A 9.0-volt battery whose internal resistance r t is 0.50 ? is connected to the c
ID: 1397481 • Letter: A
Question
A 9.0-volt battery whose internal resistance rt is 0.50 ? is connected to the circuit below.
Component
Value (?)
R1
10.0
R2
6,0
R3
8.0
R4
4.0
R5
5.0
rt
0.50
a. How much current is drawn from the battery?
b. What is the terminal voltage of the battery?
c. What is the current in the 6-? resistor?
I'm looking for a clear explanation with formulas, and how you substituted in values, and why you did what you did. I'm having a hard time with resistors and learn best from examples and repitition. I appreciate the help. I will rate the best answer.
Component
Value (?)
R1
10.0
R2
6,0
R3
8.0
R4
4.0
R5
5.0
rt
0.50
10.0 8.0 6.0 4.0 t- 5.0 r= 0.50 =9.0 VExplanation / Answer
as we can see from the figure, 8 ohm and 4 ohm resistor are parallel
1/equivalant resistor of parallel resistor = 1/R1 + 1/R2
1/Req1 = 1/8 + 1/4
Req1 = 2.67 ohm
Req1 and 6 ohm resistor are in series
equivalent resistance of series resistance = R1 + R2
Req2 = Req1 + 6 ohm
Req2 = 2.67 + 6
Req2 = 8.67 ohm
Req2 is parallel with 10 ohm resistor
1/Req3 = 1/Req2 + 1/10
1/Req3 = 1/8.67 + 1/10
Req3 = 4.644 ohm
Req3, 5 ohm and 0.50 ohm resistors are series
Req4 = Req3 + 5 + 0.5
Req4 = 4.644 + 5 + 0.5
Req4 = 10.144 ohm
by ohm's law
V = I * R
9 = I * 10.144 ohm
I = 0.887 A
a) Current drawn from the battery = 0.887 A
since battery also has an internal resistance of 0.5 ohm so its terminal voltage will be less than 9 V
terminal voltage = 9 - I * r
terminal voltage = 9 - 0.887 * 0.5
b) terminal voltage of the battery = 8.5565 V
current in the 10 ohm resistor = 8.5565 /10
current in the 10 ohm resistor = 0.85565 A
current in the 6 ohm resistor = 0.887 - 0.85565
c) current in the 6 ohm resistor = 0.03135 A
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