Saved Bookmarks
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.
| 1651. |
A fuse wire with a circular cross-section and having a diam eter of 0.4mm blows with a current of 3 ampere. The value of current for which another fuse wire made of the same material but having circular co rss-section with diameter of 0.6mm will blow is:-A. 3AB. `3xxsqrt((3)/(2))A`C. `3xx((3)/(2))^((3)/(2))A`D. `3xx((3)/(2))A` |
| Answer» For fuse `i alphar^(3//2)` | |
| 1652. |
`AB` is a wire of uniform resistance. The galvanometer `G` shows no deflection when the length `AC = 20 cm` and `CB = 80 cm`. The resistance `R` is equal to. . |
| Answer» `6/R= 20/(60-20) = 20/40 = 1/2 or R =12 Omega` | |
| 1653. |
What is the safest voltage you can put across a `98 Omega-0.5 W` resistor ? |
| Answer» `P = V^(2)/R or V = sqrt(PR) = sqrt(0.5 xx 98) = 7V` | |
| 1654. |
A current in a circuit having constant resistance is tripled. How does this affect the power dissipation? |
|
Answer» The power dissipated in a circuit is `P = I^(2)R`. Thus, `P prop I^(2)`, when R is constant. When I becomes 3 times, P becomes 9 times, i.e., Power dissipation becomes 9 times. |
|
| 1655. |
What is the difference between heater wire and fuse wire ? |
| Answer» The melting point of heater wire is hogh but melting point of fuse wire is low. | |
| 1656. |
Two identical heaters rated `220 V , 1000 W` are paced in series with each other across `220 V` line , then the combined power is |
|
Answer» For series combination, the total power (P) is `1/P = 1/P_(1) + 1/P_(2)` or `P = (P_(1)+P_(2))/(P_(1) + P_(2)) = (1500 xx 1500)/(1500+1500)= 750W` |
|
| 1657. |
The resistance of wire in a heater at roo temeperature is `65Oemga`. When the heater is connected to a `220V` supply the current settles after a few seconds to 2.8A. What is the steady temperature of the wire. (Temperature coefficient of resistance `alpha=1.70xx10^(-4) ^(@).C^(-)`)A. `955^(@)C`B. `1055^(@)C`C. `1155^(@)C`D. `1258^(@)C` |
|
Answer» Correct Answer - D `"Here", T_(1)=27^(@)C, R_(1)=65Omega` `R_(2)=("Supply voltage")/("Steady current")=(220)/(2.8)=78.6Omega` Now using the relation `R_(2)=R_(1)[1+alpha(T_(2)-T_(1)) ` `therefore T_(2)-T_(1)=(R_(2)-R_(1))/(R_(1))xx(1)/(alpha)=(78.6-65)/(65)xx(1)/(1.7xx10^(-4))` `T_(2)-T_(1)=1231` `T_(2)=1231+T_(1)=1231+27=1258^(@)C` |
|
| 1658. |
One filament takes 10min to heat a kettle and another takes 15 min. If connected in parallel they combindly take………min to heat the same kettle:A. 6B. 12.5C. 25D. 7.5 |
|
Answer» Correct Answer - A For first filament `H=(V^(2))/(R_(1)) t_(1)rArr R_(1)=(V^(2))/H t_(1).....(i)` For second filament `H=(V^(2))/(R_(2))t_(2)rArr R_(2)=(V^(2))/H T_(2).....(ii)` When place din parallel `H=(V^(2))/(R_(P)) t_(2) rArr R_(p)=(V^(2))/H t_(p)....(iii)` From eqs. (i),(ii),(iii) we get `1/(R_(P))=1/(R_(1))+1/(R_(2))` `rArr H/(V^(2)t_(p))=H/(V^(2)t_(1))+H/(V^(2)t_(2))` `rArr 1/(t_(p)) =1/10 +1/15 rArr t_(p)` =6 min |
|
| 1659. |
The potential difference between the point A and B in figure will be A. `(2)/(3) V`B. `(8)/(9) V`C. `(4)/(3) V`D. 2V |
|
Answer» Correct Answer - A `i_(1)=i_(2)=2/15 amp rArr V_(A)+2/15xx5-2/15xx5-2/15xx5=V_(B) rArr V_(A)-V_(B) =10/15 =2/3 V` |
|
| 1660. |
Two cells, having the same emf, are connected in series through an external resistance `R`. Cells have internal resistance `r_(1)` and `r_(2) (r_(1) gt r_(2))` respectively. When the circuit is closed, the potentail difference across the first cell is zero the value of `R` isA. `r_(1)+r_(2)`B. `r_(1)-r_(2)`C. `(r_(1)+r_(2))/(2)`D. `(r_(1)-r_(2))/(2)` |
