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Pset 1 Solution

# Pset 1 Solution - Pset 1 Solution 1 The current through the...

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Pset 1 Solution 1. The current through the shunt resistor is A R V I 5 . 0 4 2 = = = . The same through the 40 Ohm resistor is the same. Therefore, the voltage of the DC source is V I V s 22 ) 4 40 ( = + = . If the shunt is absent, the current would have been A V I s 55 . 0 40 0 = = . 2. (a) 2 R and L R are in parallel. L L eq R R R R R + = 2 2 . Use voltage divider formula: eq eq o R R e + = 100 0 . 12 . Knowing o e , solve the above to get = 3 . 63 eq R . That is 3 . 63 2 2 = + L L R R R R . Knowing 7 10 = L R , solve the above, we get = 3 . 63 2 R . (b). 2 R now is in parallel with 1 k resistor (the input impedance of the voltmeter). The equivalent resistance is now = + × = + 53 . 59 1000 3 . 63 1000 3 . 63 2 2 L L R R R R Therefore, the voltmeter will read, according to voltage divider formula V V 48 . 4 100 53 . 59 53 . 59 12 = + 3. (a) 0 4 / 0 2 / 4 / 2 2 0 2 2 0 2 / 0 2 2 0 2 0 2 0 2 0 2 584 . 0 ) 4 1 4 )( 2 ( )] 4 ( 2 1 ) 2 1 4 ( 2 1 )[ 2 ( ) ( ) 4 ( sin [ ) ( ) ( sin )[ 2 ( ) ( ) ( 4 ) 2 1 ( ) ( ) ( 2 1 ) 2 ( ) ( ) ( ) ( ) ( 1 E t d E t d t E t d t e t d t e t d t e T t d t e dt t e T V T T T rms = = + = ∫∫ + = = = = = = π ω ππ (b) Power dissipated by the resistor Watt R V P rms 82 . 21 2 = =

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4. × = = T dt t v T V T 1 ) ( 1 0 Area of the triangle= 2 A . For 1 = α : 3 3 3 1 ) ( 1 ) ( 1 1 2 0 2 2 0 2 0 2 A A dt t T A T dt t T A T dt v T V T T T rms = = = = = 3 2 ~ 2 2 A V V V rms = = For 2 / 1 = : 3 3 3 1 ) 2 ( 2 ) 2 ( 2 2 1 2 2 / 0 2 2 2 / 0 2 2 / 0 2 0 2 A A dt t T A T
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Pset 1 Solution - Pset 1 Solution 1 The current through the...

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