2492 + Interview Questions in CURRENT-ELECTRICITY IN GENERAL AWARENESS Page 50 InterviewSolution

2451.	Assertion : The switch `S` shown in the figure is closed at `t = 0`. Initial current flowing through battery is `(E)/(R + r)` Reason : Initially capacitor was uncharged, so resistance offered by capacitor at `t = 0` is zero A. Statement-1 is True,Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True,Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1C. Statement -1 is True, Statement-2 is false.D. Statement-1 is False, Statement-2 is True.
Answer» Correct Answer - A Charge on capacitor `q=CE(1-e^(-t//CR_(eq)))` `:. I=(dq)/(dt)=E/(R_(eq))e^(-t//CR)` at `t=0,I=E/(R_(eq))=E/(R+r)implies` resistance offered by capacitor is zero,

Discussion

2452.	In the arrangement shown in figure when the switch `S_2` is open, the galvanometer shown no deflection for `l=L//2`. When the switch `S_2` is closed, the galvanometer shown no deflection for `l=5L//12`. The internal resistance `(r)` of `6V` cell, and the emf `E` of the other battery are respectively A. `3 Omega, 8 V`B. `2 Omega, 12 V`C. `2 Omega, 24 V`D. `3 Omega, 12 V`
Answer» Correct Answer - B (b) When `S_(2)` open Assume resistance of `AB = R` Resistance of wire per unit length. `x = (R )/(L)` `I = (E)/(R )` Now in `AC` `(E)/(R ) xx (R )/(L) xx (L)/(2) = 6` `E = 12 V` When `S_(2)` closed `V_(1) = (E)/(R ) xx (R )/(L) xx (5L)/(12) = (5R)/(12) = (5 xx 12)/(12) = 5V` `implies 6 - I_(1) r = 5 implies 6 - ((5)/(10)) r = 5 implies r = 2 Omega`

Discussion

2453.	Assertion : The switch `S` shown in the figure is closed at `t = 0`. Initial current flowing through battery is `(E)/(R + r)` Reason : Initially capacitor was uncharged, so resistance offered by capacitor at `t = 0` is zero A. If both assertion and reason are ture and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false.
Answer» Correct Answer - A (a) Charge on capacitor `q = CE (1 - e^(-t//CR_(eq)))` `:. I = (dq)/(dt) = (E)/(R_(eq)) e^(-t//CR)` at `t = 0` `I = (E)/(R_(eq)) = (E)/(R + r) implies` Resistance offered by capacitor is zero.

Discussion

2454.	S.I. unit of potential gradient isA. V cmB. V/cmC. V mD. V/m
Answer» Correct Answer - D

Discussion

2455.	In the infinite gride. If the value of `r = 2 (sqrt(5 - 1)) Omega`. Find the Equivalent resistance between the point `a` and `c`. A. `6 (sqrt(5 - 1)) Omega`B. `4 (sqrt(5 - 1)) Omega`C. `4 Omega`D. none of these
Answer» Correct Answer - C (c ) Due to symmetry, the line of resistors passing through `b` will be useless `r_(eq) = 2r + (r r_(eq))/(2 r + r_(eq)) implies r_(eq) = (1 + sqrt5) r` `= (sqrt( 5 + 1)) 2 (sqrt(5 - 1)) = 4 Omega`

Discussion

2456.	In comparing e.m.f. of two cells by a potentiometer balance point is obtained on `5^(th)` wire, the current flowing through the wire is takenA. from both cellsB. from one cellC. from the main battery of circuitD. none of these
Answer» Correct Answer - C

