Class 12 Exam > Class 12 Videos > L22 : Numericals - Alternating current, Physics, Class 12
L22 : Numericals - Alternating current, Physics, Class 12 Video Lecture
FAQs on L22 : Numericals - Alternating current, Physics, Class 12 Video Lecture
| 1. What is alternating current (AC)? |  |
Ans. Alternating current (AC) is an electric current that periodically changes direction. It flows in both positive and negative cycles, constantly reversing its polarity. AC is commonly used in households and industries to power various electrical devices.
| 2. How is alternating current different from direct current (DC)? | |
Ans. The main difference between alternating current (AC) and direct current (DC) is the direction of flow. AC constantly changes direction, while DC flows in only one direction. Additionally, AC has a sinusoidal waveform, while DC has a constant voltage level.
| 3. What is the frequency of an alternating current? | |
Ans. The frequency of an alternating current refers to the number of complete cycles it completes in one second. It is measured in hertz (Hz). In most countries, the standard frequency of AC is 50 Hz or 60 Hz, depending on the electrical grid.
| 4. How is alternating current generated? | |
Ans. Alternating current is generated using an electrical generator or alternator. The generator consists of a rotating magnetic field that induces an alternating voltage in the coils of wire. This voltage is then transmitted through power lines to provide electricity to various devices.
| 5. Why is alternating current preferred over direct current for long-distance transmission? | |
Ans. Alternating current is preferred over direct current for long-distance transmission due to its ability to be easily stepped up or down in voltage using transformers. This allows for efficient transmission over long distances with minimal power loss. In contrast, direct current requires higher voltages for long-distance transmission, leading to higher energy losses.
Text Transcript from Video
hello friends this video on alternating
currents part 22 is brought to you by
exam feel calm no more fear from exam
please make sure that you have watched
all the videos 10 part 21 before going
ahead with part 22 so now here again it
says that obtain the answers to e and B
if the circuit is connected to 110 volts
12 kilo Hertz supply
hence explained the statement that a
capacitor is a conductor at very high
frequencies so by solving these three
four problems we are getting to know new
facts we got to know in our previous
problem that if we use a high frequency
power supply then an inductor can behave
like a open circuit now here we will see
that in if we use a high frequency power
supply then a capacitor can behave like
a conductor so let us look at it now in
this case what are the values that are
given it some values are already given
in the previous problem like the
capacitance is given as hundred into 10
to the power minus 6 pallets the
resistance is given as 40 ohms RMS value
of voltage is given as hundred ten volts
frequency is given as 12 kilo Hertz
which is equal to 12 into 10 to the
power 3 Hertz so what would be the mean
value of voltage the mean the peak value
of voltage would be equal to root 2
times the RMS value of voltage right
that is equal to 1 1 0 2 root 2 volts
and God would be the angular frequency
that would be equal to 2 pi nu which is
equal to 2 pi into 12 into 10 to the
power 3 Hertz that is equal to 24 5 into
10 to the power 3 Hertz right so here
again we will calculate the maximum
value of current order
peak value of current which is equal to
VM divided by Z and what is Z Z is R
square plus XC square so this will be
equal to VM divided by PI root over R
square plus 1 by Omega square C square
so this comes out to be hundred and two
divided by root over R square which is
sixteen hundred plus one by Omega square
that is 24 pi into 10 to the power three
square into C square that is 10 to the
power minus four square therefore this
maximum value of current comes out to be
three point nine amperes now if you
compare this value of current with the
value of current we obtain in during
lower frequency power supply in the
previous problem is almost the same
there is no marked difference between
them so let us try to find out the
second part and the second part again we
have to find out the time lag and time
lag is equal to Phi by Omega so we have
to calculate the value of Phi we know
that tan Phi is equal to XC divided by I
which is equal to 1 by Omega C R that is
equal to 1 divided by 24 pi into 10 to
the power 2 3 into 10 to the power minus
4 into 40 so this is equal to 1 by 96 pi
so from this we can see that Phi comes
out to be 0.2 degrees that is equal to
zero point 2 into 5 divided by 180
therefore time lag is equal to Phi by
Omega that is equal to zero point 2 pi
divided by 180 into 1 by 24 pi into 10
to the power 3 so Phi and Phi will get
cancelled and this comes out to be 0.04
microseconds so what do we observe
this we observe that the value of Phi is
almost zero at high frequencies we
