2492 + Interview Questions in CURRENT-ELECTRICITY IN GENERAL AWARENESS Page 33 InterviewSolution

1601.	Find the equivalent resistance of the networks shown in figure between the points a and b.
Answer» Correct Answer - A::B::C::D b. Three resistors are in parallel. Then, one resistor in series with this combination. c. Balanced wheatstone bridge. Hence, two resistors in vertical wire can be removed. d. All four resistors are in parallel e. A balanced Wheatstone bridge.

Discussion

1602.	In the circuit shown in figure AB is a uniform wire of length `L = 5m`. It has a resistance of `2 Omega//m`. When `AC = 2.0 m`, it was found that the galvanometer shows zero reading when switch s is placed in either of the two positions 1 or 2. find the emf `E_(1)`.
Answer» Correct Answer - 150 V

Discussion

1603.	In the circuit shown, after the switch is shifted to position 2 the heat generated in `50 Omega` resistance is `6 J`. find the emf (V) of the cell.
Answer» Correct Answer - 179 V

Discussion

1604.	Two conductors made of the same material have lengths `L` and `2L` but have equal resistance. The two are connected in series in a circuit which current is flowing. Which of the following is/are correct?A. The potential difference across the two conductors is the sameB. the drift speed is larger inte conductor of lengthLC. The electric field in the first conductor is twice that in the secondD. the electric field in the second conductor is twice that in the first
Answer» Correct Answer - A::B::C `a. V=iR` in series i is same. Hence, `V` is also same as `R` is given same. b. `R=(rhol)/A` `R` is same. Hence, `AS` should be smaller in first wire. Secondly `v_d=1/("ne"A)` or `v_dprop 1/A` `A` of first wise is less. Hence, its drift velocity should be more. `c. E=V/l or Eprop 1/l` (`Vrarr` same)

Discussion

1605.	In the following star circuit diagram (figure), the equivalent resistance between the points A and H will be A. `1.944 r`B. `0.973 r`C. `0.486 r`D. `0.243 r`
Answer» Correct Answer - B

Discussion

1606.	Following figure shows cross-section through three long conductors of the same length and material, with square cross-section of edge lengths as shown. Conductor `B` will snugly within conductor `B`. Relationship between their end to end resistance is A. `R_(A) = R_(B) = R_(C )`B. `R_(A) gt R_(B) gt R_(C )`C. `R_(A) lt R_(B) lt R_(C )`D. Information is not sufficient
Answer» Correct Answer - A (a) All the conductors have equal lengths. Area of cross section of `A` is `{(sqrt3 a)^(2) - (sqrt2 a)^(2)} = a^(2)` Similarly area of cross-section of `B` = area of cross section of `C = a^(2)` Hence according to formula `R = rho (l)/(A)`, resistance all the conductors are equal i.e., `R_(A) = R_(B) = R_(C)`

Discussion

1607.	Following figure shows cross-sections through three long conductors of the same length and material, with square cross-section of edge lengths as shown. Conductor B will fit snugly within conductor A , and conductor C will fit snugly within conductor B . Relationship between their end to end resistance is A. `R = R = R`B. `R gt R gt R`C. `R lt R lt R`D. Information is not sufficient
Answer» Correct Answer - A

Discussion

1608.	Find the ammeter reading in the cirucit shown in Fig. 4.65.
Answer» Correct Answer - `3 A`

Discussion

1609.	An ammeter reads a current of 30 A when it is connected across the terminals of a cell of emf 2 V. Neglecting the meter resistance, find the amount of heat (in calories) in cell in 20 seconds.
Answer» Here, `I = 30A, V = 2V, t=20s`. Let r be the internal resistance of a cell. Then `r = epsilon/I = 2.0/30 = 1/15 Omega` Heat produced in cell, `H = (I^(2)rt)/(J)` `= ((30)^(2) xx (1//15)xx 20)/4.2 = 285.7` cals

Discussion

1610.	In the circuit shown in Fig. 4.64, the battery has an emf of 12.0 V and an internal resistance of `5R//11`. If the ammeter reads 2.0 A, what is the value of R ?
Answer» Correct Answer - `6 Omega`