Discussion

2457.	Battery shown in figure has e.m.f. `E` and internal resistance `r`. Current in the circuit can be varied by sliding the contact `J`. If at any instant current flowing through the circuit is `I`, potential difference between terminals of the cells is `V`, thermal power generated in the cell is equal to `eta` fraction of total electrical power generated in it, then which of the following graphs is correct ? A. B. C. D. Both (a) and (b) are correct
Answer» Correct Answer - D (d) Terminal voltage `V = E - Ir`. Hence, the graph between `V` and `i` will be a straigh line having negative slopw and positive intercept Thermal power generated in the exteranl circuit `P = EI - I^(2) r`. Hence, graph between `P` and `I` will be a parabola passing through origin. Also at an instant, thermal power generated in the cell ` = i^(2)r` and total electrical power generated in the cell `=E i`. Hence the fraction `eta = (I^(2) r)/(EI) = ((r)/(E)) I`, so `eta prop I`. If means graph between `eta` and `I` will be a straight line

Discussion

2458.	In the circuit shown, which way would you move the sliding contact, to the left or to the right, in order to increase current through resistance `R_(1)` ? What will happen to current through `R_(2)` as you move the contact?
Answer» Correct Answer - To left

Discussion

2459.	If the length of potentiometer wire is increased, then the length of the previously obtained balance willA. IncreaseB. DecreaseC. Remain unchangedD. Become two times
Answer» Correct Answer - A

Discussion

2460.	An ammeter A, a voltmeter V and a resistance R are connected as shown in the figure. If the voltmeter reading is `1.6V` and the ammeter reading is `0.4`A, then R is A. equal to `4Omega`B. greater than `4Omega`C. less than `4Omega`D. `between `3Omega` and `4Omega`
Answer» Correct Answer - B

Discussion

2461.	In a potentiometer, balance point is obtained, whenA. The e.m.f. of the battery becomes equal to the e.m.f. of the experimental cellB. The p.d. of the wire between the + ve end to jockey becomes equal to the e.m.f. of the experimental cellC. The p.d. of the wire between + ve point and jockey becomes equal to the e.m.f. of the batteryD. The p.d. across the potentiometer wire becomes equal to the e.m.f. of the battery
Answer» Correct Answer - B

Discussion

2462.	In the experiment of potentiometer, at balance, there is no current in theA. Main circuitB. Galvanometer circuitC. Potentiometer circuitD. Both main and galvanometer circuits
Answer» Correct Answer - B

Discussion

2463.	In the diagram shown, the reading of voltmeter is 20 V and that of ammeter is 4 A . The value of R should be (Consider given ammeter and voltmeter are not ideal) A. Equal to `5 Omega`B. Greater from `5Omega`C. Less than `5Omega`D. Greater or less than `5Omega` depends on the material of R
Answer» Correct Answer - C

Discussion

2464.	How amny joules of energy is equivalent to 1 k Wh?
Answer» Energy `= 1 kWh = 100 W xx (60 xx 60) s` `= 3.6 xx 10^(6)J`

Discussion

2465.	To use the milliammeter of the previous question as a voltmeter of range `10 V`, a resistance `R` is placed in series with it. The value of `R` isA. `9 Omega`B. `99 Omega`C. `999 Omega`D. `1000 Omega`
Answer» Correct Answer - C `V = i_g (R + G)` `10 = 10 xx 10^-3(R + 0.9) rArr R = 999 Omega`.

Discussion

2466.	The resistance of a galvanometer is 90 ohm s. If only 10 percent of the main current may flow through the galvanometer, in which way and of what value, a resistor is to be usedA. 10 ohms in seriesB. 10 ohms in parallelC. 810 ohms in seriesD. 810 ohms in parallel
Answer» Correct Answer - B

Discussion

2467.	Which of the following statement is wrongA. Voltmeter should have high resistanceB. Ammeter should have low resistanceC. Ammeter is placed in parallel across the conductor in a circuitD. Voltmeter is placed in parallel across the conductor in a circuit
Answer» Correct Answer - C

Discussion

2468.	What is the meaning of 1 unit electric energy in domestic use?
Answer» 1 unit of electrical energy = 1 kWh. If an electrical appliance of power 1000 watt is operated on mains for 1 hour, it will consume 1 unit of electricity.

