Physics Solution Manual for 1100 and 2101

# 19 and equation 1 1 1 1 1 1 1 1 1 1 1 1 cs

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: right end of the resistor is + and the left end is –. The potential drops and rises for the two cases are: Potential drops Potential rises Part a IR V Part b V IR SOLUTION Since the current I goes from left-to-right through the 3.0-Ω and 4.0-Ω resistors, the left end of each resistor is + and the right end is –. The current goes through the 5.0-Ω resistor from right-to-left, so the right end is + and the left end is –. Starting at 1106 ELECTRIC CIRCUITS the upper left corner of the circuit, and proceeding clockwise around it, Kirchhoff’s loop rule is written as ( 3.0 Ω ) I + 12 V + 244444443 = ( 4.0 Ω ) I + ( 5.0 Ω ) I 14444444 4 4 36 V { Potential rises Potential drops Solving this equation for the current gives I = 2.0 A ______________________________________________________________________________ 81. REASONING First, we draw a current I1 (directed to the right) in the 6.00-Ω resistor. We can express I1 in terms of the other currents in the circuit, I and 3.00 A, by applying the junction rule to the junction on the left; the sum of the currents into the junction must equal the sum of the currents out of the junction. I { Current into junction on left = I1 + 3.00 A 14244 4 3 or I1 = I − 3.00 A Current out of junction on left In order to obtain values for I and V we apply the loop rule to the top and bottom loops of the circuit. SOLUTION Applying the loop rule to the top loop (going clockwise around the loop), we have 24.0 V + ( I − ( 3.00 A ) ( 4.00 Ω ) + ( 3.00 A ) (8.00 Ω ) = 1444443.00 A ) ( 6.00 Ω ) 1444444 24444444 4 3 4 2444444 3 Potential drops Potential rises This equation can be solved directly for the current; I = 5.00 A . Applying the loop rule to the bottom loop (going counterclockwise around the loop), we have V ( I − 3.00 A ) ( 6.00 Ω ) + 24.0 V + I ( 2.00 Ω ) = { 1444444442444444443 4 4 Potential drops Potential rises Substituting I = 5.00 A into this equation and solving for V gives V = 46.0 V . ______________________________________________________________________________ 82. REASONING Because all currents in the diagram flow from left to right (see the drawing), currents I1 and I2 both flow into junction A. Therefore, the current I in resistor R, which flows out of junction A, is, by Kirchhoff’s junction rule, equal to the sum of the other two currents: R1 = 2.70 Ω I I1 = 3.00 A I2 R2 = 4.40 Ω A R Chapter 20 Problems I = I1 + I 2 1107 (1) The current I1 is given. To determine I2, we note that the resistors R1 and R2 are connected in parallel, and therefore must have the same potential difference V12 = V1 = V2 across them. We will use Ohm’s law V = IR (Equation 20.2) to determine the potential difference V12 and then the current I2. SOLUTION Using V = IR (Equation 20.2), we express the voltage V12 across the resistors R1 and R2 in terms of currents and resistances as V12 = I1R1 = I 2 R2 (2) Solving Equation (2) for I2 yields I2 = I1R1 (3) R2 Substituting Equation (3) into Equation (1), we obtain R I1R1 2.70 Ω = I1 1 + 1 = ( 3.00 A ) 1 + = 4.84 A R R2 4.40 Ω 2 I = I1 + I 2 = I1 + ______________________________________________________________________________ 83. REASONING This problem can be solved by using Kirchhoff’s loop rule. We begin by drawing a current through each resistor. The drawing shows the directions chosen for the currents. The directions are arbitrary, and if any one of them is incorrect, then the analysis will show that the corresponding value for the current is negative. V1 = 4.0 V A− + F E R1 = 8.0 Ω − + R2 = 2.0 Ω − + I2 V2 = 12 V +− B I1 C D We mark the two ends of each resistor with plus and minus signs that serve as an aid in identifying the potential drops and rises for the loop rule, recalling that conventional current is always directed from a higher potential (+) toward a lower potential (–). Thus, given the 1108 ELECTRIC CIRCUITS directions chosen for I1 and I 2 , the plus and minus signs must be those shown in the drawing. We then apply Kirchhoff's loop rule to the top loop (ABCF) and to the bottom loop (FCDE) to determine values for the currents I1 and I2. SOLUTION Applying Kirchhoff’s loop rule to the top loop (ABCF) gives V1 + I R2 = I1R1 14 2 3 { 24 Potential rises Potential drop (1) Similarly, for the bottom loop (FCDE), V2 = I 2 R2 { 13 2 Potential rise Potential drop (2) Solving Equation (2) for I2 gives I2 = V2 12 V = = 6.0 A R2 2.0 Ω Since I2 is a positive number, the current in the resistor R2 goes from left to right , as shown in the drawing. Solving Equation (1) for I1 and substituting I2 = V2/R2 into the resulting expression yields V V1 + 2 R2 V +I R V +V 4.0 V + 12V R2 I1 = 1 2 2 = = 1 2= = 2.0 A R1 R1 R1 8.0 Ω Since I1 is a positive number, the current in the resistor R1 goes from left to right , as shown in the drawing. ______________________________________________________________________________ 84. REASONING In preparation for applying Kirchhoff’s rules, we now choose the currents in each res...
View Full Document

## This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

Ask a homework question - tutors are online