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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.
| 1051. |
A step-down transformer is connected to main supply `200V` to operate a `6V`, 30W` bulb. The current in primary isA. `3A`B. `1.5A`C. `0.3A`D. `0.15A` |
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Answer» Correct Answer - D `V_(p)=200V,V_(s)=6V` `P_(out)=V_(s)i_(s)implies30=6xxi_(s)impliesi_(s)=5A` From `(V_(s))/(V_(p))=(i_(p))/(i_(s))implies(6)/(200)=(i_(p))/(5)impliesi_(p)=0.15A` |
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| 1052. |
A step-down transformer is connected to main supply `200V` to operate a `6V`, `30W` bulb. The current in primary isA. `3A`B. 1.5 AC. 0.3 AD. 0.15 A |
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Answer» Correct Answer - D `P_(P)=P_(S)` `I_(P)*e_(P)=P_(S)` `I_(P)=(P_(S))/(e_(P))=(30)/(200)=0.15A` |
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| 1053. |
An a.c. voltage source of `E = 150 sin 100 t` is used to run device, which offers a resistance of 20 ohm and restricts the flow of current in one direction only. Calcualte the average and rms value of current in the circuit. |
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Answer» Here, `E = 150 sin 100 t, R = 20 ohm` `:. E_(0) = 150 V, I_(0) = (E_(0))/(R ) = (150)/(20) = 7.5 A` As the flow current is restricted in one direction only, therefore, average value of current, `I_(av) = (I_(0))/(pi)` (instead of `2I_(0)//pi`) `= (7.5)/(3.14) = 3.93 A` and rms value of current `I_(rms) = (I_(0))/(sqrt2 sqrt2)` (instead of `I_(0)//sqrt2`) `= (7.5)/(2) = 3.75 A` |
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| 1054. |
An a.c. source of voltage `V = 100 sin 100 pi t`. Is connected to a resistor of `(25)/(sqrt2) ohm`. What is the r.m.s value of current (in ampere) through the resistor ? |
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Answer» Correct Answer - 4 Here, `V = 100 sin pi t, R = (25)/(sqrt2) ohm`. Compare with the stantard form `V = E_(0) sin omega t` `E_(0) = 100`, Volt , `E_(upsilon) = (E_(0))/(sqrt2) = (100)/(sqrt2)` volt `I_(upsilon) = (E_(upsilon))/(R ) = (100 // sqrt2)/(25 // sqrt2) = 4 A` |
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| 1055. |
In the given circuit the capacitor (C) may be charged through resistance R by a battery V by closing switch `(S_1)`. Also when `(S_1)` is opend and `(S_2)` is closed the capacitor is connected in series with inductor (L). Given taht the total charge stored in the LC circuit is `(Q_0)`. for `Tge0, the charge on the capacitor isA. the charge on the capacitor is `Q=Q_(0) cos""((pi)/(2)+(t)/(sqrt(LC)))`B. the charge on the capacitor is `Q=Q_(0) cos""((pi)/(2)-(t)/(sqrt(LC)))`C. the charge on the capacitor is `Q= -LC(d^(2)Q)/(dt^(2))`D. the charge on the capacitor is `Q= -(1)/(sqrt(LC))(D^(2)Q)/(dt^(2))` |
| Answer» Correct Answer - C | |
| 1056. |
Initially the `900muF` capacitor is charged to 100 V and the `100muF` capacitor is uncharged in the figure shown. Then the switch `S_(2)` is closed for a time `t_(1)`, after which it is opened and at the same instant switch `S_(1)` is closed for a time `t_(2)` and then opened. It is now found that the `100muF` capacitor is charged to 300 V. If `t_(1) and t_(2)` minimum possible values of the time intervals, then findout `(t_(1))/(t_(2))` |
| Answer» Correct Answer - C | |
| 1057. |
In an LCR circuit as shown below both switches are open initially. Now switch `S_(1)` kept open. (q is charge on the capacitor and `tau = RC` is Capacitive time constant). Which of the following statement is correct? A. Work done by the battery is half of the energy dissipated in the resistorB. At `t=tau,q=CV//2`C. At `t=2tau,q=CV(1-e^(-2))`D. At `t=(tau)/(2),q=CV(1-e^(-1))` |
| Answer» Correct Answer - C | |
| 1058. |
