Answer - D | |
| 495. |
In the circuit shown in the figure, the current drawn from the battery is `4A`. If `10 Omega` resistor is replaced by `20 Omega` resistor, then current drawn from the circuit will be A. 1AB. 2AC. 3AD. 0A |
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Answer» Correct Answer - D |
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| 496. |
Thirteen resistances each of resistance `R` ohm are connected in the circuit as shown in the figure below. The effective resistance between `A` and `B` is A. `2R`B. `(4R)/(3)`C. `(2R)/(3)`D. R |
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Answer» Correct Answer - C |
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| 497. |
Calculate the equivalent resistance between A and B A. `(9)/(2) Omega`B. `3 Omega`C. `6 Omega`D. `(5)/(3)Omega` |
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Answer» Correct Answer - A |
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| 498. |
Thirteen resistances each of resistance `R` ohm are connected in the circuit as shown in the figure below. The effective resistance between `A` and `B` is A. `2R Omega`B. `(4R)/(3)Omega`C. `(2R)/(3)Omega`D. `R Omega` |
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Answer» Correct Answer - C |
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| 499. |
A resistor circuit is constructed such that 12 resistor are arranged to form a cube as shown in fig. Each resistor has a resistance of `2Omega.` The potential difference of 30V is applied across two of the opposing points as shown The points having the same potential are i. B,D,E ,ii. C,F,H ,iii. C,E |
| Answer» Correct Answer - D | |
| 500. |
The equivalent resistance between P and Q in the given figure is A. `50 Omega`B. `40 Omega`C. `30 Omega`D. `20 Omega` |
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Answer» Correct Answer - D |
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