Discussion

1611.	`N` identical cells, each emf `E` and internal resistance `r` are joined in series. Out of `N` cells, `n` cells are wrongly connected i.e., their terminals are connected in reverse of the required for series connection `(n lt (N)/(2))`. Let `E_(0)` be the emf of resulting battery and `r_(0)` be its internal resistance. ThenA. `E_(0) = (N - n) E, r_(0) = (N - n) r`B. `E_(0) = (N - 2n) E, r_(0) = (N - 2n) r`C. `E_(0) = (N - 2n) E, r_(0) = N r`D. `E_(0) = (N - n) E, r_(0) = N r`
Answer» Correct Answer - C (c ) For every cell that is wrongly connected, the emf decreases by `2 E`. However, internal resistance does not depend on the direction and therefore remains same for all cells.

Discussion

1612.	Two cells of emf of 1.5 V and 2.0 V having internal resistances of `1 Omega` and `2 Omega`, respectively are connected in parallel so as to send current in the same direction through an external resistance of `5 Omega`. Find the current in the external circuit.
Answer» Correct Answer - `(5)/(17)A`

Discussion

1613.	When a resistance of `2 Omega` is connected across the terminals of a cell, the current is 0.5 A. When resistance is increased to `5 Omega`, the current is 0.25 A. the emf of the cell isA. 1VB. 1.5 VC. 2 VD. 2.5 V
Answer» Correct Answer - B E=i(R+r) E=1.5V

Discussion

1614.	Twelve identical wires each of resistance `6 Omega` are arranged to from a skelence cube. A current of `40 m A` is led cube at the current and out at the diagonally opposite corner. Calculate the potential difference development across these current and the effective resistance of the network
Answer» Correct Answer - `0.2 V` Follow example `10` the effective resistance between two two point , which are diagonally opposite cornes of a skellon cube of `12 ` wires `= 5 r//6 = 5 xx 6//6 = 5 Omega` pot diff developed = resistance `x` current `5 xx (40 xx 10^(-3)) = 0.2 V`

Discussion

1615.	Twelve identical wires each of resistance `6 Omega` are joined to from a skeleton cube. Find the resistance between the current of the same edge of the cube
Answer» Correct Answer - `3.5 Omega`

Discussion

1616.	A sample of mono-atomic ideal gas is taken through a cyclic process ABCDA. A graph is plotted in which heat exchanged by the gas upto an instant is shown on x-axis and work done by the gas upto that instant is shown on y-axis. The given cyclic process can be represented on a pressure-volume graph as:A. B. C. D.
Answer» Correct Answer - A

Discussion

1617.	Five identical cells are connected in parallel. Now, polarity of one of the cells is reversed. Percentage change in equivalent emf will beA. -0.1B. -0.2C. -0.4D. -0.6
Answer» Correct Answer - C

Discussion

1618.	It is desired to make a `20.0Omega` coil of wire whose temperature coefficient of resistance is zero. To do this, a carbon resistor of resistance `R_(1)` is placed in series with an iron resistor of resistance `R_(2)`. The proportion of iron and carbon are so chosen that `R_(1) +R_(2)= 20 Omega` for all temperatures near `20^@C`. Find the values of `R_(1)` and `R_(2)`. Temperature coefficient of resistance for carbon, `alpha_(C)= - 0.5 xx 10^(-3)//^(@)C` and that of iron is `alpha_(Fe) =5 xx 10^(-3)//^(@)C`.
Answer» Let `Delta t^@C` be the rise in temperature. As per question we need, `R_(1) ( 1 +alpha_(C) Delta t) +R_(2) (1 + alpha_(Fe) Delta t)=20` Since, `R_(1) +R_(2) = 20`, therefore, `R_(1) alpha_(C) Delta t +R_(2) alpha_(Fe) Delta t=0` or `R_(1) +alpha_(C) = - R_(2) 1 + alpha_(Fe)` or `R_(1) xx (+0.5 xx10^(-3)) = - R_(2) xx 5xx10^(-3)` or `R_(1) = 10 R_(2)` So, `10 R_(2) + R_(2) = 20 or 11 R_(2) = 20` or `R_(2) = 20//11=1.82 Omega` and `R_(1)=10 xx 1.82 = 18.2 Omega`

