Thursday, March 4, 1999
Announcements:

Lecture notes:

• In this case, the current going through R₁ is the same as that going through R₂.
• Hence,  x - iR₁ - iR₂ = 0
•  x = i (R₁ + R₂)
• Therefore. R_eq = R₁ + R₂
• If there are 3 resistors in series, then R_eq = R₁ + R₂ + R₃.
• If there are n resistors in series, then R_eq = R₁ + R₂ + R₃ +...R_n.

• There is an equal emf (potential) across R₁ and R₂.
• x = i₁R₁ = i₂R₂   (Equation 1), but  i = i₁ + 1₂ (Equation 2)
• From equation 1:  i₁ =  x/ R₁ and  i₂ =  x/ R₂
• From equation 2:  i = i₁ + 1₂  =   (x/ R₁ ) + ( x/ R₂ ) =  x [(1/ R₁ ) +  (1/ R₂ )]
• For equivalent circuit, x = i₁R_eq or  i =  x/ R_eq.
• Therefore, x/ R_eq=  x [(1/ R₁ ) +  (1/ R₂ )]
• or similarly,  (1/R_eq) =   [(1/ R₁ ) +  (1/ R₂ )]
• If there are more than 2 resistors in parallel, then  (1/R_eq) =   [(1/ R₁ ) +  (1/ R₂ ) + (1/R₃) +.... (1/R_n)]
• See problem 39 from Chapter 28 for an example

• An ammeter is a device that measures current that is flowing through a part of a circuit and the meter.
• A voltmeter is a device that measures potentials across two points on a circuit.
• See figure 28-12 on page 685 of the textbook for a diagram showing how to connect an ammeter and voltmeter.
• Note: 1) A good ammeter must be one with very low resistance otherwise the insertion of the ammeter into the circuit changes the amount of current flowing though the circuit
• 2) A good voltmeter must be one with very high resistance. If the resistance is low, it will divert too much current from the circuit  and change the potential across the points of interest.

• Example taken from Chapter 28, problem 43

• R₁ =  1.0 W ; R₂ = 2.0 W
• x₁ = 2.0 V ; x₂ = x₃ = 4.0 V
• A) Calculate the current though each ideal battery (emf device)
• B) Calculate V_a - V_b
• Procedure to solve multiloop circuits
• Define current at different sections of the circuit... for example
 

• Follow the "Junction Rule". Namely, the sum of the currents entering any junction must be equal to the sum of the currents leaving that junction.
• Therefore,  i₁ = i₂ + i₃  ( Notice that this equation is satisfied at both point a and point b.)
• Let this be equation 1
• Follow the "Loop Rule". That is, the algebraic sum of the changes in potential encountered in a complete traversal of a complete loop of a circuit must be zero.
• Therefore, in the loop on the left, starting from point b,  - i₁R₁ + x₁ - i₁R₁ - i₂R₂ - x₂ = 0
• Let this be equation 2
• In the loop on the right, starting from point a, -i₃R₁ - x₃ - i₃R₁ + x₂ + i₂R₂ = 0
• Let this be equation 3
• In the outer loop, starting from point a,  -i₃R₁ - x₃ - i₃R₁ - i₁R₁ + x₁ - i₁R₁ = 0
• Let this be equation 4
• So we have four possible equations and three unknowns ( i₁ , i₂ , and i₃), One equation is redundant.
•   i₁ = i₂ + i₃ .... (equation 1)
• - i₁(1) + (2) - i₁(1) - i₂(2) - (4) = 0 .... (equation 2 with appropriate variable substitution)
• -i₃(1) - (4) - i₃(1) + (4) + i₂(2) = 0..... (equation 3 with appropriate variable substitution)
• From these equations we can find that....
• i₁ = i₂ + i₃
• i₁ + i₂ = -1
• -2i₃ + 2i₂ = 0
• Solution?
• i₁ = 2i₂ = 2i₃ = -2/3 amp
• i₁ = -2/3 amp and i₂ = i₃ = -1/3 amp
• What do the minus signs mean?
• We need to reverse the direction of all the currents in the figure.

• Part b..... V_a - V_b = ?
•  = x₂  - i₂R₂ = 4 - (1/3) (2) = 3.333V
• Look through sample problems 28-3, 28-4, 28-5 and problem #35.