|
Answer» Correct Answer - B |
|
| 1661. |
Two cells each of emf E and internal resistacne `r_(1)" and "r_(2)` respectively are conneted in series with an external resistance R. The potential defference between the terminats of the first cell will be zero when R equal toA. `(r_(1)+r_(2))/2`B. `sqrt(r_(1)^(2)r_(1)^(2))`C. `r_(1)-r_(2)`D. `(r_(1)r_(2))/(r_(1)+r_(2))` |
| Answer» Correct Answer - C | |
| 1662. |
Two cells, having the same emf, are connected in series through an external resistance `R`. Cells have internal resistance `r_(1)` and `r_(2) (r_(1) gt r_(2))` respectively. When the circuit is closed, the potentail difference across the first cell is zero the value of `R` isA. `sqrt(r_(1)r_(2))`B. `r_(1)+r_(2)`C. `r_(1)-r_(2)`D. `(r_(1)+r_(2))/(2)` |
|
Answer» Correct Answer - C |
|
| 1663. |
Two cells, having the same emf, are connected in series through an external resistance `R`. Cells have internal resistance `r_(1)` and `r_(2) (r_(1) gt r_(2))` respectively. When the circuit is closed, the potentail difference across the first cell is zero the value of `R` isA. `r_(1)-r_(2)`B. `(r_(1)+r_(2))/(2)`C. `(r_(1)-r_(2))/(2)`D. `r_(1)+r_(2)` |
| Answer» Correct Answer - A | |
| 1664. |
A wire of diameter 0.02 metre contains 1028 free electrons per cubic metre. For an electrical current of 100 A , the drift velocity of the free electrons in the wire is nearlyA. `1 xx 10^(-19) m//s`B. `5 xx 10^(-10) m//s`C. `2 xx 10^(-4) m//s`D. `8 xx10^(3) m//s` |
|
Answer» Correct Answer - C |
|
| 1665. |
A 100 W bulb `B_1`, and two 60 W bulb `B_2` and `B_3`, are connected to a 250 V source, as shown in figure. Now `W_1, W_2 and W_3` are the output powers of the bulbs `B_1, B_2 and B_3`, respectively. Then A. `W_1 gt W_2=W_3`B. `W_1gtW_2gtW_3`C. `W_1 lt W_2=W_3`D. `W_1 lt W_2 lt W_3` |
|
Answer» `P=V^2/R` so, `R=V^2/P` `:. R_1=V^2/100` and `R_2=R_3=V^2/60` Now, `W_1=((250)^2)/((R_1+R_2)^2)R_1` `W_2=((250)^2)/((R_1+R_2)^.R_2` and `W_2=((250)^2)/R_3` `W_1:W_2:W_3=15:25:64 or W_1gtW_2gtW_3` `:.` The correct option is d. |
|
| 1666. |
In the circuit shown the cells are ideal and of equal emfs, the capacitance of the capacitor is `C` and the resistance of the resistor is `R`. `X` is first joined to `Y`and then to `Z`. After a long time, the total heat produced in the resistor will be A. equal to the energy finally stored in the capacityB. half of the energy finally stored in the capacitorC. twice the energy finally stored in the capacitorD. 3 times the energy finally stored in the capacitor |
|
Answer» Correct Answer - D (d) Word done by cell `= 2 q.V` `= 4 ((1)/(2) q.V)` `= 4` (energy stored in capacitor) |
|
| 1667. |
In above question, the charge flown is :A. `2 Q`B. `Q`C. `3 Q`D. `(3 Q)/(2)` |
|
Answer» Correct Answer - D (d) On closing the key, the inner charges willget neutralised, So a charge of `3Q//2` will flow from left to right. |
|
| 1668. |
`B_(1), B_2 and B_3` are the three identical bulbs connected to a battery of steady emf with key `K` closed. What happens to the brightness of the bulbs `B_(1)` and `B_(2)` when the key is opened? A. Brightness of the bulb `B_(1)` increases and that of `B_(2)` decreasesB. Brightness of the bulbs `B_(1)` and `B_(2)` increasesC. Brightness of the bulb `B_(1)` decreases and `B_(2)` increasesD. Brightness of the bulbs `B_(1)` and `B_(2)` decreases |
|