observe in the observation so we will
list down the observation from this
problem that is Phi tends to zero at
very high frequency for 0.2 degrees can
be considered as almost near to 0 so Phi
tends to 0 at very high frequency what
do you mean by Phi tends to 0 Phi tends
to 0 means there is no phase difference
between voltage and current that means
voltage and current Phi tends to zero
means voltage and current are in same
phase so that means we are talking about
a conductor with a fixed resistance
there so we are talking about a
resistive circuit that means we can see
that the capacitor behaves like
conductor like and that is why that
sends it behaves like a conductor
therefore the maximum amount of current
that can flow through the circuit also
did not decrease thank you please visit
exam comm to watch free educational
videos try 3 online tests get the best
quality study materials study from the
best tutors and mentors and much more
thank you once again
currents part 22 is brought to you by
exam feel calm no more fear from exam
please make sure that you have watched
all the videos 10 part 21 before going
ahead with part 22 so now here again it
says that obtain the answers to e and B
if the circuit is connected to 110 volts
12 kilo Hertz supply
hence explained the statement that a
capacitor is a conductor at very high
frequencies so by solving these three
four problems we are getting to know new
facts we got to know in our previous
problem that if we use a high frequency
power supply then an inductor can behave
like a open circuit now here we will see
that in if we use a high frequency power
supply then a capacitor can behave like
a conductor so let us look at it now in
this case what are the values that are
given it some values are already given
in the previous problem like the
capacitance is given as hundred into 10
to the power minus 6 pallets the
resistance is given as 40 ohms RMS value
of voltage is given as hundred ten volts
frequency is given as 12 kilo Hertz
which is equal to 12 into 10 to the
power 3 Hertz so what would be the mean
value of voltage the mean the peak value
of voltage would be equal to root 2
times the RMS value of voltage right
that is equal to 1 1 0 2 root 2 volts
and God would be the angular frequency
that would be equal to 2 pi nu which is
equal to 2 pi into 12 into 10 to the
power 3 Hertz that is equal to 24 5 into
10 to the power 3 Hertz right so here
again we will calculate the maximum
value of current order
peak value of current which is equal to
VM divided by Z and what is Z Z is R
square plus XC square so this will be
equal to VM divided by PI root over R
square plus 1 by Omega square C square
so this comes out to be hundred and two
divided by root over R square which is
sixteen hundred plus one by Omega square
that is 24 pi into 10 to the power three
square into C square that is 10 to the
power minus four square therefore this
maximum value of current comes out to be
three point nine amperes now if you
compare this value of current with the
value of current we obtain in during
lower frequency power supply in the
previous problem is almost the same
there is no marked difference between
them so let us try to find out the
second part and the second part again we
have to find out the time lag and time
lag is equal to Phi by Omega so we have
to calculate the value of Phi we know
that tan Phi is equal to XC divided by I
which is equal to 1 by Omega C R that is
equal to 1 divided by 24 pi into 10 to
the power 2 3 into 10 to the power minus
4 into 40 so this is equal to 1 by 96 pi
so from this we can see that Phi comes
out to be 0.2 degrees that is equal to
zero point 2 into 5 divided by 180
therefore time lag is equal to Phi by
Omega that is equal to zero point 2 pi
divided by 180 into 1 by 24 pi into 10
to the power 3 so Phi and Phi will get
cancelled and this comes out to be 0.04
microseconds so what do we observe
this we observe that the value of Phi is
almost zero at high frequencies we
observe in the observation so we will
list down the observation from this
problem that is Phi tends to zero at
very high frequency for 0.2 degrees can
be considered as almost near to 0 so Phi
tends to 0 at very high frequency what
do you mean by Phi tends to 0 Phi tends
to 0 means there is no phase difference
between voltage and current that means
voltage and current Phi tends to zero
means voltage and current are in same
phase so that means we are talking about
a conductor with a fixed resistance
there so we are talking about a
resistive circuit that means we can see
that the capacitor behaves like
conductor like and that is why that
sends it behaves like a conductor
therefore the maximum amount of current
that can flow through the circuit also
did not decrease thank you please visit
exam comm to watch free educational
videos try 3 online tests get the best
quality study materials study from the
best tutors and mentors and much more
thank you once again
About this Video
Oct 18, 2025
Last updated
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