In an LCR circuit as shown below both switches are open initially. Now switch `S_(1)` kept open. (q is charge on the capacitor and `tau = RC` is Capacitive time constant). Which of the following statement is correct? A. Work done by the battery is half of the energy disslpaled in the resistorB. At `t=tau, q=CV//2`C. At `t=2tau,q=CV(1-e^(-2))`D. At `t=tau/2,q=CV(1-e^(-1))` |
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Answer» Correct Answer - C `q=CV(1-e^(t//tau))` at `t=2tau` `q=CV(1-e^(-2))` |
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| 1059. |
In the circuit shown, the capacitor is initially charged with a 12V battery, when switch `S_(1)` is open and switch `S_(2)` is closed. `S_(1)` is then closed and, at the same time, `S_(2)` is opened. The maximum value of current in the circuit is A. after time interval `tau`, charge on the capacitor is CV/2B. after time interval `2tau` , charge on the capacitor is `CV(1-e^(-2))`C. the work done by the voltage source will be half of the heat dissipated when the capacitor is fully charged.D. after time interval `2tau`, charge on the capacitor is `CV(1-e^(-1))` |
| Answer» Correct Answer - B | |
| 1060. |
An indcutor having self inductance L with its coil resistance R is connected across a battery of emf `elipson`. When the circuit is in steady state t = 0, an iron rod is inserted into the inductor due to which its inductance becomes nL (ngt1). When again circuit is in steady state, the current in it isA. `I lt epsi//R`B. `I gt epsi//R`C. `I=epsi//R`D. none of these |
| Answer» Correct Answer - C | |
| 1061. |
An indcutor having self inductance L with its coil resistance R is connected across a battery of emf `elipson`. When the circuit is in steady state t = 0, an iron rod is inserted into the inductor due to which its inductance becomes nL (ngt1). When again circuit is in steady state, the current in it isA. `I lt epsilon//R`B. `I gt epsilon//R`C. `I = epsilon//R`D. None of these |
| Answer» Correct Answer - C | |
| 1062. |
An indcutor having self inductance L with its coil resistance R is connected across a battery of emf `elipson`. When the circuit is in steady state t = 0, an iron rod is inserted into the inductor due to which its inductance becomes nL (ngt1). After insertion of rod which of the following quantities will change with time? (1) Potential difference across terminals A and B (2) Inductance (3) Rate of heat produced in coilA. only(1)B. (1) & (3)C. Only (3)D. (1),(2) & (3) |
| Answer» Correct Answer - C | |
| 1063. |
An indcutor having self inductance L with its coil resistance R is connected across a battery of emf `elipson`. When the circuit is in steady state t = 0, an iron rod is inserted into the inductor due to which its inductance becomes nL (ngt1). after insertion of rod, current in the circuit:A. Increases with timeB. Decreases with timeC. Remains constant with timeD. First decrease with time then becomes constant |
| Answer» Correct Answer - A | |
| 1064. |
The coils of a step down transformer have `500` and `5000` turns. In the primary coil an ac of `4` ampere at `2200` volts is sent. The value of the current and potential difference in the secondary coil will beA. `20A,220V`B. `0.4A,22000 v`C. `40 A,220V`D. `40A,22000V` |
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Answer» Correct Answer - C `(N_(p))/(N_(s))=(V_(p))/(v_(s))=(i_(s))/(i_(p))`. The transformer is step-down type, so primary coil will have more turns. Hence `(5000)/(500)=(2200)/(V_(s))=(i_(s))/(4)impliesV_(s)=220V,i_(s)=40 amp` |
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| 1065. |
A loss free transformer has `500` turns on its primary winding and `2500` in secondary. The meters of the secondary indicate `200` volts at `8` amperes under these condition. The voltage and current in the primary isA. 100 V, 16 SAB. 40 V, 40 AC. 160 V, 10 AD. 80 V, 20 A |
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Answer» Correct Answer - B `(e_(P))/(e_(S))=(n_(P))/(n_(S))` `therefore e_(P)=(n_(P)*e_(S))/(n_(S))=(500xx200)/(2500)=40V` `I_(P)=(n_(S))/(n_(P))xxI_(S)=(2500)/(500)xx8=40A` |