Discussion

1619.	A black body calorimeter filled with hot water cools from `60^(@)C` to `50^(@)C` in `4 min` and `40^(@)C` to `30^(@)C` in `8min`. The approximate temperature of surrounding is :A. `10^(@)`B. `15^(@)C`C. `20^(@)C`D. `25^(@)C`
Answer» Correct Answer - B

Discussion

1620.	Figure shows a potentiometer circuit for comparision of two resistances. The balance point with a standard resistor `R=10.0Omega` is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. the value of X isA. `11.75Omega`B. `12.55Omega`C. `9.55Omega`D. `12.75Omega`
Answer» Correct Answer - A `(E_(1))/(E_(2))=(l_(1))/(l_(2)),(IR)/(IX)=(l_(1))/(l_(2))`

Discussion

1621.	Figure shows a potentiometer circuit for comprasion of two resistances. The balance point with a standard resistor `R = 10.0 Omega` is found to be 58.3 cm, while that with the unknown resistance X is 68.5 Cm. determine the value of X. What might you do if you failed to find a balance point with the given cell of emf `epsilon` ?
Answer» Here `L_(1) = 58.3 cm, L_(2) = 68.5 cm, R = 10 Omega, 1 X = ?` Let bet the current in the potentiometer wire and `epsilon_(1)` and `epsilon_(2)` be the potential drops across R and X respectively when connected in circuit by closing respective key. Then `(epsilon_(2))/(epsilon_(1)) = (1X)/(1R) = (x)/(R )` or `X = (epsilon_(2))/(epsilon_(1)) R` But `(epsilon_(2))/(epsilon_(1)) = (L_(2))/(L_(1))` From (i) `X = (L_(2))/(L_(1)) R = (68.3)/(58.3) xx 10.0 = 11.75 Omega` If there is no balance point with given cell of emf it means potential drop across R or X greater than the potential drop across the potentiometer wire. AB in order to obtain the balance point, the potential drops across R and X are to reduced which is possilbe by reducing the current in R and X for that either suitable resistance should be put in series with R and X or cell of smaller emf E should be used. Another possilbe way is to increase the potential drop across the potentiometer wire by increasing the voltage of dirver cell.

Discussion

1622.	A galvanometer of resistance `50Omega` is connected to a b attery of 3V alongwith a resistance of `2950Omega` in series. A full scale deflection of 30 division is obtained in the galvanometer in order to reduce this deflection to 20 division. The resistance in sereis. should be:-A. `6050Omega`B. `4450Omega`C. `5050Omega`D. `5550Omega`
Answer» Correct Answer - 2 `I_(2)=(3)/(50+2950)prop30I_(g)=(3)/(50+R) prop20` `rArr(50+R)/(50+2 950)=(3)/(2)rArr50+R=4500` `rArrR=4450Omega`

Discussion

1623.	A` lmuF `capacitor is connected in the circuit shown below. The EMF of the cell is 3V and internal resistance is `0.50Omega`. The resistors `R_(1) and R_(2)` have values `4Omega and 1Omega` respectively. The charge on the capacitor in steady state is : A. `1muC`B. `2muC`C. `1.33mu C`D. Zero
Answer» Correct Answer - B

Discussion

1624.	A galvanometer of resistance `50Omega` is connected to a b attery of 3V alongwith a resistance of `2950Omega` in series. A full scale deflection of 30 division is obtained in the galvanometer in order to reduce this deflection to 20 division. The resistance in sereis. should be:-A. `5050Omega`B. `550Omega`C. `6050Omega`D. `4450Omega`
Answer» Correct Answer - D Current through the galvanomete, `l=(3)/((50+2950))=10^(-3)A` Current for 30 divisions `= 10^(-3)A` Current for 20 divisions `= (10^(-3))/(30)xx20 =(2)/(3)xx10^(-3)A` For the same deflection to obtain for 20 divisions, let resistance added be R `therefore " " (2)/(3)xx10^(-3)=(3)/((50+1R))` `rArr " " R = 4450 Omega`

