## Practice

### practice problem 1

- the equivalent resistance,
- the total current from the power supply,
- the current through each resistor,
- the voltage drop across each resistor, and
- the power dissipated in each resistor.

#### solution

- Follow the rules for series circuits.
- Resistances in series add up.
*R*=_{T}*R*_{1}+ *R*_{2}+ *R*_{3}*R*=_{T}20 Ω + 30 Ω + 50 Ω *R*= 100 Ω_{T} - Total current is determined by the voltage of the power supply and the equivalent resistance of the circuit.
*I*=_{T}*V*/_{T}*R*_{T}*I*= 125 V/100 Ω_{T}*I*= 1.25 A_{T} - Current is constant through resistors in series.
*I*=_{T}*I*_{1}=*I*_{2}=*I*_{3}= 1.25 A - The voltage drops can be found using Ohm's law.
*V*_{1}=*I*_{1}*R*_{1}*V*_{1}= (1.25 A)(20 Ω)*V*_{1}= 25.0 V*V*_{2}=*I*_{2}*R*_{2}*V*_{2}= (1.25 A)(30 Ω)*V*_{2}= 37.5 V*V*_{3}=*I*_{3}*R*_{3}*V*_{3}= (1.25 A)(50 Ω)*V*_{3}= 62.5 VVerify your calculations by adding the voltage drops. On a series circuit they should equal the voltage increase of the power supply.

*V*_{T}= *V*_{1}+ *V*_{2}+ *V*_{3}125 V = 25.0 V + 37.5 V + 62.5 V 125 V = 125 V - There are three equations for determining power. Since we have three resistors, let's apply a different equation to each as an exercise.
*P*_{1}=*V*_{1}*I*_{1}*P*_{1}= (25.0 V)(1.25 A)*P*_{1}= 31.250 W*P*_{2}=*I*_{2}^{2}*R*_{2}*P*_{2}= (1.25 A)^{2}(30 Ω)*P*_{2}= 46.875 W*P*_{3}=*V*_{3}^{2}/*R*_{3}*P*_{3}= (62.5 V)^{2}/(50 Ω)*P*_{3}= 78.125 W

- Resistances in series add up.
- Follow the rules for parallel circuits.
- Resistances in parallel combine according to the sum-of-inverses rule.
1 = 1 + 1 + 1 *R*_{T}*R*_{1}*R*_{2}*R*_{3}1 = 1 + 1 + 1 *R*_{T}20 Ω 100 Ω 50 Ω 1 = 5 + 1 + 2 *R*_{T}100 Ω 100 Ω 100 Ω 1 = 8 *R*_{T}100 Ω *R*_{T}= 100 Ω = 12.5 Ω 8 - Total current is determined by the voltage of the power supply and the equivalent resistance of the circuit.
*I*=_{T}*V*/_{T}*R*_{T}*I*= 125 V/12.5 Ω_{T}*I*= 10 A_{T} - (Note: we'll answer part iv before part iii.) On a parallel circuit, each branch experiences the same voltage drop.
*V*=_{T}*V*_{1}=*V*_{2}=*V*_{3}*V*= 125 V_{T} - The current in each branch can be found using Ohm's law.
*I*_{1}=*V*_{1}/*R*_{1}*I*_{1}= (125 V)/(20 Ω)*I*_{1}= 6.25 A*I*_{2}=*V*_{2}/*R*_{2}*I*_{2}= (125 V)/(100 Ω)*I*_{2}= 1.25 A*I*_{3}=*V*_{3}/*R*_{3}*I*_{3}= (125 V)/(50 Ω)*I*_{3}= 2.50 AVerify your calculations by adding the currents. On a parallel circuit they should add up to the current from the power supply.

*I*_{T}= *I*_{1}+ *I*_{2}+ *I*_{3}10 A = 6.25 A + 1.25 A + 2.50 A 10 A = 10 A - Again as an exercise, use a different equation to determine the electric power of each resistor.
*P*_{1}=*V*_{1}*I*_{1}*P*_{1}= (125 V)(6.25 A)*P*_{1}= 781.25 W*P*_{2}=*I*_{2}^{2}*R*_{2}*P*_{2}= (1.25 A)^{2}(100 Ω)*P*_{2}= 156.25 W*P*_{3}=*V*_{3}^{2}/*R*_{3}*P*_{3}= (125 V)^{2}/(50 Ω)*P*_{3}= 312.50 W

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

### practice problem 2

- Draw a schematic diagram of this circuit.
- Which of these appliances can be operated simultaneously without tripping the circuit breaker?

