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IN AN AC CIRCUIT THE PEAK VALUE OF THE P.D IS 180V. WHAT IS THE INSTATANEOUS P.D WHEN IT HAS REACH
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Jan 23, 2025
IN AN AC CIRCUIT THE PEAK VALUE OF THE P.D IS 180V. WHAT IS THE INSTATANEOUS P.D WHEN IT HAS REACH (1/8)TH OF A CYCLE? (WAEC 1989)
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in an a
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circuit the P value of the potential
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difference is 180
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volts what is the instantaneous
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potential difference when it has reached
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one over8 of a
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ccle
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1989 in an easy
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cute the P value of the potential
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difference is 180 volts
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therefore our step one
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data we have F value which is B not is
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equals to
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180
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volts what is the instantaneous
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potential
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difference when it has reached 1/8 of a
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cycle what is
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B when it has reached 1/8 of a cycle in
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this case they said a
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cycle remember that a cycle is
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360° therefore in this case now this
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360° represent Theta and Theta is the
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face angle which is Omega
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T and this is equals to 1 /
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8 of this
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cycle therefore now we now see the P
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angle
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Theta is equal
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to 360 which is the
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cycle because it is 1/8 of this cycle
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which is 360 * 1/
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8 and Theta which is space angle which
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is Omega T is + 2 360 * 1/ 8 will give
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you 360 / 8 and this is going to give
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you 360 / 8al
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45° therefore this is our face angle now
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the next step is to find the formula
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which formula is suitable for finding
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the answer to this problem we now say B
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the instance ous potential difference B
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isal to
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v
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s Omega
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T this is the formula we going to use
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and therefore we'll now move to step
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three which is to find the solution and
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this is page one of the solution
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therefore we now say V is equals to B
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not here is 180 volts therefore we say8
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T
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Time s of Omega T Omega
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T is Omega T here and this is
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45° therefore the P angle which is Omega
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T
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is s of
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45° now there's something I want to show
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you s of
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45° is equal to
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uh I think
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it's let me check the trick table
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trigonometric
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table
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uh is
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uh square root of
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2un 2/ two that's from trick table trick
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table
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from trick
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square root of s 45°
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isun 2 therefore in this case now we now
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say B is equals
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to
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180
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180
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* s of
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45° s of 45 here is < TK 2 2
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therefore < TK 2 / 2 2 into 2 1 2 into
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18 180 90 therefore B which is the
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instantaneous potential difference is
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equal
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to
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90 Square < TK of
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2 therefore in this case we will now say
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that b is equal to 90
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Square < TK of 2 volts and this is our
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final answer therefore you can see now
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this is what we did B not is equal to
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180
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volts which is the B voltage from the
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question instantaneous value of the
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potential difference we don't know it so
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we try to find it the Theta is the P
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angle which is Omega T and the P angle
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is a cycle which is
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360° therefore and in this case or from
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the question it says 1 / 8 of 360 1 / 8
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of 360 is 1 8 * 360 which is 360 / 8 and
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it will give us 45° the formula we
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supposed to use to find instantaneous
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potential difference is B sin Omega T
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and from the solution you can see B is
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180 s of
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45° and from page two of the solution
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you can now see that s of 45° to root of
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of 2 / 2 therefore when you multiply it
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by 180° it is going to give
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you B = 90 < TK 2 and this is our final
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answer
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