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$$i_1 = sin(wt +45^0) $$ $$ i_2 = -sin(wt +60^0)$$

Question : what is the phase difference dows i1 lead i2

My method

1 ) Eliminate minus sign that is i2 = sin(wt + 60 + 180 ) = sin(wt + 240 )

My answer is that i1 leads i2 by 45 - 240 = -195

but correct answer is i1 leads i2 by +165 deg

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  • \$\begingroup\$ Try a couple of trig identities. Either of these should help. \$\endgroup\$ Commented Jan 14, 2014 at 8:18
  • \$\begingroup\$ $$\sin \left( -x \right)\; =\; -\sin \left( x \right)$$ $$\sin \left( x+\pi \right)\; =\; -\sin \left( x \right)$$ \$\endgroup\$ Commented Jan 14, 2014 at 8:24

1 Answer 1

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It is the same answer, -195 degrees = 165 degrees on the circle.

To convert negative degrees to positive, add 360.

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  • \$\begingroup\$ adding - 180 to eliminate the minus sign in i2 yields , 165 phase difference . Then I tried both of these methods ( adding -180 or adding +180 ) to other problems and answers were the same (not 180 ) Why the answers are same in these Q \$\endgroup\$ Commented Jan 14, 2014 at 7:00

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