An alternating current can be defined as a current that changes its magnitude and polarity at regular intervals of time. It can also be defined as an electrical current that repeatedly changes or reverses its direction opposite to that of a Direct Current. Alternating current can be produced or generated by using devices that are known as alternators. AC is the form of current that are mostly used in different appliances

**Question 1) The inductance of two LR circuits are placed next to each other, as shown in the figure. The value of the self-inductance of the inductors, resistance, mutual-inductance and applied voltages are specified in the given circuit. After both the switches are closed simultaneously the total work done by the batteries against the induced EMF in the inductors by the time the currents reach their steady-state values is ______ mJ.**

**Answer: 55 mJ**

**Solution:**

Given,

Mutual Inductance, M = 5mH

L_{1} = 10 mH

V_{1} = 5 V

L_{2} = 20 mH

V_{2} = 20 V

I_{1} = V_{1}/R_{1} = 5/5 = 1A

I_{2}= V_{2}/R_{2} = 20/10 = 2A

After both the switches are closed simultaneously, the total work done by the batteries against the induced EMF = Increase in the magnetic energy

Therefore, W = Î”U = (Â½) L_{1}I_{1}^{2} + (Â½) L_{2}I_{2}^{2 }+ MI_{1}I_{2 }

= (Â½)(10 x 10^{-3})1^{2} + (Â½)(20 x 10^{-3})(2^{2}) + (5 x 10^{-3}) x 1 x 2

= (5 + 40 + 10) x 10^{-3}

= 55 mJ

**Question 2) A 20 Henry inductor coil is connected to a 10 ohm resistance in series as shown in the figure. The time at which rate of dissipation of energy (Jouleâ€™s heat) across the resistance is equal to the rate at which magnetic energy is stored in the inductor, is**

(A) 2/ln 2

(B) (Â½) ln2

(C) 2ln 2

(D) ln 2

Answer: (C) 2ln 2

Solution:

i = i_{0} (1 – e^{-t/Ï„})

di/dt = (i_{0}/Ï„) e^{-t/Ï„}

di/dt = [E/(10 x 2)] e^{-t/2 }——–(1) [since Ï„ = L/R = 20/10 = 2]
[L(di/dt)i] = i^{2}R

â‡’ L(di/dt) = iR ———(2)

From equation (1) and (2) we get

L [E/(10 x 2)] e^{-t/2 }= iR

(L/R)[E/(20)] e^{-t/2 }= i_{0} (1 – e^{-t/Ï„}) [since Ï„ = L/R = 20/10 = 2]
[E/10] e^{-t/2 }= [E/10) (1 – e^{-t/Ï„})

e^{-t/2 }= Â½

t/2 = ln 2

**Question 3) In the figure shown, a circuit contains two identical resistors with resistance R = 5 ohm and inductance with L = 2 mH. An ideal battery of 15 V is connected in the circuit. What will be the current through the battery long after the switch is closed?**

(A) 5.5 A

(B) 7.5 A

(C) 3 A

(D) 6 A

**Answer: (D) 6 A**

**Solution:**

For a long time after the switch is closed, the inductor will be idle so the equivalent diagram will be

I = Îµ/[(R x R)/(R + R)]

= 2Îµ/R

= (2 x 15)/5

= 6 A

**Question 4)An AC voltage source of variable angular frequency and fixed amplitude V _{0} is connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When Ï‰ is increased **

(A) the bulb glows dimmer

(B) the bulb glows brighter

(C) total impedance of the circuit is unchanged

(D) total impedance of the circuit increases

**Answer: (B) the bulb glows brighter **

**Solution:**

Impedance

As Ï‰ increases, Z decreases

Hence, the bulb will glow brighter

**Question 5) A series R-C combination is connected to an AC voltage of angular frequency Ï‰=500 radian/s. If the impedance of the R-C circuit is Râˆš1.25 the time constant (in millisecond) of the circuit is?**

(A) 1

(B) 2

(C) 3

(D) 4

Answer: (D) 4

Solution:

Given, Ï‰=500 radian/s

The capacitance of the capacitor is C

X_{c} = 1/Ï‰C = 1/500C

Impedance of the circuit, Z = Râˆš1.25

Using Z^{2} = R^{2} + X_{c}^{2}

1.25R^{2} = R^{2} + 1/(500)^{2}C^{2}

0.25R^{2} = 1/(0.25 x 10^{6})C^{2}

R^{2}C^{2} = 10^{-6}/(0.25)^{2}

â‡’ RC = 10^{-3}/0.25

= 0.004 s

= 4 ms

**Question 6) In the given circuit, the AC source has Ï‰ = 100 rad/s. Considering the inductor and capacitor to be ideal, the correct choice (s) is(are) **

(A) The current through the circuit, I is 0.3 A.

