Chapter5 Flashcards
In the Series Parallel circuit below, what is the total resistance and how is it calculated?
Answer : 35 ohms
This one is straight forward.
Simply add the series resistance (R3) to the total parallel resistance (R1 & R2).
In the previous question, we determined the total parallel resistance of the two 40 ohm resitors is 20 ohms.
So, series resistance (15 ohm) + Parallel resistance (20 ohm) = 35 ohm
15 Ω + 20 Ω = 35 Ω Total Resistance
A circuit operates with 5 Amperes at a voltage of 120 AC Volts. What is the power of the circuit?
Answer : 1.5A
P = I x E
P = 5A x 120V = 600 VA (VA means Volts x Amps)
A 3 phase circuit operates at 208 AC Volts and with a current of 9 A. What is the power in watts?
Answer : 3186 W
P = I x E x 1.7 =
P = 9A x (208V x 1.7) =
P = 9A x 354V = 3186 W (also apparent power VA)
How many Amperes is 1,200 mA equal to?
Answer 1.2A
We want to convert mA (milliamperes) to A (Amperes)
1,200 mA, move the decimal point, 3 places to the left.
1.2 A
In the circuit below, what is the parallel resistance R1 + R2?
Answer : 30 ohm
Lets start with the parallel resistance.
Notice, R1 and R2 are in parallel and they have different resistance values.
Let me show you how to easily calculate the total parallel resistance of R1 and R2.
12 Ohm (R1 + R2) + 18 Ohm (R3) = 30 Ohm
In the circuit below, will all the Ampere meters read: the same current or will each meter read differently?
Answer : all amp meters will read the same.
In the circuit below, how much is the Voltage drop across R3?
Answer : 72V
The circuit current (I) was calculated, using the following formula:
I = E / R I = 120V / 30 Ohms = 4A
Now, lets proceed and calculate the voltage drop (IR) drop across R3.
Using the EIR circle again, cover what you want to find (E). What remains, is formula for voltage drop.
I x R = E
4A x 18 ohms = 72V (voltage across R3
What is the resistance of 1,200 ft of stranded #6 AWG aluminum wire?
Answer : 0.97 ohms
Chapter 9, Table 8
1,000 ft (1kFT) - #6 AWG stranded aluminum wire - has 0.808 ohms of resistance per 1,000 feet (1kFT)
Therefore, 1200 ft, of stranded #6 AWG (aluminum), the wire resistance is calculated as follows:.
0.808 ohms x 1.2 (1,200 ft) = 0.97 ohms (rounded off)
What is the resistance of 600 ft of uncoated #10 AWG solid wire?
Answer : 0.73 ohms Chapter 9 Table 8
Chapter 9 Table 8
Remember, all wire is automatically copper, unless we are specifically told otherwise.
1,000 ft (1kFT) - #10 AWG solid uncoated copper wire - has 1.21 ohms of resistance per 1,000 feet (1kFT).
Therefore, 600 ft, of #10 AWG (solid, uncoated copper), the wire resistance is calculated as follows:
1.21 ohms x .6 (600 ft length) = 0.73 ohms (rounded up to hundredth (2) decimal places)
Two 40 Ω (ohm) resistors are in parallel. What is the total resistance?
Answer : 20 ohms