Answer» Correct Answer - C |
|
| 1669. |
A capacitor of capacitance `3 mu F` is first charged by connecting it across a `10 V` battery by closing key `K_(1)`, then it is allowed to get discharged through `2 Omega` and `4 Omega` resistors by opening `K_(1)` and closing the key `K_(2)`. The total energy dissipated in the `2 Omega` resistor is equal to A. `0.5 mJ`B. `0.05 mJ`C. `0.15 mJ`D. none of these |
|
Answer» Correct Answer - B (b) Initial energy stored on capacitor : `U = (1)/(2) xx 3 xx 10^(-6) xx (10)^(2) = 150 xx 10^(-6) J` When capacitor is allowed to charge, the whole energy stored on capacitor will be dissiapted as heat in `2 Omega` and `4 Omega` resistors. Heat dissipated in either resistor will be directly proportional to the value of the resistor. So heat dissiapted in `2 Omega` resistor is `H = (2 U)/(2 + 4) = (U)/(3) = 50 xx 10^(-6) J` `H = 0.050 mJ` |
|
| 1670. |
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 and then decreases |
|
Answer» Correct Answer - B (b) From the given situation `i_(1)= (V)/(R ) (e^(-t//3 RC))` and `i_(2) = (V)/(R ) (e)^((2t)/(3RC))` `(i_(1))/(i_(2)) = (e)^((2t)/(3RC))` Which increases with time. |
|
| 1671. |
A battery having 12V emf and internal resistance `3Omega` is connected to a resistor. If the current in the circuit is 1A, then the resistance of resistor and lost voltage of the battery when circuit is closed will beA. `7Omega,7V`B. `8Omega,8V`C. `9Omega,9V`D. `9Omega,10V` |
|
Answer» Correct Answer - C Here, `epsi=12V, r=3Omega, I=1A,V=IR=epsi-Ir` `therefore R=(epsi-Ir)/(I)=(12-1xx3)/(1)=12-3=9Omega` and `V=IR=1xx9=9V` |
|
| 1672. |
A heater coil rated at 1000 W s connected to a 110 V mains. How much time will take to melt 625 gm of ice at `0^(@)C`. (for ice `L=80(Cal)/(gm)`)A. `100s`B. `150s`C. `200s`D. `210s` |
|
Answer» Correct Answer - D `JQ=Pxxt` `JxxmL=Pxxt` `1xx625xx10^(-3)xx(80xx4.2)/(10^(-3))=1000xxt` `t=210s` |
|
| 1673. |
n identical calls are joined in series with its two cells `A` and `B` in loop with reversed polarties. `EMF` of each shell is `E` and internal resistance `r`. Potential difference across cell `A` or `B` is (here `n gt 4`)A. `(2E)/(n)`B. `2E (1 - (1)/(n))`C. `(4E)/(n)`D. `2E (1 - (2)/(n))` |
|
Answer» Correct Answer - D (d) The two opposite cells `A` and `B` will cancel two more other cells, so net emf will be `n - 4`. So current `I = ((n - 4)E)/(nr)`, Now p.d. across `A` or `B` `=E + Ir` (as they will be in charging state) `= E + ((n - 4)E)/(n) = 2 E (1 - (2)/(n))` |
|
| 1674. |
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 potential difference across each cell, except A, isA. `(2 epsilon)/(n)`B. `((n-1)/(n)) epsilon`C. `((n-2)/(n)) epsilon`D. `((2n)/(n -2)) epsilon` |
|
Answer» Correct Answer - A `i = ((n - 2)E)/(nr)` `V = E - ir = E - ((n-2)E)/(nr).r = (2E)/(n)`. |
|
| 1675. |
A prism shaped network of resistors has been shown in the figure. Each arm (like AB, AC, CD, DF ...) has resistance R. Find the equivalent resistance of the network between (a) A and B (b) C and D |
|
Answer» Correct Answer - (a) `(8R)/(15)` (b) `(3R)/(5)` |
|
| 1676. |
In identical calls are joined in series with its two cells `A` and `B` in loop with reversed polarties. `EMF` of each shell is `E` and internal resistance `r`. Potential difference across cell `A` or `B` is (here `n gt 4`) |
|