Discussion

2469.	A voltmeter has a resistance `G` and range `V`. Calculate the resistance to be used in series with it to extend its range to `nV`.A. `nG`B. `(n-1)G`C. `(G)/(n)`D. `(G)/((n-1))`
Answer» Correct Answer - B

Discussion

2470.	A voltmeter has a resistance `G` and range `V`. Calculate the resistance to be used in series with it to extend its range to `nV`.A. `n G`B. `( n - 1) G`C. `(n + 1) G`D. `G`
Answer» Correct Answer - B `V = i_g G rArr i_g = V//G` `nV = i_g (G + R)` `nV = (V)/(G) (G + R)` `R = (n-1) G`.

Discussion

2471.	A heating element is marked 210 V, 630 W. What is the current drawn by the element when connected to a 210 V dc mains ? What is the resistance of an element ?
Answer» Here, V = 210 V , P = 630 W, Current drawn by element, `I = P/V = 630/210 = 3A` Resistance of the element, `R = V/I = 210/3 = 70 Omega`

Discussion

2472.	Write an expression for the heat produced when an electric current is passed through it.
Answer» `H = Vit = I^(2) R t = V^(2)/R t` in energy units (joules). or `H = (Vit)/(J) = (I^(2)Rt)/(J) = (V^(2)t)/(JR)` in heat units (cal).

Discussion

2473.	Two resistors of `2 Omega` and `4 Omega` are connected in parallel to a constant d.c. voltage. In which case more heat is produced ?
Answer» Heat produced, `H = (V^(2)t)/(R) or H prop 1/R` Thus more heat is produced in `2 Omega` resistor than that of `4 Omega` resistor.

Discussion

2474.	A heating element is marked 210 V, 630 W. What is the value of the current drawn by the element when connected to a 210 V dc source ?
Answer» Here, `V = 210 V, P = 630 W`, Current `I = P/A = 630/210 = 3A`

Discussion

2475.	Assertion : Resistance of an ammeter is less than the resistance of a milliammeter. Reason : Value of shunt required in case of ammeter is more than a milliammeter.A. If both Assertion and Reason are true and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are true and Reason is not correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true.
Answer» Correct Answer - B

Discussion

2476.	Distinguish between kilowatt and kilowatt hour.
Answer» Kilowatt is the unit of power. 1 kilowatt = 1000 watt. Kilowatt hour is the unit of electric energy, where 1 Kilowatt hour = 1000 watt `xx` 1 hour = 1000 watt `xx` 60 `xx` 60 `xx` second `= 3.6xx10^(6)J`

Discussion

2477.	Assertion : If by mistake, a voltmeter is connected in series, then circuit will burn. Reason : Current will drastically decrease in the circuit.A. If both Assertion and Reason are true and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are true and Reason is not correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true.
Answer» Correct Answer - D

Discussion

2478.	Assertion : Kilovol-ampere (kVA) and kilowatt-hour have the same dimensions. Reason : Both are the units of energy.A. If both Assertion and Reason are true and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are true and Reason is not correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If both Assertion and Reason are false.
Answer» Correct Answer - D

Discussion

2479.	Why does the cord of an electric heater not glow while the heating element does ?
Answer» Correct Answer - Nichrome

Discussion

2480.	Of what substance is a fuse wire made of?
Answer» Correct Answer - 63% tin + 37% lead.

Discussion

2481.	State the characteristics of fuse wire.
Answer» A fuse wire should have (i) high resistivity (ii) low melting point and (iii) of suitable current rating corresponding to the load in the circuit.