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| 1066. |
An ideal inductor when connected in a.c. circuit does not produce heating effect though it reduces the current in the circuit. Explain why ? |
| Answer» A ideal inductor is coil having some inductance (L) but no ohmic resistance R. Amount of heat produced in time `t = I^(2) R t`. As `R = 0`, therefore, heat produced = 0. However, the inductor offers inductive reactance `X_(L) = omega L = 2 pi v L` to the a.c. Therefore, the current is reduced. | |
| 1067. |
An alternating e.m.f. of peak value 350 V is applied across an ammeter of resistance 100 ohm. What will be the reading of ammeter ? |
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Answer» Here, `E_(0) = 350 V, R = 100 Omega, I_(v) = ?` `I_(v) = (E_(v))/(R ) = (E_(0))/(sqrt2 R) = (350)/(1.414 xx 100) = 2.47 A` |
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| 1068. |
A cylindrical bar magnet is rotated about its axis (Figure). A wire is connect from the axis and is made to touch the cylindrical surface through a contact. Then A. a direct current is flows in the ammeter AB. no current flows through the ammeter AC. an alternating sunusoidal current flows through the ammeter A with a time period `T = 2 pi//omega`D. a time varying non-sinosoidal current flows through the ammeter A |
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Answer» Correct Answer - B As there is no change in magnetic flux associated with the circuit, no current is induced in the circuit. The ammeter A shows no deflection. Choice (b) is correct. |
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| 1069. |
A cylindrical bar magnet is rotated about its axis (Figure). A wire is connect from the axis and is made to touch the cylindrical surface through a contact. Then |
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Answer» Correct Answer - b AS there is no change in magnetic flux associated with the circuit, no current, is induced in the circuit.The ammete A shows no deflection. |
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| 1070. |
An alternating e.m.f. of angular frequency `omega` is applied across an inductance. The instantaneous power developed in the circuit has an angular frequencyA. `2 omega`B. `omega`C. `(omega)/4`D. `(omega)/2` |
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Answer» Correct Answer - A For a pure inductance, the current lags behind the voltage by `(pi)/2`. `:.` For the circuit, `E=E_(0)I_(0) sin omega t sin(omega t = (pi)/2)` `= E_(0)I_(0) sin omega t (-cos omega t)` `=E_(0)I_(0) sin omega t cos omega t` `1/2 E_(0)I_(0)(sin 2 omegat)` Thus the angular frequency of the instantaneous power is `2 omega`. |
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| 1071. |
The self inductance of the motor of an electric fan is 10H. In order to impart maximum power at 50 Hz, it should be connected to a capacitance ofA. `4 mu F`B. `2 mu F`C. `8 mu F`D. `1 mu F` |
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Answer» Correct Answer - D For maximum power, `L omega = 1/(C omega)` `:. C = 1/(L omega^(2)) = (1)/(L cdot 4 pi^(2)n^(2))` `= (1)/(10 xx 4 xx 10 xx 2500) [pi^(2)-:10 and n=50]` `:. C = 1/(10^6) = 10^(-6)F = 1 mu F`. |
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| 1072. |
The self inductance of the motor of an electric fan is 10H. In order to impart maximum power at 50 Hz, it should be connected to a capacitance ofA. `8 mu F`B. `4 `mu F`C. `2 mu F`D. `1 mu F` |
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Answer» Correct Answer - D (d) For maximum power, `(X_L)=X_(C), ` which yields `C=(1)/((2pi n)^(2)L)=(1)/(4 pi ^(2)xx50xx50xx10)` `:. C=0.1xx10^(-5)=1 (mu)F`. |
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| 1073. |
The self inductance of the motor of an electric fan is 10H. In order to impart maximum power at 50 Hz, it should be connected to a capacitance ofA. `1 mu F`B. `2 mu F`C. `3 mu F`D. `4 mu F` |
