HC Verma Questions and Solutions: Chapter 39: Alternating Current- 2 | HC Verma Solutions - JEE PDF Download

Q.1. Find the time required for a 50 Hz alternating current to change its value from zero to the rms value.

Frequency of alternating current, f = 50 Hz
Alternation current ( i ) is given by,
 i = i₀sinωt   ...(1)
Here, i₀ = peak value of current
Root mean square value of current (i_rms) is given by,

On substituting the value of the root mean square value of current in place of alternating current in equation (1), we get:

= 2.5 ms

Q.2. The household supply of electricity is at 220 V (rms value) and 50 Hz. Find the peak voltage and the least possible time in which the voltage can change from the rms value to zero.

RMS value of voltage, Erms = 220 V,
Frequency of alternating current, f = 50 Hz
(a) Peak value of voltage (E₀) is given by,

where E_rms = root mean square value of voltage

(b) Voltage (E) is given by,
E = E₀sin ωt,
where E₀ = peak value of voltage
Time taken for the current to reach zero from the rms value = Time taken for the current to reach the rms value from zero
In one complete cycle, current starts from zero and again reaches zero.
So, first we need to find the time taken for the current to reach the rms value from zero.
As E = E₀/√2



⇒ t = 2.5 ms
Thus, the least possible time in which voltage can change from the rms value to zero is 2.5 ms.

Q.3. A bulb rated 60 W at 220 V is connected across a household supply of alternating voltage of 220 V. Calculate the maximum instantaneous current through the filament.

Power of the bulb, P = 60 W
Voltage at the bulb, V = 220 V
RMS value of alternating voltage, E_rms = 220 V
P = V²R,
where R = resistance of the bulb

= 806.67
Peak value of voltage (E_0) is given by,

Q.4. An electric bulb is designed to operate at 12 volts DC. If this bulb is connected to an AC source and gives normal brightness, what would be the peak voltage of the source?

Voltage across the electric bulb, E = 12 volts
Let E₀ be the peak value of voltage.
We know that heat produced by passing an alternating current ( i ) through a resistor is equal to heat produced by passing a constant current (i_rms) through the same resistor. If R is the resistance of the electric bulb and T is the temperature, then





=16.97 = 17 ⇒ V
Thus, peak value of voltage is 17 V.

Q.5. The peak power consumed by a resistive coil, when connected to an AC source, is 80 W. Find the energy consumed by the coil in 100 seconds, which is many times larger than the time period of the source.

Peak power of the resistive coil, P0 = 80W
Time, t = 100 s
RMS value of power (P_rms) is given by,

where P₀ = Peak value of power
 
Energy consumed (E) is given by,
is given by,
E = P_rms × t
= 40 × 100
= 4000 J = 4.0 kJ

Q.6. The dielectric strength of air is 3.0 × 10⁶ V/m. A parallel-plate air-capacitor has area 20 cm² and plate separation 0.10 mm. Find the maximum rms voltage of an AC source that can be safely connected to this capacitor.

Given:
Area of parallel-plate air-capacitor, A = 20 cm²
Separation between the plates, d = 0.1 mm
Dielectric strength of air, E= 3 × 10⁶ V/m
E = V/d,
where V = potential difference across the capacitor

Thus, peak value of voltage is 300 V.
Maximum rms value of voltage (V_rms) is given by,

Q.7. The current in a discharging LR circuit is given by i = i₀ e^−t/τ , where τ is the time constant of the circuit. Calculate the rms current for the period t = 0 to t = τ.

As per the question,

We need to find the rms current. So, taking the average of i within the limits 0 to τ and then dividing by the given time period τ, we get:

Q.8. A capacitor of capacitance 10 μF is connected to an oscillator with output voltage ε = (10 V) sin ωt. Find the peak currents in the circuit for ω = 10 s⁻¹, 100 s⁻¹, 500 s⁻¹ and 1000 s⁻¹.

