Resistors in Circuits

1. Follow the rules for series circuits.

   1. Resistances in series add up.

      RT =	R1	+	R2	+	R3
      RT =	20 Ω	+	30 Ω	+	50 Ω
      RT = 100 Ω

   2. Total current is determined by the voltage of the power supply and the equivalent resistance of the circuit.

      IT = VT/RT IT = 125 V/100 Ω IT = 1.25 A

   3. Current is constant through resistors in series.

      I_T = I₁ = I₂ = I₃ = 1.25 A

   4. The voltage drops can be found using Ohm's law.

      V1 = I1R1 V1 = (1.25 A)(20 Ω) V1 = 25.0 V

      V2 = I2R2 V2 = (1.25 A)(30 Ω) V2 = 37.5 V

      V3 = I3R3 V3 = (1.25 A)(50 Ω) V3 = 62.5 V

      Verify your calculations by adding the voltage drops. On a series circuit they should equal the voltage increase of the power supply.

      VT	=	V1	+	V2	+	V3
      125 V	=	25.0 V	+	37.5 V	+	62.5 V
      125 V	=	125 V

      We're good, so let's finish.

   5. There are three equations for determining power. Since we have three resistors, let's apply a different equation to each as an exercise.

      P1 = V1 I1 P1 = (25.0 V)(1.25 A) P1 =  31.250 W

      P2 = I22R2 P2 = (1.25 A)2(30 Ω) P2 =  46.875 W

      P3 = V32/R3 P3 = (62.5 V)2/(50 Ω) P3 =  78.125 W

      In a series circuit, the element with the greatest resistance consumes the most power.

2. Follow the rules for parallel circuits.

   1. Resistances in parallel combine according to the sum-of-inverses rule.

      1	=	1	+	1	+	1
      RT		R1		R2		R3
      1	=	1	+	1	+	1
      RT		20 Ω		100 Ω		50 Ω
      1	=	5	+	1	+	2
      RT		100 Ω		100 Ω		100 Ω
      1	=	8
      RT		100 Ω
      RT	=	100 Ω	= 12.5 Ω
      		8

   2. Total current is determined by the voltage of the power supply and the equivalent resistance of the circuit.

      IT = VT/RT IT = 125 V/12.5 Ω IT = 10 A

   4. (Note: we'll answer part iv before part iii.) On a parallel circuit, each branch experiences the same voltage drop.

      V_T = V₁ = V₂ = V₃ = 125 V

   3. The current in each branch can be found using Ohm's law.

      I1 = V1/R1 I1 = (125 V)/(20 Ω) I1 = 6.25 A

      I2 = V2/R2 I2 = (125 V)/(100 Ω) I2 = 1.25 A

      I3 = V3/R3 I3 = (125 V)/(50 Ω) I3 = 2.50 A

      Verify your calculations by adding the currents. On a parallel circuit they should add up to the current from the power supply.

      IT	=	I1	+	I2	+	I3
      10 A	=	6.25 A	+	1.25 A	+	2.50 A
      10 A	=	10 A

      Good, it works.

   5. Again as an exercise, use a different equation to determine the electric power of each resistor.

      P1 = V1I1 P1 = (125 V)(6.25 A) P1 = 781.25 W

      P2 = I22R2 P2 = (1.25 A)2(100 Ω) P2 = 156.25 W

      P3 = V32/R3 P3 = (125 V)2/(50 Ω) P3 = 312.50 W

      In a parallel circuit, the element with the least resistance consumes the most power.

1. Outlets are wired in parallel so that the appliances on a circuit are independent of one another. Turning the coffee maker off will not result in the toaster turning off (assuming both were on at the same time). Each appliance will also get the same regulated voltage, which simplifies the design of electrical devices. The downside to this scheme is that the parallel currents can add up to dangerously high levels. A circuit breaker in series before the parallel branches can prevent overloads by automatically opening the circuit.

   

2. A 15 A circuit operating at 120 V consumes 1,800 W of total power.

   P = VI = (120 V)(15 A) = 1,800 W

   Total power in a parallel circuit is the sum of the power consumed on the individual branches.

   coffee maker	+	microwave oven
   850 W	+	1200 W
   2050 W
   microwave oven	+	toaster
   1200 W	+	900 W
   2100 W
   toaster	+	coffee maker
   900 W	+	850 W
   1750 W

   On this circuit, only the coffee maker and toaster can be operated simultaneously. All other combinations will trigger the circuit breaker to open.