Discussion

2482.	Assume that the batteries in figure have negligible internal resistance. Find a. the current in the circuit, (b) the power dissipated in each resistor and (c) the power of each battery, stating whether energy is supplied by or absorbed by it
Answer» Correct Answer - A::B::C::D a. `i=(12-6)?(4+8)=0.5A` b. `P_(R_1)=i^2R_1=1W` `rarr P_(R_2)=i^2R_2=2W` c. Power supplied by `E_1=E_1i=6W` and power consumed by `E_2=E_1i=3W`

Discussion

2483.	In the circuit find the potential difference between point `A and B`. Assume that both the batteries have zero internal resistance
Answer» Correct Answer - `11 V` Here, `epsilon_(1) = 20 V, epsilon_(2) = 8 V, R_(1) = 3 Omega , R_(2) = 1Omega` As the batteries are sending the current in opposite direction so effective emf `= 20 - 8 = 12 V` The internal resistance of batteries is zero `R_(1),R_(2)` are in series in the circuit .Therefore Total resistance `R = R_(1)+ R_(2)= 3 + 1= 4 Omega` Effective current `1 = 12 //4 = 3A` `V_(AB) = epsilon_(1) - IR_(1) = 20 - 3 xx 3 = 11V` Also `V_(AB) = epsilon_(2) - IR_(2) = 8 + 3 xx 1 = 11V`

Discussion

2484.	A circuit is connected as shown in the figure with the switch S open. When the switch is closed, the total amount of charge that flows from Y to X is .
Answer» Correct Answer - A,B,D

Discussion

2485.	If `theta_i`, is the inversion temperature, `theta_n` is the neutral temperature, `theta_c` is the temperature of the cold junction, thenA. `theta_(i)+theta_(c)=theta_(n)`B. `theta_(i)-theta_(c)=2 theta_(n)`C. `(theta_(1)+theta_(2))/2 =theta_(n)`D. `theta_(c)-theta_(1)=2 theta_(n)`
Answer» Correct Answer - C In Seeback (thermoelectric) effect, the temperature of hot junction at which the thermo emf is maximum is called the neutral temperature `(theta_(n))`, and the temperature at which thermo emf changes its sign is called inversion temperature `(theta_(i))`. if `theta_(c)` is the temperature of cold junction then these three are related by the expression. `theta_(n)-theta_(c)=theta_(i)-theta_(n)` `:, theta_(n)=(theta_(i)+theta_(c))/2`

Discussion

2486.	If `theta_i`, is the inversion temperature, `theta_n` is the neutral temperature, `theta_c` is the temperature of the cold junction, thenA. `theta_i + theta_c = theta_n`B. `theta_i - theta_c = 2theta_n`C. `(theta_i + theta_c)/2 = theta_n`D. `theta_c - theta_i = 2theta_n`
Answer» Correct Answer - C (c) `theta_m = (theta_i + theta_c)/2`

Discussion

2487.	If in the circuit, power dissipation is 150W, the R is A. `2 Omega`B. `6 Omega`C. `5 Omega`D. `4 Omega`
Answer» Correct Answer - B (b) The equivalent resistance is `R_(eq) = (2xxR)/(2+R)` `:. Power dissipation P = V^2/R_(eq)` `:. 15 0 = (15 xx 15)/R_(eq) :. R_(eq) = 15/10 = 3/2` `rArr (2R)/(2+R) = 3/2 rArr 4R = 6+3R rArr R = 6 Omega`.

Discussion

2488.	A current of `2 A` flows through a `2 Omega` resistor when connected across a battery. The same battery supplies a current of `0.5 A` when connected across a `9 Omega` resistor. The internal resistance of the battery isA. `0.5Omega`B. `1//3 Omega`C. `1//4Omega`D. `1Omega`
Answer» Correct Answer - 2 Let internal resistance of b attery be r then according to question `(E)/(2+r)=2` and `(E)/(9+r)=0.5` `rArr(2)/(0.5)=(9+r)/(2+r)rArrr=(1)/(3)Omega`

Discussion

2489.	The temperature coefficent of resistance of a wire is 0.00125 per `.^(@)C`, At 300 K, its resistance is `1Omega`. At what temperature will its resistance become `2 Omega` ?
Answer» Correct Answer - `1173 K`

