Class 12-science H C VERMA Solutions Physics Chapter 17 - Alternating Current
Alternating Current Exercise 330
Solution 1
Equation of alternating current
After time t, it reaches rms value 
ms
Solution 2
Peak voltage, 
Volt
Equation of alternating voltage
Time taken from
rms value to zero is equal to time taken from zero to rms value
.
ms
Solution 3
Resistance of the bulb
Now,
Solution 4
Since, brightness is same. So, heat consumed in DC and AC will be same.
volt
Solution 5
RMS value of power
Energy consumed
Solution 6
V/m
Potential difference across the capacitor
volt
Rms value of voltage which can be connected
volt
Solution 7
Solution 8
Capacitive reactance, 
Peak voltage, V0 = 10 Volt
(a) At 
(b) At 
(c) At 
(d) At 
Solution 9
Inductive
reactance, 
Peak voltage, V0 = 10 Volt
Now,
(a)
(b)
(c)
Solution 10
Impedance of the circuit
Power consumed=
W
Solution 11
Energy dissipated as heat
J
Solution 12
Hz
Impedance of the circuit
Peak current
A
Average power dissipated
W
Solution 13
Resistance of the bulb
Potential drop
across resistance 
volt
Potential drop across inductor VL
Applied potential
volt
I=0.5A
So, current through inductor is I=0.5A
Now, for inductor
H
Solution 14
Hz
R=300
Impedance 
(a) Rms current in the circuit
A
(b) 
volt
volt
volt
Solution 15
Energy stored in the capacitor
mJ
Energy stored in the Inductor
mJ
Solution 16
(a) Current is maximum when circuit is in resonance condition
Hz
(b) Impedance
Z=R=10kΩ
Now,
V=IZ
I=2mA
Solution 17
Impedance of the circuit is equal to resistance of circuit as current is maximum when impedance is minimum.
V=IZ
24=6.Z
Z=4Ω
Now, when battery is connected,
Current in circuit
V=IR
12=I (8)
I=1.5A
Solution 18
Impedance of the circuit
Current in the circuit
Potential drop across capacitor
(a) f=10kHz
mV
(b) f=100kHz
mV
(c) f=
kHz
mV
(d) f=
kHz
V
Solution 19
Zero voltage across secondary coil as there is no change in flux.