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Answer» Correct Answer - A Power is maximum at resonant frequency `(f_R)` `f_(R) = 1/(2 pisqrt(LC)) :. F_(R)^(2) = (1)/(4 pi^(2)LC)` `:. C = 1/(4pi^(2)f_(R)^(2)xxL)` `1/(4 xx 10 xx 50 xx 50 xx 10) = 10^(-6)F = 1mu F`. |
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| 1074. |
An ideal coil of 10H is connected in series with a resistance of `5(Omega)` and a battery of 5V. 2second after the connections is made, the current flowing in ampere in the circuit isA. `(1 - e^(-1))`B. `(1 - e)`C. eD. `e^(-1)` |
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Answer» Correct Answer - A Here,`L = 10 h, R = 5 Omega, E = 5 V` Max. current, `I_(0) = (E)/(R ) = (5)/(5) = 1 A` During growth of current `I = I_(0) (1 - e^((-R )/(L) t))` `:. I = 1 (1 - e^((-5)/(10) xx 2)) = (1 - e^(-1))` |
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| 1075. |
An ideal coil of 10H is connected in series with a resistance of `5(Omega)` and a battery of 5V. 2second after the connections is made, the current flowing in ampere in the circuit isA. `(1-e^(-1))`B. `(1-e)`C. `e`D. `e^(-1)` |
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Answer» Correct Answer - A (a)`I=(I_0)(1-e^(-(R/L)t))` (when current is in growth in LR circuit) `(E)/(R)(1-e^(-(R/L)t))=5/5(1-e^(-(5/10)xx2))=(1-e^(-1))` |
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| 1076. |
A constant voltage is applied to a series `R-L` circuit by closing the switch. The voltage across inductor `(L=2H)` is `20V` at `t=0` and drops of `5V` and `20 ms`. The value of `R` in `Omega` isA. `100 In 2Omega`B. `100(1-In2)Omega`C. `100 In 4Omega`D. `100(1-In4)` |
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Answer» Correct Answer - C |
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| 1077. |
The series resonant circuit is called asA. acceptor circuitB. rejector circuitC. rectifier circuitD. transfer circuit |
| Answer» Correct Answer - A | |
| 1078. |
Resonant frequency is given by (for a series L-C-R circuit)A. `f=(1)/(2pisqrt(LC))`B. `f=(1)/(sqrt(LC))`C. `f=(2pi)/(sqrt(LC))`D. `f=2pisqrt(LC)` |
| Answer» Correct Answer - A | |
| 1079. |
At resonance series/parallel resonant circuit acts as purelyA. resistive circuitB. inductive circuitC. capacitive circuitD. none of these |
| Answer» Correct Answer - A | |
| 1080. |
In parallel resonant circuit current through condenser leads the source voltage4 byA. `0^(@)`B. `90^(@)`C. `170^(@)`D. `180^(@)` |
| Answer» Correct Answer - B | |
| 1081. |
In parallel resonant circuit, the current and voltage at resonance areA. both maximumB. both minimumC. maximum and minimum respectivelyD. minimum and maximum respectively |
| Answer» Correct Answer - D | |
| 1082. |
The current flowing in two branches of parallel resonant circuit at resonance areA. in phaseB. out of phase by `pi`C. differ in phase by `pi//2`D. differ in phase by `pi//4` |
| Answer» Correct Answer - B | |
| 1083. |
In parallel resonant circuit, at resonanceA. current is maximum and impedance is maximumB. current is maximum and impedance is minimumC. current is minimum and impedance is maximumD. current is minimum and pmpedance is minimum |
| Answer» Correct Answer - C | |
| 1084. |
A resistor, a capacitor and inductor are connected in series with a source of ac which of the following statement is true? The current in resistor lags behind theA. current in capacitorB. current in inductorC. voltage across capacitorD. voltage across inductor |
| Answer» Correct Answer - D | |
| 1085. |
In series resonant circuit, at resonance the phase difference between the voltage across inductor and voltage across condenser isA. `3pi//2`B. `pi//2`C. `0`D. `pi` |
| Answer» Correct Answer - D | |
| 1086. |
In the circuit of Fig. (1) and (2) are ammeters. Just after the key `K` is pressed to complete the circuit, the reading is A. Maximum in both 1 and 2B. Zero in both 1 and 2C. Zero in 1, minimum in 2D. Maximum in 1,zero in 2 |