Capacitance of the capacitor, C = 10 μF = 10 × 10⁻⁶ F = 10⁻⁵ F
Output voltage of the oscillator, ε = (10 V)sinωt
On comparing the output voltage of the oscillator with
ε = ε₀, we get:
Peak voltage ε₀ = 10 V
For a capacitive circuit,
Reactance,
Here, ω = angular frequency
C = capacitor of capacitance
 Peak current,
(a) At ω = 10 s⁻¹:
Peak current,



= 1 × 10⁻³ A
(b)  At ω = 100 s⁻¹:
Peak current,

= 0.01 A
(c) At ω = 500 s⁻¹:
Peak current, I₀ =


(d) At ω = 1000 s⁻¹:
Peak current,

Q.9. A coil of inductance 5.0 mH and negligible resistance is connected to the oscillator of the previous problem. Find the peak currents in the circuit for ω = 100 s⁻¹, 500 s⁻¹, 1000 s⁻¹.

Given:
Inductance of the coil, L = 5.0mH = 0.005H
(a) At  ω = 100 s⁻¹:
Reactance of coil (X_L) is given by,


peak current,
(c) ω = 1000 s⁻¹:

Q.10. A coil has a resistance of 10 Ω and an inductance of 0.4 henry. It is connected to an AC source of 6.5 V, (30/π)Hz. Find the average power consumed in the circuit.

Given:
Resistance of coil, R = 10 Ω
Inductance of coil, L = 0.4 Henry
Voltage of AC source, Erms = 6.5 V
Frequency of AC source,
Reactance of resistance-inductance circuit (Z) is given by,

Here, R = resistance of the circuit
X_L = Reactance of the pure inductive circuit

Average power consumed in the circuit (P) is given by,

Q.11. A resistor of resistance 100 Ω is connected to an AC source ε = (12 V) sin (250 π s⁻¹)t. Find the energy dissipated as heat during t = 0 to t = 1.0 ms.

Given:
Peak voltage of AC source, E0 = 12 V
Angular frequency, ω = 250πs−1
Resistance of resistor, R = 100 Ω
Energy dissipated as heat (H) is given by,

Here, Erms = RMS value of voltage
R = Resistance of the resistor
T = Temperature
Energy dissipated as heat during t = 0 to t = 1.0 ms,

Q.12. In a series RC circuit with an AC source, R = 300 Ω, C = 25 μF, ε₀ = 50 V and ν = 50/π Hz. Find the peak current and the average power dissipated in the circuit.

Given:
Resistance of the series RC circuit, R = 300 Ω
Capacitance of the series RC circuit, C = 25 μF
Peak value of voltage, ε₀ = 50 V
Frequency of the AC source, ν = 50/π Hz
Capacitive reactance (Xc) is given by,

Here, ω = angular frequency of AC source
C = capacitive reactance of capacitance

Net reactance of the series RC circuit

(a) Peak value of current (I₀) is given by,

(b) Average power dissipated in the circuit (P) is given by,

 

Q.13. An electric bulb is designed to consume 55 W when operated at 110 volts. It is connected to a 220 V, 50 Hz line through a choke coil in series. What should be the inductance of the coil for which the bulb gets correct voltage?

Power consumed by the electric bulb, P = 55 W
Voltage at which the bulb is operated, V= 110 V
Voltage of the line, V = 220 V
Frequency of the source, v = 50 Hz
P = V²/R
where R = resistance of electric bulb
∴ R = V²/P

If L is the inductance of the coil, then total reactance of the circuit (Z) is given by,

Here, ω = angular frequency of the circuit
Now, current through the bulb, I = V/Z
∴ Voltage drop across the bulb, V =
As per question,

Q.14. In a series LCR circuit with an AC source, R = 300 Ω, C = 20 μF, L = 1.0 henry, ε_rms = 50 V and ν = 50/π Hz. Find (a) the rms current in the circuit and (b) the rms potential difference across the capacitor, the resistor and the inductor. Note that the sum of the rms potential differences across the three elements is greater than the rms voltage of the source.