Answer» Correct Answer - `2E (1 - (2)/(n)) In a series combination of identical cells each of emf `E` if one cell is connected with reverse polarity, if will reduce the effective emf with of all cell by `2E` but total internal resistance of cell will remain unchanged As two cells are connected with reverse polarity in the combination of cell , their effective emf `= n E - 4E = (n - 4) E` Total internal resistance of the cell `= nr` Current in the circuit `I = ((n - 4)E)/(nr)` potential difference across `A ` or `B` is `V =E + I r = E = ((n - 4)Er)/(nr) = 2E (1 - (2)/(n))` |
|
| 1677. |
Two cells A andB ofEMF 1.3V and 1.5V respectively are arranged as shown in figure-3 .322. The voltmeter connected in circuit is ideal and it reads 1.45V. Which cell has the higher internal resistance and how many times that of the other? A. `r_(1)=2r_(2)`B. `r_(1)=3r_(2)`C. `r_(2)=2r_(1)`D. `r_(2)=3r_(1)` |
|
Answer» Correct Answer - B |
|
| 1678. |
The resistance of each arm of the wheat stone bridge is `10Omega`. A resistance of `10Omega` is connected in series with galvanometer then the equivalent resistance across the battery will be:-A. `10Omega`B. `15Omega`C. `20Omega`D. `40Omega` |
| Answer» Correct Answer - 1 | |
| 1679. |
The current in the arm CD of the circuit will be A. `i_(1) +i_(2)`B. `i_(2) +i_(3)`C. `i_(1) +i_(3)`D. `i_(1) -i_(2)+i_(3)` |
|
Answer» Correct Answer - B |
|
| 1680. |
Two cells `a` and `B` of emf `1.3 V` amd `1.5V `respectively are arranged as shown in figure. The voltmeter reds `1.45V`. The voltmeter is assumed to be ideal. Then A. `r_1=2r_2`B. `r_1=3r-2`C. `r_2=2r_1`D. `r_2=3r_1` |
|
Answer» Correct Answer - B No current will flow through voltmeter. As it is ideal (infinite resistance). Current through two batteries `i=(1.5-1.3)/(r_1+r_2)=0.2/(r_1+r_2)` Now, `V=E_2-ir_2` `:. 1.45=1.5-(0.2/(r_1+r_2))(r_2)` Solving this equation we get `r_1=3r_2` |
|
| 1681. |
Find the minimum number of 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`. |
|
Answer» Correct Answer - `120` Here, `epsilon= 1.5 V , r = 1.0 Omega R = 30 Omega , 1 = 1.5 A` As `r` comparable in `R`, so cells are to be connected in mixed grouping for maximum current .Let then be `n` cell in series in each row and `m` each rows in parallel Current in the mixed grouping of cell is `I = (mn epsilon)/(mR + nr) :. 1.5 = (mn xx 1.5)/(m xx 30 + n xx 1)` or `45m + 1.5 n = 1.5 m n `.....(i) For maximum current `R = (nr)/(m)` so, `30 = (n xx 1)/(m) or n = 30m`.....(ii) we have `45 m + 1.5 xx 30 m` or `45 m + 45m = 45 m^(2)` or `90 45 m or m = 2` From (ii) `m n = 2 xx 60 = 120` |
|
| 1682. |
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.A. 2B. 4C. 5D. 10 |
|
Answer» Correct Answer - A |
|
| 1683. |
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.A. 2B. 3C. 50D. 40 |
|
Answer» Correct Answer - A Let n be the number of cells in a row and m be the number of rows. Now, maximum current is obtained for `r=R`. In mixed combination total resistance of the circuit is given by, `R=(nr)/(m) :.25(n)/(m) r :. (n)/(m)=25` `:.n=25m` ....(i) In mixed combination, we have, `nm=100` ...(ii) From equation (i) and (ii) we have, `mxx25m=100` `25m^(2)=100` `m^(2)=4:.m=2 ` and `n=50` |
|
| 1684. |
A gas undergoes a change of state during which 100J of heat is supplied to it and it does 20J of work. The system is brough back to its original state through a process during which 20 J of heat is released by the gas. What is the work done by the gas in the second process?A. 60 JB. 40 JC. 80 JD. 20 J |
|
Answer» Correct Answer - A |
|
| 1685. |