Discussion

1625.	The resistance of an ammeter is `13 Omega` and its scale is graduated for a current upto `100 A`. After an additional shunt has been connected to this ammeter it becomes possible to measure currents upto `750 A` by this meter. The value of shunt resistance isA. `20 Omega`B. `2Omega`C. `0.2Omega`D. `2kOmega`
Answer» Correct Answer - B Let `i_(a)` is the current flowing through ammeter and i is the total current. So, a current `i-i_(a)` will flow through shunt resistance. Potential difference across ammeter and shunt resistance is same. i.e., `i_(a)xx R=(i-i_(a))xxS` or `" " S=(i_(a)R)/(i-i_(a))` ....(i) Give, `i_(a)=100 A,i=750 A, R=13 Omega` Hence, `S=(100xx13)/(750-100)=2 Omega`

Discussion

1626.	A heater coil is cut into two parts of equal length and one of them is used in the leader. The ratio of the heat procued by this half coil to that by the original coil isA. `2 :1`B. `1 : 2`C. `1 : 4`D. `4 : 1`
Answer» Correct Answer - A (a) `P = (V^(2))/(R ) implies P prop (1)/(R )` and `R = prop l` `:. P prop (1)/(l) implies (P_(1))/(P_(2)) = (l_(1))/(l_(2)) = (2)/(1)`

Discussion

1627.	`2A` cell having an emf `epsilon` and internal resistance `r` is connected across a variable external resistance `R`. As the resistance `R` is increased, the plot of potential difference `V` across `R` is given byA. B. C. D.
Answer» Correct Answer - C (c ) Here, `E = I (R + r) implies E = IR + Ir` and `E = V + Ir` `E = V + (Er)/(R + r)` `V = E - (E)/(R + r) xx r = E [1 - (1)/(1 + R//r)]` This equaiton represents option `(c )`.

Discussion

1628.	The total power dissipated in watts in the circuit shown here is A. 16B. 40C. 54D. 4
Answer» Correct Answer - C (c ) The resistance of `6 Omega` and `3 Omega` are in parallel in the given circuit, their equivalent resistance is `(1)/(R_(1)) = (1)/(6) + (1)/(3) = (1 + 2)/(6) = (1)/(2)` or `R_(1) = 2 Omega` Again, `R_(1)` is in series with `4 Omega` resistance, hence ltbgt `R = R_(1) + 4 = 2 + 4 = 6 Omega` Thus, the total power dissipatesin the circuit `P = (V^(2))/(R )` Here, `V = 18 V, R = 6 Omega` Thus, `P = ((18)^(2))/(6) = 54 W`

Discussion

1629.	A copper refining cell consists of two parallel copper plate electrodes 5 cm apart and 1 metre squre, immersed in a copper sulphate solution of resistivity `12xx10^(-2)Omegam`. Then the potential difference which must be established between the plates to provide a constant current to deposit `0.66kg` of copper on cathode in one hour is nearly `(Z=3.33xx10^(-7)kg//C)`A. 0.6VB. 0.33VC. 3.0VD. 15V
Answer» `m=Zit=Z((V)/(R))t` `rARr V=(mR)/(Zt)=(mrhol)/(Zat)` `=(0.66xx1.2xx10^(-2)xx5xx10^(-2))/(3.33xx10^(-7)xx1xx60xx60)` =0.33 volt

Discussion

1630.	The total power dissipated in watts in the circuit shown here is A. 4WB. 16 WC. 40 WD. 54 W
Answer» Correct Answer - D The resistance of `6 Omega` and `3 Omega` are in parallel in the given circuit, their equivalent resistance is `1/(R_(1))=1/6+1/3=(1+2)/6=1/2` or `R_(1)=2Omega` Again `R_(1)` is in series with `4 Omega` resistance hence `R=R_(1)+4=2+4 =6Omega` Thus, the total dissipated in the circuit `p=(V^(2))/R` Here `V=18 V, R=6 Omega` Thus, `p=((18)^(2))/6=54W`