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Answer» Correct Answer - D |
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| 1087. |
In a LCR series circuit, if the phase difference between applied voltage and current is zero thenA. `X_(L)=X_(C)`B. `X_(L) gt X_(C)`C. `X_(L) lt X_(C)`D. none of these |
| Answer» Correct Answer - A | |
| 1088. |
In series resonant circuit, at resonance,A. `Z=sqrt(R^(2)+(X_(L)-X_(C))^(2))`B. Z=RC. `Z=X_(L)-X_(C)`D. `Z=X_(C)` |
| Answer» Correct Answer - B | |
| 1089. |
The series `RLC` circuit in resonance is called:A. series resonantB. parallel resonantC. reactive circuitD. none of these |
| Answer» Correct Answer - A | |
| 1090. |
Three series capacitors of capacitances 2.0, 3.0 and `6.0 mu F` are charged by a 60 V vattery. Find the total energy stored. |
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Answer» Here, `C_(1) = 2.0 mu F, C_(2) = 3.0 mu F`, `C_(3) = 6.0 mu F, V = 60 V`, As `(1)/(C_(s)) = (1)/(C_(1)) + (1)/(C_(2)) + (1)/(C_(3)) = (1)/(C_(2)) + (1)/(C_(3)) + (1)/(6) = 1` `:. C_(s) = 1 mu F = 10^(-6) F` `U = (1)/(2) C_(s) V^(2) = (1)/(2) xx 10^(-6) (60)^(2) = 1.8 xx 10^(-3) J` |
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| 1091. |
In series RLC AC circuit, at resonance, the current is:A. Always in phase with the genrator voltage.B. Always lags the generator voltageC. Always leads the generator voltageD. May lead or lag behind the generator voltage. |
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Answer» Correct Answer - B |
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| 1092. |
Two capacitors of capacitances `2C` and `C` are connected in series with an inductor of inductance `L`. Initially, capacitors have charge such that `V_B - V_A = 4V_0` and `V_C- V_D = V_0`. Initial current in the circuit is zero. Find (a) maximum current that will flow in the circuit, (b) potential difference across each capacitor at that instant, (c) equation of current flowing towards left in the inductor. |
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Answer» Correct Answer - `(a) 2.40 A//s (b) 0.80 A//s (c )0.413 A (d) 0.750 A` |
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| 1093. |
An LCR circuit has L = 10 mH, `R = 3 Omega` and `C = 1 mu F` and is connected in series to a source of `(20 sin omega t)` volt. Calculate the current amplitude at a frequency 20 % lower than the resonance frequency of the circuit. |
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Answer» Here, `L = 10 mH = 10^(2) H, R = 3 Omega` `C = 1 mu F = 10^(-6) F`, `E = 20 sin omega t = E_(0) sin omega t` `:. E_(0) = 20 V, I_(0) = ?` Resonant frequency, `v_(0) = (1)/(2 pi sqrt(LC))` `= (1)/(2 pi sqrt(10^(-2) xx 10^(-6))) = (10^(4))/(2 pi)` Frequency at which current amplitude is required `= v = 80% v_(0) = (80)/(100) xx (10^(4))/(2 pi) = (8 xx 10^(-3))/(2 pi) Hz` `X_(L) = omega L = 2 pi v L` `= 2 pi xx (8 xx 10^(3))/(2 pi) xx 10^(-2) = 80 ohm` `X_(C ) = (1)/(omega C) = (1)/(2 pi v C)` `= (2 pi)/(2 pi xx (8 xx 10^(3)) xx 10^(-6))` `= (10^(3))/(8) = 125 ohm` `Z = sqrt(R^(2) + (X_(L) - X_(C ))^(2))` `= sqrt(3^(2) + (80 - 125)^(2)) = 45.1 ohm` `I_(0) = (E_(0))/(Z) = (20)/(45.1) = 0.443 A` |
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| 1094. |
An inductance of `0.4/(pi)` henry and a resistace of `30 Omega` are joined in series. If an alternating e.m.f. of 200 V, 50 Hz is applied to their combination, then the impedance of the circuit will beA. `50 Omega`B. `40 Omega`C. `100 Omega`D. `10 Omega` |
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Answer» Correct Answer - A `=sqrt(30^(2)+(2 pixx50xx4/(10pi))^(2)) = 50 Omega` |
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| 1095. |
A 70 turn coil with average diameter of 0.02 m is placed perpendicular to magnetic field of 9000 T. If the magnetic field is changed to 6000 T is 3s, what is the magnitude of induced emf. |
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Answer» Correct Answer - 22 V |