An unknown resistance `R_(1)` is connected is series with a resistance of `10 Omega`. This combination is connected to one gap of a meter bridge, while other gap is connected to another resistance `R_(2)`. The balance point is at `50 cm` Now , when the `10 Omega` resistance is removed, the balanced point shifts to `40 cm` Then the value of `R_(1)` is.A. `10 Omega`B. `20 Omega`C. `40 Omega`D. `60 Omega` |
|
Answer» Correct Answer - b For first case `(R_(1) + 10)/(R_(2)) = (50)/(100 - 50) = 1` or `R_(2) = R_(1) + 10`…..(i) When `10 Omega ` resistance the balance point will shift toward lower length `:.` New balancing length `= 50 - 10 = 40 cm` for second case `(R_(1))/(R_(2)) = (40)/(100 - 40) = (40)/(60) = (2)/(3) :. R_(2) = (3)/(2)R_(1)` From (i) `(3)/(2)R_(1) = R_(1) = 10` or `R_(1) = 20 Omega ` |
|
| 1686. |
If the voltmeter reads 0.2 V and the ammeter reads 0.101A, the resistance of the voltmeter is (in ohm)A. 500B. 1000C. 200D. 400 |
|
Answer» Correct Answer - C `[(2x)/(2+x)](0.101)=0.2` solve for `x` |
|
| 1687. |
A uniform wire resistance `20 Omega` having resistance `1 Omega ` is bent in the forms of a circuit if the equal - valent resistance between A and B is `1.8 Omega ` then length of the shorter section is A. `1.8 m`B. `2 m`C. `3.6 m`D. `4 m` |
|
Answer» Correct Answer - b Let resistance of shorter part `AB` be `x` the total resistance is `20 Omega` the resistance of longer `AB` part is `(20- x)` between point A and B the two resistance `x` and their equivalent resistance in `R_(eq) = ((20 - x) xx x)/((20 - x) xx x) = 1.8` On solving get `x = 2 Omega` Thus, length of the shorter section `= 2 xx 1 = 2m` |
|
| 1688. |
A uniform wire resistance `20 Omega` having resistance `1 Omega per meter, ` is bent in the forms of a circuit if the equal - valent resistance between A and B is `1.8 Omega ` then length of the shorter section is A. 2mB. 5mC. 1.8 mD. 18m |
|
Answer» Correct Answer - A Let the resistance of shorter part MN be x. Then resistance of longer part is `(20-x)Omega` `R_(eq)=((20-x)x)/(20-x+x)=1.8Omega` solving we get `x=2Omega` So length of shorter part `=2m` |
|
| 1689. |
Which of the follwing characteristies of electrons determines the current in a conductor?A. drift velocity aloneB. thermal velocity aloneC. both drift velocity and thermal velocityD. neither drift nor thermal velocity |
|
Answer» The relationship between current and drift speed is given by `I=n eAv_(d)` Here `I` is the curret and `v_(d)` is the drift velocity. So, `IpropV_(d)` Thus, only drift velocity determines the current conductor. |
|
| 1690. |
If voltage is applied between terminals 1 and 2 when terminals 3 and 4 are open the power liberated is `P_(1)=40W` and when terminals 3 and 4 are connected, the power liberated is `P_(2)=80W`. If the same source is connected to the terminals 3 and 4, the power liberated in the circuit when terminals 1 and 2 are open in `P_(3)=20W`. Determine the power `P_(4)` consumed in the circuit when the terminals 1 and 2 are connected and the same voltage is applied between the terminals 3 and 4. |
|
Answer» Correct Answer - 40w When 1 and 2 are connected to voltage suppy V and 3 and are open `P_(1)=(R_(1)+R_(3))((V)/(R_(1)+R_(3)))^(2)=(V^(2))/(R_(1)+R_(3))` When 3 and 4 are closed and 1 and 2 are connected to `V` `R_(eq)=R_(1)+(R_(2)R_(3))/(R_(2)+R_(3))becauseP_(2)=(V^(2)(R_(2)+R_(3)))/(R_(1)R_(2)+R_(2)R_(3)+R_(3)R_(1))` When 3 amd 4 are connected to V and 1 and are open `P_(3)=(V^(2))/(R_(2)+R_(3))` When 3 and 4 are connected to V and 1 and 2 are closed: `R_(eq)=R_(2)+(R_(3)R_(1))/(R_(3)+R_(1))=(R_(1)R_(2)+R_(2)R_(3)+R_(3)R_(1))/(R_(3)+R_(1))` `thereforeP_(4)=(V^(2)(R_(3)+R_(1)))/(R_(1)R_(2)+R_(2)R_(3)+R_(3)R_(1))` Now, `(P_(1))/(P_(3))=(R_(2)+R_(3))/(R_(1)+R_(3))` and `(P_(2))/(P_(4))=(R_(2)+R_(3))/(R_(1)+R_(3))` `implies(P_(1))/(P_(3))=(P_(2))/(P_(4))` or `P_(4)=(P_(2))/(P_(1))xxP_(3)=(80xx20)/(40)=40W` |
