6bcircuits2 - Circuits Problem 1 The circuit shown below...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Circuits: Problem 1: The circuit shown below consists of a battery connected in series to a resistor R 1 , a resistor R 4 , and a parallel structure consisting of resistors R 2 and R 3 . The battery is maintained at a voltage E 0 . 1: Find the effective resistance of the circuit. 2: Find the voltage drop across each resistor. 3: Find the current through each resistor. 4: Find the power supplied by the battery. ± ² ³ ´ ± ² Solution: Resistors R 2 and R 3 are in parallel. Therefore, the effective resistance of the combination R 23 is given by 1 R 23 = 1 R 2 + 1 R 3 Solving for R 23 , we find R 23 = R 2 R 3 R 2 + R 3 Resistors R 1 , R 23 , and R 4 are connected in series. Therefore, the equivalent resistance of all four resistors is 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
R eq = R 1 + R 23 + R 4 = R 1 + R 2 R 3 R 2 + R 3 + R 4 = ( R 1 + R 4 )( R 2 + R 3 ) + R 2 R 3 R 2 + R 3 This is the equivalent resistance of the circuit. We now turn to parts 2 and 3 . The total current through the circuit is I eq = E 0 R eq = E 0 ( R 2 + R 3 ) ( R 1 + R 4 )( R 2 + R 3 ) + R 2 R 3 Because resistors R 1 , R 23 , and R 4 are connected in series, this is also the current through each of these resistances. That is, I 1 = I 23 = I 4 = I eq = E 0 R eq We can now compute the potential drop across each resistance: V 1 = R 1 I 1 = R 1 I eq = E 0 R 1 R eq V 23 = R 23 I 23 = R 23 I eq = E 0 R 23 R eq V 4 = R 4 I 4 = R 4 I eq = E 0 R 4 R eq Because resistors R 2 and R 3 are connected in parallel, the potential drop across each resistor is equal to the potential drop V 23 of the parallel combination. V 2 = V 3 = V 23 = E 0 R 23 R eq We can now find the current through resistors R 2 and R 3 . I 2 = V 2 R 2 = E 0 R 23 R 2 R eq I 3 = V 3 R 3 = E 0 R 23 R 3 R eq Inserting our expressions for R 23 and R eq , we have for the potential drops 2
Background image of page 2
V 1 = E 0 R 1 R eq = E 0 R 1 ( R 2 + R 3 ) ( R 1 + R 4 )( R 2 + R 3 ) + R 2 R 3 V 2 = E 0 R 23 R eq = E 0 R 2 R 3 ( R 1 + R 4 )( R 2 + R 3 ) + R 2 R 3 V 3 = E 0 R 23 R eq = E 0 R 2 R 3 ( R 1 + R 4 )( R 2 + R 3 ) + R 2 R 3 V 4 = E 0 R 4 R eq = E 0 R 4 ( R 2 + R 3 ) ( R 1 + R 4 )( R 2 + R 3 ) + R 2 R 3 For the current through each resistor, we have
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/29/2010 for the course PHYSICS 6B 318036810 taught by Professor Waung during the Spring '09 term at UCLA.

Page1 / 9

6bcircuits2 - Circuits Problem 1 The circuit shown below...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online