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An alternating voltage connected to an RLC series circuit has a peak voltage of 4.0v
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Jan 19, 2025
An alternating voltage connected to an RLC series circuit has a peak voltage of 4.0v. If the resistance is 10 ohms, the inductor is 0.5mH and the capacitor is 8µF. Determine the resonance frequency of the circuit.
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0:00
an alternating
0:02
voltage connected to an RLC series
0:06
circuit has a PE voltage of 4.0
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volts if the resistance is 10 ohms the
0:13
inductor is 0.5
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MH and the capacitor is 8
0:20
micro determine the resonance
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frequency of the
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circut as you can
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see this is the circuit
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the first thing we need to do is
0:34
to draft our data which is step one now
0:39
an alternating voltage connected to an
0:42
connected to an RLC series
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circuit has a p voltage of 4.0 volts
0:51
therefore the P
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voltage B not is equal to 4.0
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BT if the resistance is 10 ohms
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therefore we have
1:08
resistance x c is equal to
1:14
10
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ohms
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10s the
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inductor
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is 0.5
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M the
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inductor
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L is equal to
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05
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M and remember
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that
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0.5 mli is equal to *
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10^ - 3 then you write unit which is
2:04
Henry and the
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capacitor is 8
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microf we also have an have a
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capacitor C isal to
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8 micro
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par and therefore C is equals to 8 *
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micro is the same thing as * 10^ minus
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six
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par so now we've gotten our data
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determine the resonance
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frequency of the circuit we are to
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determine the resonance frequency which
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is
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f
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now the next step is to find the formula
2:55
that is suitable for this problem
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we know
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that resonance frequency is equal to 1
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over
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2 square root
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of L
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C so this is the formula we're going to
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use now we move to the next step which
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is the solution part we'll use this
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equation to solve this problem we know
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that f is equals to which is from our
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formula
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1 all
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over 2
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5 then we move square root
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of square root
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of l l here is the
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indal which
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is 0.5 * 10us 3 therefore we write
3:59
instead of
4:00
we write
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0 5 * 10 4 -
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3 then we also have C then we say times
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instead of C we'll write the capacitance
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which is * 8 8 * 10^ - 6 therefore we
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have
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8
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* 10 power - 6
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therefore this is the
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frequency we simplify this problem by
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saying f is = 1 all
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over 2 * 5 is =
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3142 therefore we have
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3142 then
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times square root of
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when you use
5:01
calculator and multiply 0.5 * 10us 3 by
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this you're going to
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have
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four time 10
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powerus
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11 this is what you're going to have
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therefore by means of uh by simplifying
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this problem
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F isal to when you use your calculator
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and multiply everything you still have
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1 this times this time this will give
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you
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6 3 9 7 and 10^ -
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5 * 10 -
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5 so when you divide
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this one by this you will end up
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having f is equals
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to 2 5
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1 6
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8 HZ you remember that unit of frequency
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is in Herz sorry F isal to
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251 68 HZ therefore by means of standard
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if we are to take this
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one 2 3 therefore you going to have f is
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equals
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to
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because
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kilo is represents
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1,000 therefore we are now to we now
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going to see this is
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25 this will represents
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kilo kilo
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HZ therefore our resonance frequency
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f is 25
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Kilz therefore as you can see from what
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we have
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here the data B not is equal to 40 volt
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XC which is the resistance is 10 OHS the
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inductor or the inductance is 0.5 * 10^
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- 3 hry the capacitance is 8 * 10^ - 6
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and we have to find the resonance
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frequency which is f therefore we apply
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the formula f = 1 /
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2un s root of
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LC so by using this formula we now
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substituted the values of L and C to get
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what we
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have under this solution in page one
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which is 1
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2un 0.5 * 10- 3 * 8 * 10^ - 6 we now
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move to f
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= 2 * 3142 because 5 =
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3142 * < TK 4 * 10 - 11 after
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simplification we ended up having 1/
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6397 * 10 4 - 5
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and then
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dividing using
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calculator we divided 1 by 6397 * 10^ -
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5 and we got
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25 168
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Herz and then we now converted it kilo
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Herz which is f which gives us f equals
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to 25 Kilz
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