Discussion

2490.	Three equal resistances connected is series across a source of e.m.f consume `20` watt. If the same resistor are connected in parallel across the same source of e.m.f., what would be the power dissipated ?
Answer» Correct Answer - `180 watt` In series ,total resistance `R_(s) = R + R+ R= 3R` Given `V^(2)//(3R) = 20 or (V^(2))/(R ) = 60 ` In parallel , total resistance `R_(p) = R//3` power consumed = `V^(2)R_(p) = 3V^(2)//R` `= 3 xx 60 = 180 watt`

Discussion

2491.	The thermo e.m.f. of a thermo- couple is `25 muV//^@C` at room temperature. A galvanometer of 40 ohm resistance, capable of detecting current as low as `10^(-5)`A, is connected with the thermo couple. The smallest temperature difference that can be detected by this system isA. `16^(@)C`B. `12^(@)C`C. `8^(@)C`D. `20^(@)C`
Answer» Correct Answer - A Thermo-emf of thermocouple is `=25 mu V//^(@)C` `:.` After connecting the thermocouple with the galvanometer, thermo-emf `E=25 muV//^(@)Cxx(theta^(@)C)impliesE=250xx10^(-6)V` Potential drop developed across the galvanometer `=iR=10^(-5)xx40=4xx10^(-4)` volt `:. 4xx10^(-4)=250xx10^(-6)` `theta=4/25xx10^(2)=16^(@)C`

Discussion

2492.	Near room t emperature, the thermo emf of a copper constantan couple is `40 muV` per degree. What is the smallest temperature difference that can be detected with a single such couple and a galvanometer of 100 ohm resistance capable of detecting current as low `10^(-6)` ampere?
Answer» Let `0^(@)C` be the smallest temperature that can be measured. Corresponding thermo emf=`40xxthetamuV=40xxthetaxx10^(-6)V` Resistance of the galvanometer `=100Omega` Smallest current that can be detected, `I=10^(-6)A` current `=(emf)/("Resistance")` or `10^(-6)=(40xxthetaxx10^(-6))/(100)` or `theta=(10^(-6)xx100)/(40xx10^(-6)).^(@)C=2.5^(@)C`.

Discussion

Explore topic-wise InterviewSolutions in .

S.I. unit of potential gradient isA. V cmB. V/cmC. V mD. V/m

In the infinite gride. If the value of `r = 2 (sqrt(5 - 1)) Omega`. Find the Equivalent resistance between the point `a` and `c`. A. `6 (sqrt(5 - 1)) Omega`B. `4 (sqrt(5 - 1)) Omega`C. `4 Omega`D. none of these

In comparing e.m.f. of two cells by a potentiometer balance point is obtained on `5^(th)` wire, the current flowing through the wire is takenA. from both cellsB. from one cellC. from the main battery of circuitD. none of these

In the circuit shown, which way would you move the sliding contact, to the left or to the right, in order to increase current through resistance `R_(1)` ? What will happen to current through `R_(2)` as you move the contact?

If the length of potentiometer wire is increased, then the length of the previously obtained balance willA. IncreaseB. DecreaseC. Remain unchangedD. Become two times

An ammeter A, a voltmeter V and a resistance R are connected as shown in the figure. If the voltmeter reading is `1.6V` and the ammeter reading is `0.4`A, then R is A. equal to `4Omega`B. greater than `4Omega`C. less than `4Omega`D. `between `3Omega` and `4Omega`

In the experiment of potentiometer, at balance, there is no current in theA. Main circuitB. Galvanometer circuitC. Potentiometer circuitD. Both main and galvanometer circuits

In the diagram shown, the reading of voltmeter is 20 V and that of ammeter is 4 A . The value of R should be (Consider given ammeter and voltmeter are not ideal) A. Equal to `5 Omega`B. Greater from `5Omega`C. Less than `5Omega`D. Greater or less than `5Omega` depends on the material of R

How amny joules of energy is equivalent to 1 k Wh?

