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1551.

A coil of inductance 8.4 mH and resistance `6 (Omega)` is connected to a 12 V battery. The current in the coil is 1.0 A at approximately the timeA. 500sB. 25 sC. 35 sD. 1 ms

Answer» Correct Answer - D
Using `I=(I_0)(1-e^(-t//tau))`
But`(I_0)=V/R and (tau)=L/R`
`:. I=V/R (1-e^(-t//tau))=12/6[1-e^(-6t//8.4xx(10^3))]`
`=1(given)`
`:. T=0.97xx(10^(-3))s~~1 ms`.
Discussion
1552.

A bulb B a capacitor C are connected in series to an a.c. source. A dielectric slab is now introduced between the plates of the capacitor. How will the brightness of bulb change ?

Answer» Current through the bulb is
`I_(v) = (E_(v))/(Z) = (E_(v))/(sqrt(2) + (1)/(omega^(2) C^(2)))` On introducing a dielectric slab between the plates of capacitor, C increases. Z decreases, `I_(v)` increases. Therefore, glow of the bulb increases.
Discussion
1553.

A parallel plate capacitor having plate area `0.5 m^(2)` and plate separation of 5 mm is complete filled with a dielectric of dielectric constant 10. What is the instantaneous displacement current, if it is being charged at the rate of `100 V//s`? `[epsilon_(0)=8.85 xx 10^(-12)C^(2)N//m^(2)]`A. `8.85 mu A`B. `0.885 mu A`C. `0.177 mu A`D. `1.77 mu A`

Answer» Correct Answer - B
The capacity of the parallel plate capacitor is given by
`C = (Kaepsilon_0)/d = (10 xx 5 xx 10^(-1) xx 8.85 xx 10^(-12))/(5 xx 10^(-3))`
`=8.85 xx 10^(-9) F`
`:.` Displacement current `I_(d) = C(dV)/(dt)`
`=8.85 xx 10^(-9) xx 100`
`=8.85 xx 10^(-7)A`
`=0.885 xx 10^(-6) A`
`:. I_(d) = 0.88r muA`.
Discussion
1554.

The dielectric strength of air is `3.0 xx 10^6 V/m`. A parallel-plate air-capacitor has area `20 cm^2` and plate separation `0.10 mm`.Find the maximum rms voltage of an AC source which can be safely connected to this capacitor.

Answer» `:.` Maximum voltage,`V_(0) =E xx d`
`= 3 xx 10^(5) xx 10^(-3) = 300 V`
Maximum r.m.s. voltage ` = (v_(0))/(sqrt2) = (300)/(1.414)`
`= 212.2 V`
Discussion
1555.

Why does a metallic piece become very hot when it surrounded by a carrying high frequency alternaiting current ?

Answer» High frequency alternating current passed through the coil surrounding the matal piece porduces eddy currents in the meta piece. The eddy currents produce joule heating in the metal piece on account of its resistance.
Discussion
1556.

The radius of the circular conducting loop shown in is R. magnetic field is decreasing at a constant rate `alpha`. Resisitance per unit length of the loop is `rho`. Then, the current in wire `AB` is (`AB` is one of the diameters) A. `(R alpha)/(2 rho)` from A to BB. `(R alpha)/(2 rho)` from B to AC. `(2 R alpha)/(rho)` from A to BD. zero

Answer» Correct Answer - D
Discussion
1557.

A sinusoidal voltage of amplitude 5V is applied to resistance of `500 Omega`. The r.m.s. current in the circuit isA. A. `5/(sqrt(2)) mA`B. B. `10/(sqrt(2)) mA`C. C. `10sqrt(2) mA`D. D. `20sqrt(2) mA`

Answer» Correct Answer - B
`I_(0)=5/500 = 1/100 A = 10 mA :. I_(rms) = 10/(sqrt2) mA`.
Discussion
1558.

A Sinusoidal voltage V = 200 sin 314 t r is applied to a resistor of 10 ohm. Calculate (i) rms value of voltage (ii) rms current (iii) power dissipated as heat.

Answer» `V = 200 sin 314 t, R = 10 Omega`
Compare with `V = E_(0) sin omega t = E_(0) sin 2 pi v t`
`E_(0) = 200 V, E_(v) = (E_(0))/(sqrt2) = (200)/(1.414) = 141.44 V`
`I_(v) = (E_(v))/(R ) = (1.414)/(10) = 14.414 A`
`P = E_(v) I_(v) = 141.144 xx 14.414 = 2000.5 V`
Discussion
1559.

Two sinusoidal current are given by `i_(1)=20 sin (omega t +(pi)/6) and i_(2)=10 sin(omega t-(pi)/4)` The pahase difference between them isA. a. `30^(@)`B. b. `60^(@)`C. c. `75^(@)`D. d. `90^(@)`

Answer» Correct Answer - C
Phase difference = `(pi)/6 -(-(pi)/(4))= (5 pi)/12 = 75^(@)`.
Discussion
1560.

An alternating current of rms value 5A, passes through a resistance of `24 Omega`. What is the maximum P.D. across the resister?A. 17 VB. 34 VC. 170 VD. 10 V

Answer» Correct Answer - C
Max P.D. = `I_(0) xx R = I_(rms) xx sqrt(2) xxR`
`=5sqrt(2) xx 24 = 170 V`.
Discussion
1561.

The inductive reactance of a choke coil of `1/(4 pi) mH` in an A.C. Circuit of frequency 50 Hz isA. `25 Omega`B. `2.5 Omega`C. `0.025 Omega`D. `0.25 Omega`

Answer» Correct Answer - C
`X_(L) = omega L = 2 pi fL = 0.025 Omega`
Discussion
1562.

An alternating voltage `e=220sqrt(2) sin (100t)` is connected to `4 mu F` capacitor and a ammeter. The ammeter will readA. 11 mAB. 22 mAC. 44 mAD. 88 mA

Answer» Correct Answer - D
`E_(rms) = (E_0)/(sqrt2) = 220 V and omega = 100`
`I_(rms) = (E_(rms))/(1//omegaC) = E_(rms) xx omega C`
`:. I_(rms) = 220 xx 100 xx 4 xx 10^(-6) = 88 xx 10^(-3) = 88 mA`.
Discussion
1563.

The valules of resistance and inductive reactance of a choke coil are `8 Omega` and `6 Omega` respectively. What is the power factor of the coil?A. A. 0.2B. B. 0.4C. C. 0.6D. D. 0.8

Answer» Correct Answer - D
For an L-R circuit
Power factor `(cos theta) = (R)/(sqrt(R^(2)+L^(2)omega^(2)))`
`8/(sqrt(64+36)) = 8/10 = 0.8`.
Discussion
1564.

In an A.C. circuit I=10 cos(100 t) ampere and `V=20 sin(100t). The power loss in the circuit will beA. 20 wattB. 200 wattC. 0 wattD. 50 watt

Answer» Correct Answer - C
`I = 10 cos (100 t) = 10 sin (100t+(pi)/2)`
There is a phase diff. Of `(pi)/2` between I and V.
`:.` Power factor = `cos theta =0 :.` Power less =0.
Discussion