Discussion

1631.	The power dissipated in the circuit shown in the is 30 watts. The value of `R` is A. `10Omega`B. `30Omega`C. `20Omega`D. `15Omega`
Answer» `P_("Total")=(V^(2))/(R_(eq))` So `R_(eq)=(V^ (2))/( P_("total"))=(10xx10)/(30)=(10)/(3)` As `R_(eq)=(R_(1)R_(2))/(R_(1)+R_(2))` `rArr (10)/(3)=(5R)/(5+R)=10Omega`

Discussion

1632.	The power dissipated in the circuit shown in the is 30 watts. The value of `R` is A. `10Omega`B. `30Omega`C. `20Omega`D. `15Omega`
Answer» Correct Answer - 1 `P_("total")=(V^(2))/(R_(eq))` So `R_(eq)=(V^(2))/(P_("total"))=(10xx10)/(30)=(10)/(3)` As `R_(eq )=(R_(1)R_(2))/(R_(1)+R_(2))` `rArr(10)/(3)=(5R)/(5+R)rArrR=10Omega`

Discussion

1633.	The power dissipated in the circuit shown in the is 30 watts. The value of `R` is A. `20 Omega`B. `15 Omega`C. `10 Omega`D. `30 Omega`
Answer» Correct Answer - c Equivalent resistance of circuit `R_(eq) = (5R)/(5+R)` Power dissipated, `P = (V^(2))/(R_(eq)) or R_(eq) = (V^(2))/(P) = (10^(2))/(30) = (10)/(3)` `:. (10)/(3) = (5R)/(5+R) or 50 + 10R = 15 R or R = 10 Omega`

Discussion

1634.	Both terminals of a battery of emf E and internal resistance r are grounded as shown. Select the correct alternative(s). A. Potential difference across A and B is zeroB. Potential difference acorss A and B is EC. Current across AB is zeroD. Current across AB is `(E)/(r)`
Answer» Correct Answer - A::D

Discussion

1635.	The variation between V-I has shown by graph for heating filamentA. B. C. D.
Answer» Correct Answer - A

Discussion

1636.	To draw the maximum current from a combination of cells, how should be the cells be grounded ?A. ParallelB. SeriesC. Depends upon the relative values of internal and external resistanceD. Mixed grouping
Answer» Correct Answer - C

Discussion

1637.	When a current of `(2.5 +- 0.5)` ampere flows through a wire, it develops a potential difference of `(20 +- 1)` volt, the resistance of the wire isA. `(8 pm 2)Omega`B. `(8 pm 1.6)Omega`C. `(8 pm 1.5)Omega`D. `(8 pm 3)Omega`
Answer» Correct Answer - A

Discussion

1638.	A torch battery consisting of two cells of 1.45 volts and an internal resistance `0.15 Omega` , each cell sending currents through the filament of the lamps having resistance 1.5ohms. The value of current will beA. `16.11 amp`B. `1.611 amp`C. `0.1611 amp`D. `2.6 amp`
Answer» Correct Answer - B

Discussion

1639.	A battery when connected by resistance of `16Omega` gives a terminal voltage of 12 V. and when connected by a resistance of `10Omega` gives a terminal voltage of 11V. Then the emf of the battery ad its internal resistanceA. 12.8 VB. 13.7 VC. 10.7 VD. 9 V
Answer» Correct Answer - B `E=(V)/(R)(R+r)=`constant

Discussion

1640.	A potentiometer wire of length 4 cm has a some resistance. The resistance connected in series with the wire and accumulator of e.m.f. 2V is `16Omega`. If the potential gradient along wire is `10^(-3)V//cm`. The resistance of potentiometer wire isA. `0.066Omega`B. `0.022Omega`C. `0.033Omega`D. `0.044Omega`
Answer» Correct Answer - C `Irho =((E)/(R+R_(h)))(R)/(L)` `:.R+R_(h)=(ExxR)/(L xx I_(rho))=(2xxR)/(10^(-2)xx4xx10^(-1))` `:.R+16=(2R)/(4xx10^(-13))=500R` `:.R=(16)/(499)=0.033 Omega`

