Facebook
Google
GitHub
LinkedinBasic Electricity
Series DC Circuits Practice Worksheet with Answers
27 questions By Tony R. Kuphaldt
-
Question 1 of 27
In this circuit, three resistors receive the same amount of current (4 amps) from a single source. Calculate the amount of voltage “dropped” by each resistor, as well as the amount of power dissipated by each resistor:
Reveal answerE1 Ω = 4 volts
E2 Ω = 8 volts
E3 Ω = 12 volts
P1 Ω = 16 watts
P2 Ω = 32 watts
P3 Ω = 48 watts
Follow-up question: Compare the direction of current through all components in this circuit with the polarities of their respective voltage drops. What do you notice about the relationship between current direction and voltage polarity for the battery, versus for all the resistors? How does this relate to the identification of these components as either sources or loads?
Notes:The answers to this question should not create any surprises, especially when students understand electrical resistance in terms of friction: resistors with greater resistance (more friction to electron motion) require greater voltage (push) to get the same amount of current through them. Resistors with greater resistance (friction) will also dissipate more power in the form of heat, given the same amount of current.
Another purpose of this question is to instill in students’ minds the concept of components in a simple series circuit all sharing the same amount of current.
Challenge your students to recognize any mathematical patterns in the respective voltage drops and power dissipations. What can be said, mathematically, about the voltage drop across the 2 Ω resistor versus the 1 Ω resistor, for example?
-
Question 2 of 27
The brightness of a light bulb - or the power dissipated by any electrical load, for that matter - may be varied by inserting a variable resistance in the circuit, like this:
This method of electrical power control is not without its disadvantages, though. Consider an example where the circuit current is 5 amps, the variable resistance is 2 Ω, and the lamp drops 20 volts of voltage across its terminals. Calculate the power dissipated by the lamp, the power dissipated by the variable resistance, and the total power provided by the voltage source. Then, explain why this method of power control is not ideal.
Reveal answerPlamp = 100 watts
Presistance = 50 watts
Ptotal = 150 watts
Follow-up question: note how in the original question I offered a set of hypothetical values to use in figuring out why a series rheostat (variable resistance) is not an efficient means to control lamp power. Explain how the assumption of certain values is a useful problem-solving technique in cases where no values are given to you.
Notes:Discuss the concept of energy conservation: that energy can neither be created nor destroyed, but merely changed between different forms. Based on this principle, the sum of all power dissipations in a circuit must equal the total amount of power supplied by the energy source, regardless of how the components are connected together.
-
Question 3 of 27
A modern method of electrical power control involves inserting a fast-operating switch in-line with an electrical load, to switch power on and off to it very rapidly over time. Usually, a solid-state device such as a transistor is used:
This circuit has been greatly simplified from that of a real, pulse-control power circuit. Just the transistor is shown (and not the “pulse” circuit which is needed to command it to turn on and off) for simplicity. All you need to be aware of is the fact that the transistor operates like a simple, single-pole single-throw (SPST) switch, except that it is controlled by an electrical current rather than by a mechanical force, and that it is able to switch on and off millions of times per second without wear or fatigue.
If the transistor is pulsed on and off fast enough, power to the light bulb may be varied as smoothly as if controlled by a variable resistor. However, there is very little energy wasted when using a fast-switching transistor to control electrical power, unlike when a variable resistance is used for the same task. This mode of electrical power control is commonly referred to as Pulse-Width Modulation, or PWM.
Explain why PWM power control is much more efficient than controlling load power by using a series resistance.
Reveal answerWhen the transistor is on, is acts like a closed switch: passing full load current, but dropping little voltage. Thus, its “ON” power (P = I E) dissipation is minimal. Conversely, when the transistor is off, it acts like an open switch: passing no current at all. Thus, its “OFF” power dissipation (P = I E) is zero. The power dissipated by the load (the light bulb) is the time-averaged power dissipated between “ON” and “OFF” transistor cycles. Thus, load power is controlled without “wasting” power across the control device.
Notes:Students may have a hard time grasping how a light bulb may be dimmed by turning it on and off really fast. The key to understanding this concept is to realize that the transistor’s switching time must be much faster than the time it takes for the light bulb’s filament to fully heat or fully cool. The situation is analogous to throttling the speed of an automobile by rapidly “pumping” the accelerator pedal. If done slowly, the result is a varying car speed. If done rapidly enough, though, the car’s mass averages the “ON”/“OFF” cycling of the pedal and results in a nearly steady speed.
This technique is very popular in industrial power control, and is gaining popularity as an audio amplification technique (known as Class D). The benefits of minimal wasted power by the control device are many.
llaboutcircuits.com/worksheets/series-dc-circuits/. In question 5, total series resistance is sum of each resistor, not product. Correct both question and answer
Hello Everybody and thanks for your fantastic articles!
I’m having troubles solving Excercise 19 of your Series circuits problems (https://www.allaboutcircuits.com/worksheets/series-dc-circuits/): For this excercise I’m getting a power dissipation for the heater of 403 W instead of 321.1 W.
Since I’m learning electronics I’m surely doing something wrong… Here how I’m reasoning:
1. With 110 V and a power dissipated of 500 W, I can calculate the current in the circuit, that should be I=P/V 500/110 = 4.545 A
2. Knowing that the current in the circuit is 4.545 A I can use Ohm’s law to calculate the resistance of the heater: R=V/I 110/4.545 = 24 Ohm
3. Now, longing the wires it’s the same as adding 6 Ohm of resistance, for a total resistance of 30 Ohm (24 + 6) so the current in the circuit will change accordingly. To find the new amount of current I use Ohm’s law once again: I=V/R 110/30 = 3.666 A
4. Having a new current value of 3.666 A, I can now calculate power dissipation P=V*I 110*3.666 = 403 W
Could you please help me understand where I’m wrong?
Thank you so much!
Stefano.