|
| 1691. |
A battery having e.m.f. 5 V and internal resistance `0.5 Omega` is connected with a resistance of `4.5 Omega` then the voltage at the terminals of battery isA. `4.5V`B. `4V`C. `0V`D. `2V` |
|
Answer» Correct Answer - A |
|
| 1692. |
The emf of a storage battery is `90 V` before charging and `100 V` after charging. When charging began the current was `10 A`. What is the current at the end of charging if the internal resistance of the storge battery during the whole process of charging may be taken as constant and equal to `2Omega`? |
|
Answer» The voltage supplied by the charging plant is here constant which is equal to `V=E_i+i_i.r=(90)+(10)(2)` `=10V` ltbergt Let `i_f` be the current at the end of charging Then, `V=E-f+i_fr` `i_f=(V-E_f)/r` `=(110-100)/2` `=5A` |
|
| 1693. |
The potential difference across the terminals of the cell in open circuit is called asA. currentB. pot. diff. of cellC. e.m.f. of cellD. resistance |
| Answer» Correct Answer - C | |
| 1694. |
A storage cell is charged by 5 amp D . C . for 18 hours . Its strength after charging will beA. 18 AHB. 5 AHC. 90 AHD. 15AH |
|
Answer» Correct Answer - C |
|
| 1695. |
A battery consists of a variable number `n` of identical cells having internal resistance connected in series. The terminals of the battery are short circuited and the current `I` measured. Which one of the graph below shows the correct relationship between `I` and `n`?A. B. C. D. |
|
Answer» Correct Answer - D |
|
| 1696. |
What determines the e.m.f. between the two metals placed in an eletrolyte ?A. quantity of electrolyteB. Strength of electrolyteC. Distance between the metal platesD. Position of metals in electrochemical series |
| Answer» Correct Answer - D | |
| 1697. |
The e.m.f. of a cell is alwaysA. equal to terminal pot. diff. of a cellB. greater than pot. diff . of a cellC. less than pot. diff. of a cellD. internal resistance of a cell |
| Answer» Correct Answer - B | |
| 1698. |
A cell of emf 6 V and resistance 0.5 ohm is short circuited. The current in the cell isA. 3 ampB. 12 ampC. 24 ampD. 6 amp |
|
Answer» Correct Answer - B |
|
| 1699. |
Electromotive force is the force, which is able to maintain a constantA. potentential differenceB. powerC. resistanceD. current |
|
Answer» Correct Answer - A Emf is not force, it is analouge to potential difference. The energy spent in circulating a unit charge through the circuit is known as electromotive force of the cell. It is expressed as emf in brief. |
|
| 1700. |
A battery consists of a variable number `n` of identical cells having internal resistance connected in series. The terminals of the battery are short circuited and the current `I` measured. Which one of the graph below shows the correct relationship between `I` and `n`?A. B. C. D. |
|
Answer» Correct Answer - C For a two cell battery, `l =(2E)/(2r)=(E )/(r )` Similarly, for a n cell battery, `l = (nE)/(nr)=(E )/(r )` So, current in the circuit does not depend on number of cells in the battery. Hence, the correct graph will be (c ). |
|