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 (c) current i₁ < i₂

(a) The rms current is 2.8A

(d) Current i can be used where i is about 10A.

(c) The rms current is given by √((i²₁ + i²₂)/2)

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Correct option is (A) Statement-I is true, statement-II is true and statement-II is correct explanation for statement-I. 

When circuit is suddenly switched off, there will be a change in current, and it will lead to induced EMF.

\(|E| = L \left|\frac{di}{dt}\right|\)

Now for large ‘L’, E is also high.

A capacitor acts as an infinite resistance for (a) DC.

(1) In resistor, current is in phase with the voltage. 

(2) In inductor, current lags the applied voltage by 90° or π/2 

(3) In capacitor, current leads the applied voltage by 90° or π/2. 

 the value of X_L is (c) – X_g. 

(c) the meter reads not v but <v²> and is calibrated to read √< v² > . 

(a) The impedance of the circuit is 5Ω

Graph d represent the reactance of a series LC combination.

(c), (d)

(c)  Resistor and a capacitor. 

(d) Only a capacitor.

(a), (b), (d)

(a) For a given power level, there is a lower current. 
(b) Lower current implies less power loss. 

(d) It is easy to reduce the voltage at the receiving end using step-down transformers.

(a), (b), (c)

(a) Here, the power factor cos φ ≥ 0, P ≥ 0. 

(b) The driving force can give no energy to the oscillator (P = 0) in some cases. 

(c) The driving force cannot syphon out (P < 0) the energy out of oscillator. 

(a) Yes, if rms voltage in the two circuits are same then at resonance, the rms current in LCR will be same as that in R circuit.

(b) No, because  R ≤ Z, so I_a ≥ I_b.

Yes, 

No, The average power output can not be negative.

200 V, 50 Hz.

(a) A
At resonance condition X_L = X_C
∴ Point ‘A’ shows condition of resonance

(d) there may be minimum energy loss due to eddy currents

To reduce the eddy currents.

Most of the transformer have efficiency of 90-99%. This means the wastage of the power is negligible. The above analysis was for an ideal transformer. In a real transformer the energy losses are due to various reasons as detailed below.

1. Leakage of magnetic flux: Due to the air gaps in the core the amount of magnetic flux linked with the secondary is less than that of the primary as the energy is lost. It can be reduced by winding the primary and secondary coils over one another.

2. Iron loss: This is the energy lost as heat due to induced eddy currents in the iron core (alternating magnetic flux induces these currents). This loss is also called the eddy current loss. This is reduced by making core of laminated sheets of soft iron insulated from each other.

3. Copper loss: It is the heat energy lost across the resistance of copper windings used in primary and secondary coils. This loss can be minimised by using thick copper wire for the primary and secondary windings.

4. Hysteresis loss: This loss is due to heating up of core due to its repeated cycles of magnetisation and demagnetisation when an alternating emf is applied across the primary. This loss can be minimised by selecting a core material having a narrow hysteresis loop.

5. Magnetostriction: It is the loss due to the humming sound produced in the transformer.

The type of this transformer is Step up.

At the break, a large induced emf is produced. In case capacitor is not connected, sparking will take place. But when capacitor is used, the large induced emf produced at break is used up in charging the capacitor and no sparking takes place.

Yes, the applied instantaneous voltage is equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit. It is because voltages across different elements are not in phase.
It is not true for rms voltages. It is because rms voltages across different elements are not in phase with each other.

(a) Yes. The same is not true for rms voltage, because voltage across different element may not be in phase. 

(b) The high induced voltage, when the circuit is broken, is used to change the capacitor, thus avoiding sparks, etc. 

(c) For dc, impedance of L is negligible and C very high (infinite), so the D.C. signal appears across C. For frequency ac, impedance of L is high and that of C is low. So, the A.C. signal appears across L. 

(d) A choke coil reduces voltage across the tube without wasting power. A resister would waste power as heat.

Given L = 2.0 H, C = 32µF = 32 × 10^-6 F

R = 10Ω, Q = ?, ω₀ = ?

Resonant frequency

ω₀ = \(\frac{1}{\sqrt{LC}}\)

or ω₀ = \(\frac{1}{\sqrt{2\times32\times10^{-6}}}\) = 125s^-1

Q -factor, Q = \(\frac{1}{R}\sqrt\frac{L}{C}=\frac{1}{10}\sqrt\frac{2.0}{32\times10^{-6}}\) = 25

(i) Current lags by \(\frac{\pi}{2}\)

(ii) X_c = \(\frac{1}{cω}\)

(iii) R

(iv) Phase difference is zero.

(b) Depends on phase angle between V and I

(i) Curiosity.

(ii) Induced current in the aluminium ring acts in the opposite direction to than in the coil and so magnetic field of the ring repels the magnetic field due to the coil as a result of which, the ring shoots up in air.

Frequency of the alternating current is 50 Hz.

(d) increases and becomes inductive
In the condition of resonance Z = R, so impedance is not dependent on the frequency.

(c) pure capacitor
Current leads the voltage in pure capacitor circuit.

(b) zero
Power in circuit (P) = V_rms I_rms cosϕ
= V_rms I_rms Cos 90° = 0

(i) f = f_r when X_L = X_C
The circuit is purely resistive i.e., it is non-inductive. Therefore, current and voltage in the circuit are in the same phase.
(ii) f < f_r as X_L = ωL = 2πfL and X_C = \(\frac{1}{\omega C}=\frac{1}{2 \pi f C}\)
∴ When f is small, X_L is small, X_C is large. The circuit is capacitance dominated. Therefore, current in the circuit leads the voltage by phase angle ϕ.
(iii) f > f_r
So, X_L becomes large and X_C becomes small. The circuit is inductance dominated. The current lags behind the voltage by phase angle ϕ.

We know that power loss
P = I_rms V_rms cosϕ,
Small value of ϕ i. e., cos ϕ = more
∴ Power loss is more.

The phase difference between voltage of inductance and capacitance in a series LCR resonance circuit is 180°.

Between 0 to ± \(\frac{\pi}{2}\).

Power factor in circuit
cosϕ = \(\frac{R}{Z}\)
When R = Z
∴ cos ϕ = \(\frac{R}{Z}\) = 1

The reactance is 31.4 Ω.

Zero, zero and one.

(c)  X/4

Explanation :

X = \(\frac{1}{C_ω},X' =\frac{1}{4Cω}\)

∴  X' =  \(\frac{X}{4}\)

Resonant frequency = 39.79 Hz 

(i) 2000 Ω 

(ii) 100 Ω 

(iii) 2A 

(iv) 0 90 

(v)1800