Discussion

1641.	The resistance of a potentiometer wire of length `10 m` is `20 Omega`. A resistance box and a `2` volt accumulator are connected in series with it. What resistance should be introduced in the box to have a potential drop of one microvolt per millimetre of the potentiometer wire ?
Answer» Correct Answer - `980 Omega ` Here `L = 10 m , r = 20 Omega V = 2 V, R = ?` `K = 1 mu V//mm = 10^(-6) V//m = 10^(-3)V//m` Now `K = ((V)/(R + r)) (r )/(L) or 10^(-3) = (2)/((R + 20)) xx (20)/(10)` `or R + 20 = (4)/(10^(-3)) = 4000` or `R= 4000 - 20 = 3980 Omega `

Discussion

1642.	The resistance of a potentiometer wire is `10 Omega` and its length is 10m. A resistance box and 2V accumulator are placed in series with it. The value of resistance in the resistance box, if it is desired to have potential drop of 1 microvolt/mm isA. `1990Omega`B. `1890Omega`C. `1190Omega`D. `2290Omega`
Answer» Correct Answer - A `(V)/(L)=I rho=((E)/(R+R_(h)))(R)/(L)` `:.R+R_(h)=(2xx1)/(10^(-8))=2000 " " :.R_(h)=1990 Omega`

Discussion

1643.	A potentiometer having a wire of 4 m lengths is connected to the terminals of a battery with steady voltage. A leclanche cell has a null point at 1m. If the length of the potentiometer wire is increased by 1 m, the position of the null points isA. 1.5 mB. 1.25 mC. 10.05 mD. 1.31 m
Answer» Correct Answer - B `lalphaL implies(l_(1))/(l_(2))=(L_(1))/(L_(2))`

Discussion

1644.	An ammeter of `100 Omega` resistance gives full deflection for the current of `10^(-5)` amp . Now the shunt resistance required to convert it into ammeter of 1 amp . range, will beA. `10^(-4) Omega`B. `10^(-5)Omega`C. `10^(-3)Omega`D. `10^(-1)Omega`
Answer» Correct Answer - C

Discussion

1645.	A 2 volt battery, a `15 Omega` resistor and a potentiometer of `100 cm` length, all are connected in series. If the resistance resistance of potentiometer wire is `5 Omega`, then the potential gradient of the potentiometer wire isA. 0.005 V//cmB. 0.05 V//cmC. 0.02 V//cmD. 0.2 V//cm
Answer» Correct Answer - A `((2)/(15 + 5)) xx (5)/(100) = 0.005 V//cm`.

Discussion

1646.	A 2 volt battery, a `15 Omega` resistor and a potentiometer of `100 cm` length, all are connected in series. If the resistance resistance of potentiometer wire is `5 Omega`, then the potential gradient of the potentiometer wire isA. `0.005 V//cm`B. `0.05 V//cm`C. `0.02V//cm`D. `0.2V//cm`
Answer» Correct Answer - A

Discussion

1647.	An ammeter whose resistance is `180 Omega` gives full scale deflection when current is 2mA. The shunt required to convert it into an ammeter reading 20 mA (in ohms) isA. 18B. 20C. 0.1D. 10
Answer» Correct Answer - B

Discussion

1648.	A 2 volt battery, a `15 Omega` resistor and a potentiometer of `100 cm` length, all are connected in series. If the resistance resistance of potentiometer wire is `5 Omega`, then the potential gradient of the potentiometer wire isA. `0.005 V//cm`B. `0.05 V//cm`C. `0.02 V//cm`D. `0.2 V//cm`
Answer» Correct Answer - A (a) Potential gradient `= (e)/((R R_(h) + r)) = (R )/(L)` `= (2)/((15 + 5 + 0)) xx (5)/(1) = 0.5 (V)/(m) = 0.005 (V)/(cm)`

Discussion

1649.	`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. .A. `2 Omega`B. `8 Omega`C. `20 Omega`D. `40 Omega`
Answer» Correct Answer - C (c ) By Wheatstone bridge, `(R )/(80) = (AC)/(BC) = (20)/(80) implies R = 20 Omega`

Discussion

1650.	The sensitivity of a potentiom eter is increased b yA. increasing the emf of the cell.B. increasing the length of the potentiometer wire.C. decreasing the length of potentiometer wire.D. none of the above.
Answer» Correct Answer - 2

Discussion

Explore topic-wise InterviewSolutions in .