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| 1096. |
If a diamagnetic solution is poured into a U-tube and one aem of this U-tube placed between the poles of a strong magnet with the meniscus in a line with the field, then the level of the solution willA. will riseB. will fallC. will oscillateat constant amplitudeD. remain as it is |
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Answer» Correct Answer - A |
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| 1097. |
Obtain the expression for the magnetic energy stored in a solenoid in terms of magnetic field B, area A and length l of the solenoid. |
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Answer» From Equation `epsi=-L (Di)/(dT)` `U_(B)=1/2LI^(2)=1/2L((B)/(mu_(0)n))^(2) (since B=mu_(0)nI "for a solenoid")` `=1/2(mu_(0)n^(2)Al)((B)/(mu_(0)n))^(2) ["from Equation L="mu_(0)n^(2)Al]` `=1/(2mu_(0))B^(2)Al` |
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| 1098. |
An air- conred solectioid with length 30 cm,area of cross-section `25 cm^(2)` and number of turns 500, carries a curent is suddenly switched off in a brief time of `10^(-3)s`. How much is the average back emf induced across the ends of the opne switch in the circuit ? lgnore the variation inmagnetic field near the of the solenoid .A. 6.5 VB. 7.4 VC. 8.2 VD. 9.3 V |
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Answer» Correct Answer - A Given , length of solenoid l = 30 cm = `30 xx 10^(-2)` m Area of cross-section `A = 25 cm^(2) = 25 xx 10 ^(-4)m^(2)` Number of turns, N = 5000 Current ,`l_(1)= 21.5 A, l_(2) = 0` Brief time `dt = 10 ^(-3) s` Induced emf in the solenoid `e= (dphi)/(dt)= (d)/(dt)(BA)" "[thereforephi= BA]` Magnetic field induction B at point well inside the long solenoid carrying current l is `B= mu_(0) nl("where, n= number of turns per unit lenght "=(N)/(l))` `e = NA(dB)/(bt) = A(d)/(dt)(mu_(0)(N)/(l))= A (mu_(0)N)/(l).(dl)/(dt)` `e = 500 xx 25 xx 10^(-4) xx 4 xx 3.14 xx 10 ^(-7)xx(500)/(30 xx 10^(-2))xx(2.5)/(10^(-3))` |
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| 1099. |
At t = 0, switch S is closed, calculate (i) initial rate of increase of current, i.e., `(di)/(dt)"at t"=0`. (ii) `(di)/(dt)` at time when current in the circuit is `0.5A`. (iii) Current at t = `0.6` s. (iv) rate at which energy of magnetic field is increasing, rate of heat produced in resistance and rate at which energy is supplied by battery when i = 0.5 A. (v) energy stored in inductor in steady state. |
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Answer» We know that `E=iR+"L dt/dt"` `(dt)/(dt)=(E-iR)/(L)` Current in the circuit at any time t, `i=i_(0)(1-e^(-t//tau_(L)))` where, `i_(0)=E//R,tau_(L)=L//R` (i) Rate of increase of current `(di)/(dt)=(E-iR)/(L)`, when t = 0, `i` = 0 `|(di)/(dt)|_(t=0)=(E)/(L)=(12)/(2)=6A//s` (ii) At `t=0.5s,(di)/(dt)=(E-iR)/(L)=(12-0.5xx4)/(2)=(12-2)/(2)=5A//s` (iii) At t = `0.6` s, current i = `(E)/(R)(1-e^(-(Rt)/(L)))` `=(12)/(4)(1-e^((4xx0.6)/(0.5)))=3(1-e^(2.4))A` (iv) Energy stored in inductor in steady state is given by `U=(1)/(2)Li^(2)rArr(dU)/(dt)=Li(di)/(dt)` `(di)/(dt)=(E-iR)/(L)=(12-0.5xx4)/(2)=5A//s` `(dU)/(dt)=2xx0.5xx5=5J//s` Power produced per second, `P=i^(2)R=(0.5)^(2)xx4=1J//s` Power supplied by battery `=Ei=0.5xx12=6J//s` (v) We have `i_(0)=E//R=12//4=3A` Energy stored in inductor in steady state, `U=(1)/(2)Li_(0)^(2)=(1)/(2)xx2xx(3)^(2)=9J//s` |
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| 1100. |
An L-R circuit has a cell of e.m.f. E , which is switched on at time t = 0. The current in the circuit after a long time will beA. zeroB. `(E)/(R)`C. `(E)/(L)`D. `(E)/(sqrt(L^(2)+R^(2)))` |
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Answer» Correct Answer - B In case of growth of current in a L-R circuit, the current in the circuit grows exponentially with time 0 to the maximum value at, `i_(0)=E//R`. |
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