Electrical Elements word problems

Thread Starter


Joined Nov 29, 2011
I just wanna know how to solve these problems and what formula to use

1. If a capacitor is rated for 200Volts dc, what is its effective ac working voltage?
a. 50V
b. 200V
c. 100V
d. 400V

2. Suppose a 1-kΩ resistor will dissipate 1.05W, and you have many 1-W resistors of all common values. If there's room for 20 percent resistance error, the cheapest solution is to use:
A. Four 1 kΩ, 1-W resistors in series-parallel
B. Two 2.2kΩ, 1-W resistors in parallel
C. Three 3.3kΩ, 1-W resistors in parallel
D. One 1kΩ, 1-W resistor, since manufacturers allow for a 10 percent margin of safety
Last edited:

Thread Starter


Joined Nov 29, 2011
1. no luck in google
all i know is the root mean squared but the given is dc
200V * 0.707 = 141V not in choices

2. three 3.3k values in parallel equals 1.1kΩ so its still at the tolerance limit

but i dont know whats it got to do with the wattage(1.05 given) and the cheaper solution as stated in the question


Joined Mar 14, 2008
1. So you pick the closest voltage in the listed choices that gives a peak AC voltage (1.4 times RMS) no greater than 200V.

2. The cheapest solution uses the minimum number of resistors. Wattage in parallel add, thus two equal value 1-watt resistors in parallel can dissipate 2W.


Joined Nov 9, 2007
I just wanna know how to solve these problems and what formula to use
Good engineering is about attention to detail. That applies to English as well as the technicalities.

That means read properly what crutschow has bothered to tell you.

Your second question is very poorly written, but if it means that you should select from a combination of 1 or 2 or 3 or 4 1 watt resistors to meet a dissipation criterion of 1.05 watts, a single 1 watt resistor will not cut the mustard. It is not permissible to eat into safety factors as part of a design.


Joined Feb 19, 2009
i dunno but
1. is probably 200Vdc = 200Vac since ac is at rms value

2. ok its D.
The voltage rating of capacitor is the potential at which the dielectric breaks down.

What is the peak voltage of AC when the RMS value is 200V ?

Second: Components are never run at 100% or more of their capacity when designing a circuit. Any additional capacity the manufacturer may have added ( sometimes it is less, rather than more, that 10% ) is to allow the circuit to function in the real world, where voltages fluctuate at times, temperatures are taken to extremes, and mechanical vibration/shock is a continual issue.

Building a circuit to save a few cents in the manner of "wishful tolerances" will cost more in the form of refunds/recalls/lawsuits if a product you designed gets to market built with that sort of thinking.