#### solution

- 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.
- A 15 A circuit operating at 120 V consumes 1800 W of total power.
*P*=*VI*= (120 V)(15 A) = 1800 WTotal 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.

### practice problem 3

- the current through,
- the voltage drop across, and
- the power dissipated by each resistor.

#### solution

*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.

- Let's begin the process by combining resistors. There are four series pairs in this circuit.
*R*= 3 Ω + 1 Ω_{s}*R*= 4 Ω_{s}*R*= 4 Ω + 2 Ω_{s}*R*= 6 Ω_{s}*R*= 2 Ω + 3 Ω_{s}*R*= 5 Ω_{s}*R*= 1 Ω + 4 Ω_{s}*R*= 5 Ω_{s}These pairs form two parallel circuits, one on the left and one on the right.

1 = 1 + 1 *R*_{p}4 Ω 6 Ω 1 = 5 *R*_{p}12 Ω *R*=_{p}12 Ω = 2.4 Ω 5 1 = 1 + 1 *R*_{p}5 Ω 5 Ω 1 = 2 *R*_{p}5 Ω *R*=_{p}5 Ω = 2.5 Ω 2 Each gang of four resistors is in series with another.

*R*= 2.4 Ω + 0.6 Ω_{s}*R*= 3 Ω_{s}*R*= 2.5 Ω + 0.5 Ω_{s}*R*= 3 Ω_{s}The left and right halves of the circuit are parallel to each other and to the battery.

1 = 1 + 1 = 2 *R*_{p}3 Ω 3 Ω 3 Ω *R*=_{p}3 Ω = 1.5 Ω 2 Now that we have the effective resistance of the entire circuit, let's determine the total current from the power supply using Ohm's law.

*I*=_{total}*V*_{total}+ 24 V = 16 A *R*_{total}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.8 A for the

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

The branches on the left have resistances in the ratio… The branches on the right are identical, so the current splits into two equal halves.

☟

☟

☟*R*_{1&3}= 4 Ω + 2 *R*_{2&4}6 Ω 3 which means the currents will divide in the ratio… *R*_{1&3}= 3 *R*_{2&4}2 3 8A = 4.8 A 5 1 8A = 4.0 A 2 for the 1 Ω and 3 Ω

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

resistors on the right.2 8A = 3.2 A 5 1 8A = 4.0 A 2 for the 2 Ω and 4 Ω

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

resistors on the right. - Use
*V*=*IR*over and over and over again to determine the voltage drops. - Use
*P*=*VI*(or*P*=*I*^{2}*R*or*P*=*V*^{2}/*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…Left Side resistance

(Ω)current

(A)voltage

(V)power

(W)0.6 8.0 04.8 38.40 1.0 4.8 04.8 23.04 2.0 3.2 06.4 20.48 3.0 4.8 14.4 69.12 4.0 3.2 12.8 40.96 Right Side resistance

(Ω)current

(A)voltage

(V)power

(W)0.5 8 04 32 1.0 4 04 16 2.0 4 08 32 3.0 4 12 48 4.0 4 16 64

### practice problem 4

- Calculate the equivalent resistance of the circuit.
- Calculate the current through the battery.
- Graph voltage as a function of location on the circuit assuming that
*V*_{a}= 0 V at the negative terminal of the battery. - Graph current as a function of location on the circuit.

#### solution

Here are the solutions…

- The total resistance in a series circuit is the sum of the individual resistances…
*R*=_{T}*R*_{1}+*R*_{2}+*R*_{3}*R*= 3 Ω + 9 Ω + 6 Ω_{T}*R*= 18 Ω_{T} - The total current can be found from Ohm's law…
*I*=_{T}*V*/_{T}*R*_{T}*I*= (12 V)/(18 Ω) = ⅔ A_{T}*I*= 0.667 A_{T} - 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…
*V*_{1}=*I*_{1}*R*_{1}*V*_{1}= (⅔ A)(3 Ω)*V*_{1}= 2 V*V*_{2}=*I*_{2}*R*_{2}*V*_{2}= (⅔ A)(9 Ω)*V*_{2}= 6 V*V*_{3}=*I*_{3}*R*_{3}*V*_{3}= (⅔ A)(6 Ω)*V*_{3}= 4 VStarting 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.

- 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.