Find the equivalent resistance of the networks shown in figure between the points a and b.

In the circuit shown in figure AB is a uniform wire of length `L = 5m`. It has a resistance of `2 Omega//m`. When `AC = 2.0 m`, it was found that the galvanometer shows zero reading when switch s is placed in either of the two positions 1 or 2. find the emf `E_(1)`.

In the circuit shown, after the switch is shifted to position 2 the heat generated in `50 Omega` resistance is `6 J`. find the emf (V) of the cell.

In the following star circuit diagram (figure), the equivalent resistance between the points A and H will be A. `1.944 r`B. `0.973 r`C. `0.486 r`D. `0.243 r`

Find the ammeter reading in the cirucit shown in Fig. 4.65.

An ammeter reads a current of 30 A when it is connected across the terminals of a cell of emf 2 V. Neglecting the meter resistance, find the amount of heat (in calories) in cell in 20 seconds.

In the circuit shown in Fig. 4.64, the battery has an emf of 12.0 V and an internal resistance of `5R//11`. If the ammeter reads 2.0 A, what is the value of R ?

Two cells of emf of 1.5 V and 2.0 V having internal resistances of `1 Omega` and `2 Omega`, respectively are connected in parallel so as to send current in the same direction through an external resistance of `5 Omega`. Find the current in the external circuit.

When a resistance of `2 Omega` is connected across the terminals of a cell, the current is 0.5 A. When resistance is increased to `5 Omega`, the current is 0.25 A. the emf of the cell isA. 1VB. 1.5 VC. 2 VD. 2.5 V

Twelve identical wires each of resistance `6 Omega` are arranged to from a skelence cube. A current of `40 m A` is led cube at the current and out at the diagonally opposite corner. Calculate the potential difference development across these current and the effective resistance of the network

Twelve identical wires each of resistance `6 Omega` are joined to from a skeleton cube. Find the resistance between the current of the same edge of the cube

Five identical cells are connected in parallel. Now, polarity of one of the cells is reversed. Percentage change in equivalent emf will beA. -0.1B. -0.2C. -0.4D. -0.6

A black body calorimeter filled with hot water cools from `60^(@)C` to `50^(@)C` in `4 min` and `40^(@)C` to `30^(@)C` in `8min`. The approximate temperature of surrounding is :A. `10^(@)`B. `15^(@)C`C. `20^(@)C`D. `25^(@)C`

Figure shows a potentiometer circuit for comparision of two resistances. The balance point with a standard resistor `R=10.0Omega` is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. the value of X isA. `11.75Omega`B. `12.55Omega`C. `9.55Omega`D. `12.75Omega`

A` lmuF `capacitor is connected in the circuit shown below. The EMF of the cell is 3V and internal resistance is `0.50Omega`. The resistors `R_(1) and R_(2)` have values `4Omega and 1Omega` respectively. The charge on the capacitor in steady state is : A. `1muC`B. `2muC`C. `1.33mu C`D. Zero

A heater coil is cut into two parts of equal length and one of them is used in the leader. The ratio of the heat procued by this half coil to that by the original coil isA. `2 :1`B. `1 : 2`C. `1 : 4`D. `4 : 1`

`2A` cell having an emf `epsilon` and internal resistance `r` is connected across a variable external resistance `R`. As the resistance `R` is increased, the plot of potential difference `V` across `R` is given byA. B. C. D.

