Ask question

Ask question

All categories

3 years ago

14

A circuit contains four 100 ohm resistors connected in series. If you test the circuit with a DMM you should read a total resist

ance value of
A. 250 ohms.
. B. 1000 ohms
C. 400 ohms
D. 100 ohms

1 answer:

olga nikolaevna [1]3 years ago

8 0

C is the correct answer

You might be interested in

What oil specification is used to express the thickness (viscosity) of the oil?

Marianna [84]

Answer:

Viscosity is notated using the common classification “XW-XX”. The number preceding the “W” (winter) rates the oil's flow (viscosity) at zero degrees Fahrenheit (-17.8 degrees Celsius). The lower the number, the less the oil thickens in cold weather.

7 0

2 years ago

Tech A says that water soaked brake shoes can be a cause of brake fade. Tech B says that disc brakes dissipate heat faster than

alina1380 [7]

Explanation:

perturbateur ( le temps, le lieu, les personnages[description], la victime, l'enqueteur )

les peripeties

le denouement

5 0

1 year ago

Riding a bike is an example of a procedural memory.

Viefleur [7K]

Answer:

True reading a bike would be an example of procedural memory

4 0

3 years ago

Read 2 more answers

When brazing, the lowest effective brazing temperatures possible should be used to minimize the effects of heat on the base meta

nordsb [41]

Answer:

False

Explanation:

5 0

2 years ago

Rsidential Solar Solution:a. A type of photovoltaic solar panel manufactured in China receives a rating of 250W. The rating proc

aksik [14]

Answer:

a) $\eta = 13.455\%$ , b) $E_{day} = 812.716\,kJ$ , c) $C_{month. total} = 19.505\, USD$ , d) $t = 40.588\,years$

Explanation:

a) The area of the solar panel is:

$A = (20\,ft^{2})\cdot (\frac{0.3048\,m}{1\,ft} )^{2}$

$A = 1.858\,m^{2}$

The energy potential is determined herein:

$\dot E_{o} = (1000\,\frac{W}{m^{2}} )\cdot (1.858\,m^{2})$

$\dot E_{o} = 1858\,W$

The efficiency of the solar panel is:

$\eta = \frac{\dot E}{\dot E_{o}}\times 100\%$

$\eta = \frac{250\,W}{1858\,W}\times 100\%$

$\eta = 13.455\%$

b) The energy generated by the solar panel is presented below:

$E_{day} = (0.135)\cdot (150\,\frac{W}{m^{2}} )\cdot (20\,ft^{2})\cdot \left(\frac{0.3048\,m}{1\,ft} \right)^{2}\cdot (6\,h)\cdot (\frac{3600\,s}{1\,h} )\cdot (\frac{1\,kJ}{1000\,J} )$

$E_{day} = 812.716\,kJ$

c) The energy generated per month and per panel is:

$E_{month} = 30\cdot E_{day}$

$E_{month} = 30 \cdot (812.716\,kJ)\cdot \left(\frac{1\,kWh}{3600\,kJ} \right)$

$E_{month} = 6.773\,kWh$

Monthly energy savings due to the use of 18 panels are:

$C_{month, total} = 18\cdot E_{month}\cdot c$

$C_{month, total} = 18\cdot (6.773\,kWh)\cdot (\frac{0.16\,USD}{1\,kWh} )$

$C_{month. total} = 19.505\, USD$

d) The payback of the solar energy system is:

$t = \frac{9500\,USD}{12\cdot (19.505\,USD)}$

$t = 40.588\,years$

6 0

3 years ago