The way to solve a complex problem is to break it down into a series of simpler problems. Be careful not to lose sight of your goal among all the bits and pieces, however. Before beginning plot your course. In this case we'll start by finding the effective resistance of the entire circuit and the current from the battery. This sets us up to get the current in all the different segments of the circuit. (The current divides and divides again in an effort to follow the path of least resistance.) After that, it's a simple matter to calculate the voltage drops in each resistor using V = IR and the power dissipated using P = VI. No part of this problem is difficult by itself, but since the circuit is so complex we'll be quite busy for a little while.

1. Let's begin the process by combining resistors. There are four series pairs in this circuit.

   left

   Rs = 3 Ω + 1 Ω Rs = 4 Ω Rs = 4 Ω + 2 Ω Rs = 6 Ω

   right

   Rs = 2 Ω + 3 Ω Rs = 5 Ω Rs = 1 Ω + 4 Ω Rs = 5 Ω

   These pairs form two parallel circuits, one on the left and one on the right.

   left

   1  =  1  +  1 Rp 4 Ω 6 Ω

   1  =  5 Rp 12 Ω

   Rp =  12 Ω  = 2.4 Ω 5

   right

   1  =  1  +  1 Rp 5 Ω 5 Ω

   1  =  2 Rp 5 Ω

   Rp =  5 Ω  = 2.5 Ω 2

   Each gang of four resistors is in series with another.

   left

   Rs = 2.4 Ω + 0.6 Ω Rs = 3 Ω

   right

   Rs = 2.5 Ω + 0.5 Ω Rs = 3 Ω

   The left and right halves of the circuit are parallel to each other and to the battery.

   1  =  1  +  1  =  2 Rp 3 Ω 3 Ω 3 Ω

   Rp =  3 Ω  = 1.5 Ω 2

   Now that we have the effective resistance of the entire circuit, let's determine the current from the power supply using Ohm's law.

   Itotal =	Vtotal	+	24 V	= 16 A
   	Rtotal		1.5 Ω

   Now walk through the circuit (not literally of course). At each junction the current will divide with more taking the path with less resistance and less taking the path with more resistance. Since charge doesn't leak out anywhere on a complete circuit, the current will be the same for all those elements in series with one another.

   The left and right halves of the circuit are identical in overall resistance, which means the current will divide evenly between them.

   8 A for the 0.6 Ω resistor on the left.

   8 A for the 0.5 Ω resistor on the right.

   On each side the current divides again into two parallel branches.

   The branches on the left have resistances in the ratio…

   R1&3  =  4 Ω  +  2 R2&4 6 Ω 3

   which means the currents will divide in the ratio…

   R1&3  =  3 R2&4 2

   3  8A = 4.8 A 5

   for the 1 Ω and 3 Ω resistors on the left.

   2  8A = 3.2 A 5

   for the 2 Ω and 4 Ω resistors on the left.

   The branches on the right are identical, so the current splits into two equal halves.

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   1  8A = 4.0 A 2

   for the 2 Ω and 3 Ω resistors on the right.

   1  8A = 4.0 A 2

   for the 1 Ω and 4 Ω resistors on the right.

2. Use V = IR over and over and over again to determine the voltage drops. (See the tables at the end of this solution.)

3. Use P = VI (or P = I²R or P = V²/R) over and over again to determine the power dissipated. These last two tasks are so tedious you should use a spreadsheet application of some sort. Enter the resistance values given and the current values just calculated into columns and instruct your electronic device of choice to multiply appropriately. Something like this…

Here are the solutions…

1. The total resistance in a series circuit is the sum of the individual resistances…

   RT = R1 + R2 + R3 RT = 3 Ω + 9 Ω + 6 Ω RT = 18 Ω

2. The total current can be found from Ohm's law…

   IT = VT/RT IT = (12 V)/(18 Ω) = ⅔ A IT =  0.667 A

3. The voltage in a circuit rises in a battery and drops in a resistor (when we follow the flow of conventional current). The rise in the battery is given as 12 V and the drops in each resistor can be found through repeated use of Ohm's law…

   V1 = I1R1 V1 = (⅔ A)(3 Ω) V1 = 2 V

   V2 = I2R2 V2 = (⅔ A)(9 Ω) V2 = 6 V

   V3 = I3R3 V3 = (⅔ A)(6 Ω) V3 = 4 V

   Starting at zero volts on the negative terminal of the battery, the voltage goes up 12 V then drops 2 V, 6 V, and 4 V, which brings us back to zero. (We are assuming that the battery and wires have negligible resistance.) Here's how it looks when graphed.

   

   Here's how it looks when the graph is superimposed on the circuit.

   

4. Current is everywhere the same in a series circuit. We've already determined it's 0.667 A. All that remains is to draw a horizontal line at two-thirds of an amp.

   

   Here's how it looks when the graph is superimposed on the circuit.