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An alternating current of 6mA and of frequency 60/π flow pass a 3μF capacitor used
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Jan 18, 2025
An alternating current of 6mA and of frequency 60/π flow pass a 3μF capacitor used in a radio circuit. What is the voltage across the capacitor?
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0:02
an alternating current of 6
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milliamp and a frequency 60 over Pi flow
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has a 3
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micro capacitor used in a radio
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circuit what is the voltage across the
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capacitor now the first thing you need
0:21
to do is what draft your
0:26
data data
0:30
an alternating current of 6
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milliamp here you have
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six
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Milli which is the
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current
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IC is equal to 6
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mamp and 6 milliamp is equal to 6
0:53
* 10 4
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of- 3 a
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let's go back to the question again and
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the
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frequency 60 5 therefore we have our
1:11
frequency F isal
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to
1:15
60 over
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five flow fast a 3
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micro
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capacitor therefore we have a capacitor
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C which is equal to to
1:30
three
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micro far and this can be written as
1:37
this is three micro is * 10 to the power
1:42
of -
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6 then far which is the
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unit used in a radio
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circuit what is the voltage across the
1:57
capacitor what is
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B
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of
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C now we don't
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have now let's go
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back the
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formula where to
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use is to find BC equal to BC is equals
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to
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IC XC
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and we don't have
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IC and therefore we need to use the
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formula for XC which is equal to 1 /
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2
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5 f c Now we move on to the
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solution
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solution we first use this formula and
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then we have first and second step the
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first step by using Cal to
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1/
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25
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FC this is equals
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to
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1 over
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2 5 F here if we go back to the from our
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data is 60 over 5 therefore we now say
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write
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time
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60 here if we go back to the data you
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will see C is 3 * 10- 6 therefore
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* 3 * 10 4 -
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6
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now X of C is equal to after using your
4:00
calculator you five is this will go
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therefore you now going to have
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one/ 2 * 3 sorry 2 * 60 is
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120 * 3 * 10 power
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of - 6 therefore X of C is equals
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to
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we8 8 * 10^ 3
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2 8 * 10 power of 3 remember X of C is
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what is in
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ohms and therefore this is what we have
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as x c now we will now use this
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formula this formula which is the second
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formula bcal to I XC we will now
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say b
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c is equal to i
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c x
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c which is going to be I see here if you
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go back you see is 6 * 10us 3 and
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therefore we
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write 6 * 10 - 3
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time x and x here is 2.8
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*
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say 8 *
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10 and therefore if you calculate this
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by this you're going to
5:48
have
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16.8 you're going to have
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16.8
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volts which is
6:00
your voltage across the capacit and this
6:04
is
6:05
okay so you can
6:09
read
6:28
again for
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