To use the milliammeter of the previous question as a voltmeter of range `10 V`, a resistance `R` is placed in series with it. The value of `R` isA. `9 Omega`B. `99 Omega`C. `999 Omega`D. `1000 Omega`

The resistance of a galvanometer is 90 ohm s. If only 10 percent of the main current may flow through the galvanometer, in which way and of what value, a resistor is to be usedA. 10 ohms in seriesB. 10 ohms in parallelC. 810 ohms in seriesD. 810 ohms in parallel

Which of the following statement is wrongA. Voltmeter should have high resistanceB. Ammeter should have low resistanceC. Ammeter is placed in parallel across the conductor in a circuitD. Voltmeter is placed in parallel across the conductor in a circuit

What is the meaning of 1 unit electric energy in domestic use?

A voltmeter has a resistance `G` and range `V`. Calculate the resistance to be used in series with it to extend its range to `nV`.A. `nG`B. `(n-1)G`C. `(G)/(n)`D. `(G)/((n-1))`

A voltmeter has a resistance `G` and range `V`. Calculate the resistance to be used in series with it to extend its range to `nV`.A. `n G`B. `( n - 1) G`C. `(n + 1) G`D. `G`

A heating element is marked 210 V, 630 W. What is the current drawn by the element when connected to a 210 V dc mains ? What is the resistance of an element ?

Write an expression for the heat produced when an electric current is passed through it.

Two resistors of `2 Omega` and `4 Omega` are connected in parallel to a constant d.c. voltage. In which case more heat is produced ?

A heating element is marked 210 V, 630 W. What is the value of the current drawn by the element when connected to a 210 V dc source ?

Distinguish between kilowatt and kilowatt hour.

Why does the cord of an electric heater not glow while the heating element does ?

Of what substance is a fuse wire made of?

State the characteristics of fuse wire.

Assume that the batteries in figure have negligible internal resistance. Find a. the current in the circuit, (b) the power dissipated in each resistor and (c) the power of each battery, stating whether energy is supplied by or absorbed by it

In the circuit find the potential difference between point `A and B`. Assume that both the batteries have zero internal resistance

A circuit is connected as shown in the figure with the switch S open. When the switch is closed, the total amount of charge that flows from Y to X is .

If `theta_i`, is the inversion temperature, `theta_n` is the neutral temperature, `theta_c` is the temperature of the cold junction, thenA. `theta_(i)+theta_(c)=theta_(n)`B. `theta_(i)-theta_(c)=2 theta_(n)`C. `(theta_(1)+theta_(2))/2 =theta_(n)`D. `theta_(c)-theta_(1)=2 theta_(n)`

If `theta_i`, is the inversion temperature, `theta_n` is the neutral temperature, `theta_c` is the temperature of the cold junction, thenA. `theta_i + theta_c = theta_n`B. `theta_i - theta_c = 2theta_n`C. `(theta_i + theta_c)/2 = theta_n`D. `theta_c - theta_i = 2theta_n`

If in the circuit, power dissipation is 150W, the R is A. `2 Omega`B. `6 Omega`C. `5 Omega`D. `4 Omega`

A current of `2 A` flows through a `2 Omega` resistor when connected across a battery. The same battery supplies a current of `0.5 A` when connected across a `9 Omega` resistor. The internal resistance of the battery isA. `0.5Omega`B. `1//3 Omega`C. `1//4Omega`D. `1Omega`

The temperature coefficent of resistance of a wire is 0.00125 per `.^(@)C`, At 300 K, its resistance is `1Omega`. At what temperature will its resistance become `2 Omega` ?

Three equal resistances connected is series across a source of e.m.f consume `20` watt. If the same resistor are connected in parallel across the same source of e.m.f., what would be the power dissipated ?

Near room t emperature, the thermo emf of a copper constantan couple is `40 muV` per degree. What is the smallest temperature difference that can be detected with a single such couple and a galvanometer of 100 ohm resistance capable of detecting current as low `10^(-6)` ampere?