Given:
Resistance in series LCR circuit, R = 300 Ω
Capacitance in series LCR circuit, C = 20 μF= 20 × 10⁻⁶ F
Inductance in series LCR circuit, L = 1 Henry
RMS value of voltage, ε_rms   = 50 V
Frequency of source, f = 50/ πHz
Reactance of the inductor (X_L) is given by,

(a) Impedance of  an LCR circuit (Z) is given by,

RMS value of current I_rms is given

(b) Potential across the capacitor (V_C) is given by,

Potential difference across the resistor (V_R)  is given by

Potential difference across the inductor (V_L) is given by,

R.M.s potential = 50V
Net sum of all the potential drops = 50 V + 30 V + 10 V = 90 V
Sum of the potential drops > RMS potential applied

Q.15. Consider the situation of the previous problem. Find the average electric field energy stored in the capacitor and the average magnetic field energy stored in the coil.

Given:
Resistance in the LCR circuit, R = 300 Ω
Capacitance in the LCR circuit, C = 20 μF = 20 × 10⁻⁶ F
Inductance in the LCR circuit, L = 1 henry
Net impedance of the LCR circuit, Z = 500 ohm
RMS value of voltage,  = 50 V
RMS value of current, I_rms = 0.1 A
Peak current (I₀) is given by,

Electrical energy stored in capacitor  is given by,

Magnetic field energy stored in the coil (U_L) is given by,

Q.16. An inductance of 2.0 H, a capacitance of 18μF and a resistance of 10 kΩ is connected to an AC source of 20 V with adjustable frequency.
(a) What frequency should be chosen to maximize the current in the circuit?
(b) What is the value of this maximum current?

Given:
The inductance of inductor, L = 2.0 H
The capacitance of capacitor, C = 18 μF
The resistance of resistor, R = 10 kΩ
The voltage of AC source, E = 20 V
(a) In an LCR circuit, the current is maximum when reactance is minimum, which occurs at resonance, i.e. when capacitive reactance becomes equal to the inductive reactance,i.e.
X_L = X_C

(b) At resonance, reactance is minimum.
Minimum Reactance, Z = R
Maximum current (I) is given by,
I = E/R

Q.17. An inductor-coil, a capacitor and an AC source of rms voltage 24 V are connected in series. When the frequency of the source is varied, a maximum rms current of 6.0 A is observed. If this inductor coil is connected to a battery of emf 12 V and internal resistance 4.0 Ω, what will be the current?

RMS value of voltage, E_rms = 24 V
Internal resistance of battery, r = 4 Ω
RMS value of current, I_rms = 6 A
Reactance (R) is given by,

Let R' be the total resistance of the circuit. Then,
R' = R + r

Current, I =  12/8 A
= 1.5 A

Q.18. Following figure shows a typical circuit for a low-pass filter. An AC input V_i = 10 mV is applied at the left end and the output V₀ is received at the right end. Find the output voltage for ν = 10 k Hz, 1.0 MHz and 10.0 MHz. Note that as the frequency is increased the output decreases and, hence, the name low-pass filter.

Here,
Input voltage to the filter, Vi = 10 × 10⁻³ V
Resistance of the circuit, R = 1 × 10³ Ω
Capacitance of the circuit, C = 10 × 10⁻⁹ F
(a) When frequency, f = 10 kHz
A low pass filter consists of resistance and capacitance. Voltage across the capacitor is taken as the output.
Capacitive reactance (X_C) is given by,


Net impedence of the resistance-capacitance circuit (Z) is given by,

Current (I₀) is given by,

Output across the capacitor (V₀) is given by,

(b)When frequency, f = 1 MHz = 1 × 10⁶ Hz
Capacitive reactance (X_C) is given by,


 Total impedence (Z) =


Output voltage


(c) When frequency, f = 10 MHz = 10 × 10⁶ Hz = 10⁷ Hz
Capacitive reactance (X_C) is given by,

  

Q.19. A transformer has 50 turns in the primary and 100 in the secondary. If the primary is connected to a 220 V DC supply, what will be the voltage across the secondary?

A transformer works on the principle of electromagnetic induction, which is only possible in case of AC.

Hence, when DC (zero frequency) is supplied to it, the primary coil blocks the current supplied to it. Thus, the induced current in the secondary coil is zero. So, the output voltage will be zero.