Ans: I must rephrase this question into an acceptable form! One possible expression would be: What current results when 100 V is applied across a 20 ohm resistor?
Simple application of Ohm's law yields I=V/R =(100V)/(20 ohm)=5.0 A.
Ans: Let's consider the case of just two resistors.
In series, the two resistors will have the same current through each. In parallel, they will have the same voltage across each.
In series, the effective resistance is the sum of resistances; thus the effective resistance is always larger than either one. In parallel, the reciprocal of the effective resistance is the sum of the reciprocal resistances. It turns out that the effective resistance is always smaller than either one.
In series, the larger resistor will have a proportionally larger voltage across it. In parallel, the smaller resistor will have a larger current flowing through it.
In series, the two resistors have only one end tied together, with nothing else attached at that point. In parallel, the two resistors have both ends tied together, with nothing in between.
In parallel, each resistor has access to the same voltage (good for household wiring). In series, if one of the resistors "breaks" (as in a lamp "burning out"), no current flows through any resistor (bad for useful wiring).
Perhaps you can think of more...
Ans: There are two approaches that I considered in tackling this problem. One was to sum the total power requirements for all the appliances and then determine the minimum number of circuits that would suffice. The other was to consider a typical household circuit and evaluate how to distribute those among the appliances. Either approach requires knowledge of the power demand for each device. By looking at the list from the handout or the online list from the City of Newark (and referring to a catlog in my office for the soldering iron), I found: (A voltage of 115 V was assumed.)
|Appliance||Power Rating||Current Drawn|
|T.V.||100 W||0.9 A|
|space heater||1500 W||13 A|
|hairdryer||650 W -- 1500 W||5.7 A -- 13 A|
|radio||25 W (a small one!)||0.2 A|
|soldering iron||200 W (heavy duty)||1.7 A|
|microwave oven||1500 W||13 A|
The maximum current required by any of the appliances is 13 A, so circuits protected by 15 A breakers are a possibility. Two separate circuits for each bedroom will be needed, one for the outlet running the space heater and one for the hairdryer. The television has fairly modest power requirements and could be added to either of these circuits. However, if the space heater and hairdryer are plugged into the same outlet, the circuit breaker will trip!
Two 20 A circuits to each room would provide some extra capability for lighting, etc. In a pinch, a small hairdryer (650 W), space heater, and T.V. could be run from just one 20 A circuit since the total current requirement would be 19.6 A, but this is probably too close for routine operation.
Since the microwave oven requires the same amount of power as a space heater or heavy-duty hair dryer, a third circuit will be required for that room.
I liked this problem!
Ans: Again, application of Ohm's law yields I=V/R =(12V)/(8 ohm)=1.5 A.
Ans: I would rate this as an easy question, since it requires only a simple calculation from just one formula. The relationship between power, current, and voltage is P=IV; thus the power consumption is (12A)(120V)=1440W.
Ans: Evaluation of the current load for a power-rated device follows from P=IV; that is I=P/V. Thus the current drawn by an 800W toasted connected to a 115V source is I=(800W)/(115V)=7.0A. Similarly, the iron draws 8.7A and the food processor draws 6.1A. The total current drawn by the three devices sums to 21.8A; the fuse is rated for only 15A and will probably blow if the three devices are operated simultaneously.
Comments, suggestions, or requests to firstname.lastname@example.org.
Last updated Oct. 15, 1996.
Copyright George Watson, Univ. of Delaware, 1996.