Check the correct sign $pm$ in $( Ve^{-j(wt+A)})/(Ie^{-j(wt+B)})=(V/I)cdot[cos(A-B) pm jsin(A-B)]$
$begingroup$
If $v = Ve^{-j(wt+A)}, i= Ie^{-j(wt+B)}$ then show that the impedance $z = v/i$ is given by $$Z = (V/I)cdot[cos(A-B) + jsin(A-B)]$$
I get confused because when I used the correct rules, it always comes to $$(V/I)cdot[cos(A-B) - jsin(A-B)]$$ instead because of the rule that $e^{-j(theta)}=cos(theta) - sin(theta)$
proof-verification complex-numbers
edited Dec 16 '18 at 11:22
Did
248k23223460
asked Dec 16 '18 at 10:12
Mvb1Mvb1
1
$endgroup$
add a comment |
$begingroup$
If $v = Ve^{-j(wt+A)}, i= Ie^{-j(wt+B)}$ then show that the impedance $z = v/i$ is given by $$Z = (V/I)cdot[cos(A-B) + jsin(A-B)]$$
I get confused because when I used the correct rules, it always comes to $$(V/I)cdot[cos(A-B) - jsin(A-B)]$$ instead because of the rule that $e^{-j(theta)}=cos(theta) - sin(theta)$
proof-verification complex-numbers
edited Dec 16 '18 at 11:22
Did
248k23223460
asked Dec 16 '18 at 10:12
Mvb1Mvb1
1
$endgroup$
$begingroup$
If $j=sqrt{-1}$, your answer is correct (although you might want to check the rule $e^{-j(theta)}=cos(theta) - sin(theta)$ again)
$endgroup$
– Shubham Johri
Dec 16 '18 at 10:30
add a comment |
$begingroup$
If $v = Ve^{-j(wt+A)}, i= Ie^{-j(wt+B)}$ then show that the impedance $z = v/i$ is given by $$Z = (V/I)cdot[cos(A-B) + jsin(A-B)]$$
I get confused because when I used the correct rules, it always comes to $$(V/I)cdot[cos(A-B) - jsin(A-B)]$$ instead because of the rule that $e^{-j(theta)}=cos(theta) - sin(theta)$
proof-verification complex-numbers
edited Dec 16 '18 at 11:22
Did
248k23223460
asked Dec 16 '18 at 10:12
Mvb1Mvb1
1
$endgroup$
If $v = Ve^{-j(wt+A)}, i= Ie^{-j(wt+B)}$ then show that the impedance $z = v/i$ is given by $$Z = (V/I)cdot[cos(A-B) + jsin(A-B)]$$
I get confused because when I used the correct rules, it always comes to $$(V/I)cdot[cos(A-B) - jsin(A-B)]$$ instead because of the rule that $e^{-j(theta)}=cos(theta) - sin(theta)$
proof-verification complex-numbers
proof-verification complex-numbers
edited Dec 16 '18 at 11:22
Did
248k23223460
asked Dec 16 '18 at 10:12
Mvb1Mvb1
1
edited Dec 16 '18 at 11:22
Did
248k23223460
asked Dec 16 '18 at 10:12
Mvb1Mvb1
1
edited Dec 16 '18 at 11:22
Did
248k23223460
edited Dec 16 '18 at 11:22
Did
248k23223460
edited Dec 16 '18 at 11:22
Did
248k23223460
248k23223460
asked Dec 16 '18 at 10:12
Mvb1Mvb1
1
asked Dec 16 '18 at 10:12
Mvb1Mvb1
1
asked Dec 16 '18 at 10:12
Mvb1Mvb1
1
1
$begingroup$
If $j=sqrt{-1}$, your answer is correct (although you might want to check the rule $e^{-j(theta)}=cos(theta) - sin(theta)$ again)
$endgroup$
– Shubham Johri
Dec 16 '18 at 10:30
add a comment |
$begingroup$
If $j=sqrt{-1}$, your answer is correct (although you might want to check the rule $e^{-j(theta)}=cos(theta) - sin(theta)$ again)
$endgroup$
– Shubham Johri
Dec 16 '18 at 10:30
$begingroup$
If $j=sqrt{-1}$, your answer is correct (although you might want to check the rule $e^{-j(theta)}=cos(theta) - sin(theta)$ again)
$endgroup$
– Shubham Johri
Dec 16 '18 at 10:30
$begingroup$
If $j=sqrt{-1}$, your answer is correct (although you might want to check the rule $e^{-j(theta)}=cos(theta) - sin(theta)$ again)
$endgroup$
– Shubham Johri
Dec 16 '18 at 10:30
add a comment |
1 Answer
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$begingroup$
Well,
$$ frac{v}{i} = frac{V}{I}e^{-j(A-B)} = frac{V}{I} [cos(-(A-B))+jsin(-(A-B))]$$
by Euler's formula, then because cosine is even and sine is odd,
$$ frac{v}{i} = frac{V}{I} [cos(A-B) -j sin(A-B)].$$
You got it
answered Dec 16 '18 at 10:37
kevinkayakskevinkayaks
1558
$endgroup$
add a comment |
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$begingroup$
Well,
$$ frac{v}{i} = frac{V}{I}e^{-j(A-B)} = frac{V}{I} [cos(-(A-B))+jsin(-(A-B))]$$
by Euler's formula, then because cosine is even and sine is odd,
$$ frac{v}{i} = frac{V}{I} [cos(A-B) -j sin(A-B)].$$
You got it
answered Dec 16 '18 at 10:37
kevinkayakskevinkayaks
1558
$endgroup$
add a comment |
$begingroup$
Well,
$$ frac{v}{i} = frac{V}{I}e^{-j(A-B)} = frac{V}{I} [cos(-(A-B))+jsin(-(A-B))]$$
by Euler's formula, then because cosine is even and sine is odd,
$$ frac{v}{i} = frac{V}{I} [cos(A-B) -j sin(A-B)].$$
You got it
answered Dec 16 '18 at 10:37
kevinkayakskevinkayaks
1558
$endgroup$
add a comment |
$begingroup$
Well,
$$ frac{v}{i} = frac{V}{I}e^{-j(A-B)} = frac{V}{I} [cos(-(A-B))+jsin(-(A-B))]$$
by Euler's formula, then because cosine is even and sine is odd,
$$ frac{v}{i} = frac{V}{I} [cos(A-B) -j sin(A-B)].$$
You got it
answered Dec 16 '18 at 10:37
kevinkayakskevinkayaks
1558
$endgroup$
Well,
$$ frac{v}{i} = frac{V}{I}e^{-j(A-B)} = frac{V}{I} [cos(-(A-B))+jsin(-(A-B))]$$
by Euler's formula, then because cosine is even and sine is odd,
$$ frac{v}{i} = frac{V}{I} [cos(A-B) -j sin(A-B)].$$
You got it
answered Dec 16 '18 at 10:37
kevinkayakskevinkayaks
1558
answered Dec 16 '18 at 10:37
kevinkayakskevinkayaks
1558
answered Dec 16 '18 at 10:37
kevinkayakskevinkayaks
1558
answered Dec 16 '18 at 10:37
kevinkayakskevinkayaks
1558
1558
add a comment |
add a comment |
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$begingroup$
If $j=sqrt{-1}$, your answer is correct (although you might want to check the rule $e^{-j(theta)}=cos(theta) - sin(theta)$ again)
$endgroup$
– Shubham Johri
Dec 16 '18 at 10:30