The total power dissipated in watts in the circuit shown here is A. 16B. 40C. 54D. 4

The total power dissipated in watts in the circuit shown here is A. 4WB. 16 WC. 40 WD. 54 W

The power dissipated in the circuit shown in the is 30 watts. The value of `R` is A. `10Omega`B. `30Omega`C. `20Omega`D. `15Omega`

The power dissipated in the circuit shown in the is 30 watts. The value of `R` is A. `10Omega`B. `30Omega`C. `20Omega`D. `15Omega`

The power dissipated in the circuit shown in the is 30 watts. The value of `R` is A. `20 Omega`B. `15 Omega`C. `10 Omega`D. `30 Omega`

Both terminals of a battery of emf E and internal resistance r are grounded as shown. Select the correct alternative(s). A. Potential difference across A and B is zeroB. Potential difference acorss A and B is EC. Current across AB is zeroD. Current across AB is `(E)/(r)`

The variation between V-I has shown by graph for heating filamentA. B. C. D.

To draw the maximum current from a combination of cells, how should be the cells be grounded ?A. ParallelB. SeriesC. Depends upon the relative values of internal and external resistanceD. Mixed grouping

When a current of `(2.5 +- 0.5)` ampere flows through a wire, it develops a potential difference of `(20 +- 1)` volt, the resistance of the wire isA. `(8 pm 2)Omega`B. `(8 pm 1.6)Omega`C. `(8 pm 1.5)Omega`D. `(8 pm 3)Omega`

A torch battery consisting of two cells of 1.45 volts and an internal resistance `0.15 Omega` , each cell sending currents through the filament of the lamps having resistance 1.5ohms. The value of current will beA. `16.11 amp`B. `1.611 amp`C. `0.1611 amp`D. `2.6 amp`

A battery when connected by resistance of `16Omega` gives a terminal voltage of 12 V. and when connected by a resistance of `10Omega` gives a terminal voltage of 11V. Then the emf of the battery ad its internal resistanceA. 12.8 VB. 13.7 VC. 10.7 VD. 9 V

The resistance of a potentiometer wire of length `10 m` is `20 Omega`. A resistance box and a `2` volt accumulator are connected in series with it. What resistance should be introduced in the box to have a potential drop of one microvolt per millimetre of the potentiometer wire ?

A potentiometer having a wire of 4 m lengths is connected to the terminals of a battery with steady voltage. A leclanche cell has a null point at 1m. If the length of the potentiometer wire is increased by 1 m, the position of the null points isA. 1.5 mB. 1.25 mC. 10.05 mD. 1.31 m

An ammeter of `100 Omega` resistance gives full deflection for the current of `10^(-5)` amp . Now the shunt resistance required to convert it into ammeter of 1 amp . range, will beA. `10^(-4) Omega`B. `10^(-5)Omega`C. `10^(-3)Omega`D. `10^(-1)Omega`

A 2 volt battery, a `15 Omega` resistor and a potentiometer of `100 cm` length, all are connected in series. If the resistance resistance of potentiometer wire is `5 Omega`, then the potential gradient of the potentiometer wire isA. 0.005 V//cmB. 0.05 V//cmC. 0.02 V//cmD. 0.2 V//cm

A 2 volt battery, a `15 Omega` resistor and a potentiometer of `100 cm` length, all are connected in series. If the resistance resistance of potentiometer wire is `5 Omega`, then the potential gradient of the potentiometer wire isA. `0.005 V//cm`B. `0.05 V//cm`C. `0.02V//cm`D. `0.2V//cm`

An ammeter whose resistance is `180 Omega` gives full scale deflection when current is 2mA. The shunt required to convert it into an ammeter reading 20 mA (in ohms) isA. 18B. 20C. 0.1D. 10

A 2 volt battery, a `15 Omega` resistor and a potentiometer of `100 cm` length, all are connected in series. If the resistance resistance of potentiometer wire is `5 Omega`, then the potential gradient of the potentiometer wire isA. `0.005 V//cm`B. `0.05 V//cm`C. `0.02 V//cm`D. `0.2 V//cm`

`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. .A. `2 Omega`B. `8 Omega`C. `20 Omega`D. `40 Omega`

The sensitivity of a potentiom eter is increased b yA. increasing the emf of the cell.B. increasing the length of the potentiometer wire.C. decreasing the length of potentiometer wire.D. none of the above.