Resistors in Series

Resistors in Series - Physics 202 Thursday, March 4, 1999...

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Physics 202 Thursday, March 4, 1999 Announcements: Lecture notes: Resistors in Series In this case, the current going through R 1 is the same as that going through R 2 . Hence, ξ - iR 1 - iR 2 = 0 o ξ = i (R 1 + R 2 ) o Therefore. R eq = R 1 + R 2 If there are 3 resistors in series, then R eq = R 1 + R 2 + R 3 . If there are n resistors in series, then R eq = R 1 + R 2 + R 3 +...R n . Resistors in parallel
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There is an equal emf (potential) across R 1 and R 2 . ξ = i 1 R 1 = i 2 R 2 (Equation 1), but i = i 1 + 1 2 (Equation 2) o From equation 1: i 1 = ξ / R 1 and i 2 = ξ / R 2 o From equation 2: i = i 1 + 1 2 = / R 1 ) + ( ξ / R 2 ) = ξ [(1 / R 1 ) + (1 / R 2 )] For equivalent circuit, ξ = i 1 R eq or i = ξ / R eq . o Therefore, ξ / R eq = ξ [(1 / R 1 ) + (1 / R 2 )] o or similarly, (1/R eq ) = €[(1 / R 1 ) + (1 / R 2 )] If there are more than 2 resistors in parallel, then (1/R eq ) = €[(1 / R 1 ) + (1 / R 2 ) + (1/R 3 ) +. ... (1/R
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Resistors in Series - Physics 202 Thursday, March 4, 1999...

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