The letter e is the most frequently used letter in english prose, and the letter z is the least frequently used. a friend of yours doing a sociology experiment believes that this may not necessarily be true of the writings of first-year college students. to test his theory, he asks you to write a program that will take a text file and print, for each letter of the english alphabet, the number of times the letter appears in the file. you can use the sample.txt provided. try your program on other text files as well. hint: use an integer array of size 128, and use the ascii values of letters to index into the array to store and retrieve

2.72 A

7.41 V

3.95 V

Explanation:

Let's say I₁ is the current in the 3Ω resistor, I₂ is the current in the 10Ω resistor, and I₃ is the current in the 2Ω resistor.

Apply Kirchoff's voltage law to the top loop:

20 − 6 I₂ − 10 I₂ − 3 I₁ = 0

20 − 16 I₂ − 3 I₁ = 0

Apply Kirchoff's voltage law to the bottom loop:

20 − 4 I₃ − 2 I₃ − 3 I₁ = 0

20 − 6 I₃ − 3 I₁ = 0

Apply Kirchoff's current law at either junction:

I₁ − I₂ − I₃ = 0

I₁ = I₂ + I₃

Three equations, three variables.  To solve, let's multiply the first equation by 3 and the second equation by 8:

60 − 48 I₂ − 9 I₁ = 0

160 − 48 I₃ − 24 I₁ = 0

Add:

220 − 48 I₂ − 48 I₃ − 33 I₁ = 0

220 − 48 (I₂ + I₃) − 33 I₁ = 0

Substitute and solve:

220 − 48 I₁ − 33 I₁ = 0

220 − 81 I₁ = 0

I₁ = 2.72

Therefore, from the first equation:

I₂ = 0.741

And from either the second or third equation:

I₃ = 1.98

So the voltage across the 10Ω resistor is:

V = IR

V = (0.741 A) (10 Ω)

V = 7.41 V

And the voltage across the 2Ω resistor is:

V = IR

V = (1.98 A) (2 Ω)

V = 3.95 V