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lecture07 - 8 Power in electric circuits I R V W QV P = =...

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8. Power in electric circuits QV W = R V I IV t QV t W P = = I t Q = R V R I IV P 2 2 = = = IR V = [ ] ( 29 ( 29 s J s C C J A V W P / 1 / 1 / 1 1 1 = = = =
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Example: Two resistors, R 1 = 5 , R 2 = 10 , are connected in series. The battery has voltage of V = 12 V. a) Find the electric power delivered by the battery b) Find the electric power dissipated in each resistor = + = + = 15 10 5 2 1 R R R eq A 8 . 0 15 V 12 = = = eq R V I V 1 R 2 R I + I ( 29 W V R V P eq 6 . 9 15 12 2 2 = = = W A R I P R R 2 . 3 ) 5 ( ) 8 . 0 ( 2 1 2 1 1 = = = Power in the resistor R 1 : W A R I P R R 4 . 6 ) 10 ( ) 8 . 0 ( 2 2 2 2 2 = = = Power in the resistor R 2 : The total power: W W W P P P R R tot 6 . 9 4 . 6 2 . 3 2 1 = + = + = V V R R 12 10 5 2 1 = = =
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Example: Two resistors, R 1 = 5 , R 2 = 10 , are connected in parallel. The battery has voltage of V = 12 V. a) Find the electric power delivered by the battery b) Find the electric power dissipated in each resistor Power in the resistor R 1 : Power in the resistor R 2 : The total power: V V R R 12 10 5 2 1 = = = + 1 R V 2 R I I 1 I 2 I = + = + = + = 10 3 10 1 10 2 10 1 5 1 1 1 1 2 1 R
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