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QUESTION: 1

An LCR series circuit with 100Ω resistance is connected to an AC source of 200V and angular frequency 300 radians per second. When only the capacitance is removed, the current lags the voltage by 60°. When only the inductance is removed, the current leads the voltage by 60°. Then the current and power dissipated in LCR circuit are respectively.

Solution:

When capacitance is removed

when inductance is removed

From equation (1) and (2)

So it is condition of resonance.

So Z = R = 100Ω

I = *V*/*R* = 200/100= 2*A*

Power P = *I*^{2}*R* = 4 ×100 = 400 *W*

The correct answer is: 2*A*, 400*Watt*.

QUESTION: 2

In the circuit diagram shown, *X _{C}*

Solution:

QUESTION: 3

An ac source of angular frequency **ω** is fed across a resistor * R *and a capacitor

Solution:

According to given problem,

Substituting the value of I from equation (1) in (2)

QUESTION: 4

Current in an** **ac circuit is given by i = 3sin ωt + 4cos ωt, then :

Solution:

∴ Initial value of time is not given hence the mean value will be different for various time intervals.

If voltage applied is V = V_{m} sint ωt then *i* given by equation (1) indicates that it is ahead by *V* by δ where 0 < δ < 90 which indicates that the circuit contains *R* and *C*.

The correct answer is: if voltage applied is **V= V _{m}sin ωt**, then the circuit must be containing resistance and capacitance

QUESTION: 5

A current source sends a current i = i_{0} cos ωt. When connected across an unknown load gives a voltage output of across that load. Then voltage across the current source may be brought in phase with the current through it by:

Solution:

Since V lags current, an inductor can bring it in phase with current.

The correct answer is: connecting an inductor in parallel with the load

QUESTION: 6

In series *LCR* circuit voltage drop across resistance is 8*V**olt*, across inductor is 6*V**olt* and across capacitor is 12*V**olt*. Then:

Solution:

Since, (Also cos*θ* can never be greater than 1)

∴ Current will be leading

In a *LCR* circuit

*V* = 10; which is less than voltage drop across capacitor.

The correct answer is: none of these

QUESTION: 7

For a *LCR* series circuit with an *AC* source of angular frequency ω

Solution:

The circuit will have inductive nature if

Also if circuit has inductive nature will lag behind voltage.

the circuit will have resistance nature.

Power factor

Hence true.

The correct answer is: power factor of circuit will be unity if capacitive reactance equals inductive reactance

QUESTION: 8

The secondary coil of an ideal step down transformer is delivering 500W power at 12.5*A* current. If the ratio of turns in the primary to the secondary is 5 : 1, then the current flowing in the primary coil will be :

Solution:

P = VI

For an ideal transformer (100% efficient)

The correct answer is: **2.5A**

QUESTION: 9

A bulb is rated at 100*V*, 100W*,* it can be treated as a resistor. Find out the inductance of an inductor (called choke coil) that should be connected in series with the bulb to operate the bulb at its rated power with the help of an ac source of 200*V* and 50*Hz*.

Solution:

From the rating of the bulb, the resistance of the bulb can be calculated

For the bulb to be operated at its rated value the *rms* current through it should be 1*A
*

QUESTION: 10

In a black box of unknown elements (L, C or R or any other combination) an AC voltage is applied and current in the circuit was found to be Then the unknown elements in the box may be:

Solution:

If we have all *R*, *L* and *C* then *I* vs, *E* will be :

To obtain a leading phase difference of

if **X _{L} < X_{C}** and we use all

then the resultant graph will be :

which can give a leading phase difference of π/4.

Similarly if we have only resistance and capacitor then we can obtain a phase difference of π/4.

(leading) for suitable values of *I*, *X _{C}* and

The correct answer is: either capacitor, resistor and inductor or only capacitor and resistor

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