(B) The current through the circuit, is 0.3 âˆš2A

(C) The voltage across 100Î© resistor 10âˆš2 V

(D) The voltage across 50Î© resistor 10 V

**Answer: (A) and (C)**

**Solution:**

In the upper branch the net impedance, (C = 100Î¼F, R = 100Î©)

Therefore, the net impedance will be

= 100âˆš2

Current I_{1} = V/Z_{1} = 20/(100âˆš2)

cos Î¦_{1} = R/(100âˆš2)

= 100/(100âˆš2)

= 1/âˆš2

â‡’Î¦_{1} = 45^{0}

In the lower branch the net impedance (L = 0.5 H, R = 50 Î©)

Therefore, the net impedance will be

= 50âˆš2

Current, I_{2 }= V/Z_{2}

= 20/(50âˆš2)

cos Î¦_{1} = R/(50âˆš2)

= 50/(50âˆš2)

= 1/âˆš2

â‡’Î¦_{1} = 45^{0}

Thus the total current I is given by the summation of I_{1} and I_{2 }which differ by 90^{0} in phase and hence

= (1/âˆš10) A â‰ˆ 0.3 A

Voltage across 100 Î© = I_{1}R_{1} = [20/(100âˆš2)] x 100 = 10âˆš2 V

Voltage across 50 Î© = I_{2}R_{2} = [20/(50âˆš2)] x 50 = 10âˆš2 V

**Question 7) An alternating voltage v(t) = 220 sin 100 Ï€t volt is applied to a purely resistive load of 50 Î©. The time taken for the current to rise from half of the peak value to the peak value is**

(A) 5 ms

(B) 2.2 ms

(C) 7.2 ms

(D) 3.3 ms

**Answer: (D) 3.3 ms**

**Solution:**

Given,

V(t) = 220 sin 100Ï€t

I(t) = (220/50) sin 100Ï€t

Phase to be covered Î¸ = 60^{0} = Ï€/3

Time taken, t = Î¸/Ï‰

= (Ï€/3)/100Ï€

= (1/300) sec

= 3.3 ms

**Question 8) A sinusoidal voltage V(t) = 100 sin (500t) is applied across a pure inductance of L= 0.02 H. The current through the coil is**

(A) 10 cos (500 t)

(B) -10 cos (500 t)

(C) 10 sin (500 t)

(D) – 10 sin (500 t)

**Answer: (B) -10 cos (500 t)**

**Solution:**

In a pure inductive circuit current always lags behind the emf by Ï€/2

If v(t) = v_{0}sinÏ‰t

Then, I(t) = I_{0}sin(Ï‰t – Ï€/2)

Given, V(t) = 100 sin (500t)

And I_{0} = E_{0}/Ï‰L

= 100/ (500 x 0.02) = 10sin(500t – Ï€/2)

I_{0} =- 10 sin (500 t)

**Question 9) An AC circuit has R = 100 Î©, C= 2Î¼F and L= 80 mH, connected in series. The quality factor of the circuit is**

(A) 2

(B) 0.5

(C) 20

(D) 400

**Answer: (A) 2**

**Solution:**

Quality factor,

= 200/100 = 2

**Question 10) When the rms voltage V _{L}, V_{C} and V_{R} are measured respectively across the inductor L, the capacitor C and the resistor R in a series LCR circuit connected to an AC source, it is found that the ratio V_{L}: V_{C}: V_{R} = 1: 2: 3. If the rms voltage of the AC sources is 100 V, The V_{R} is close to**

(A) 50 V

(B) 70 V

(C) 90 V

(D) 100 V

**Answer: (C) 90 V**

**Solution:**

Given,

V_{L:} V_{C}: V_{R} = 1: 2: 3

V = 100 V

â‡’ V_{R} = 3K, V_{L} = K, V_{C} = 2K

We know,

100 = âˆš10 K

K =100/âˆš10

V_{R} = 3K = (3 x 100)/âˆš10

= 94.86 volts

So, V_{R} is close to 90 V

**Also Read:**

Alternating Current JEE Main Previous Year Questions With Solutions