3. Calculate the following: a) voltage drop across Rs; b) current flow through Rs; and c) power delivered on Rg. ist 5 mA R4 R2 180 ww R1 140 ww 100 R5 R7 6 KO 10 kG Ⓒ Vst 4 V 0 ww R3 2k0 112 5mA RG 410
Quantization and Resolution
Quantization is a methodology of carrying out signal modulation by the process of mapping input values from an infinitely long set of continuous values to a smaller set of finite values. Quantization forms the basic algorithm for lossy compression algorithms and represents a given analog signal into digital signals. In other words, these algorithms form the base of an analog-to-digital converter. Devices that process the algorithm of quantization are known as a quantizer. These devices aid in rounding off (approximation) the errors of an input function called the quantized value.
Probability of Error
This topic is widely taught in many undergraduate and postgraduate degree courses of:
![3. Calculate the following: a) voltage drop across Rs; b) current flow through Rg; and c) power delivered
on Rg.
ist
5mA
ww
R4
180
240
R3
2k0
ww
R1
140
112
5 mA
Re
ww
ww
100
R5
R7
10 kG
6 KO
ww
RO
4K0
Vs1
4 V](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F75150320-ecb1-43ba-9fdd-271ebc30937c%2F6127bfa1-a30e-434c-926e-4a89c8a8ba45%2F5lhchi_processed.jpeg&w=3840&q=75)
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Solved in 4 steps with 1 images
![Blurred answer](/static/compass_v2/solution-images/blurred-answer.jpg)
3. Calculate the following: a) voltage drop across R8; b) current flow through R8; and c) power delivered
on R8.
how did you get i=3.17mA, i1=2.10mA, using kvl..
the final answer are Vb=0.317V
Pb=1mW
please show me step by step solution, thank you
![i
R1
140
Vs1
4 V
i1
✓✓✓i-i1
R2
180
R5
6KD
Ist
5mA
11-5
R3
2k0
112
5mA
i-i1-5
ww
RG
4k0
84
240
ww
R7
10 k
100](https://content.bartleby.com/qna-images/question/75150320-ecb1-43ba-9fdd-271ebc30937c/1bd75488-148b-42a8-9415-9ce4b2896b6a/0f5kpgs_thumbnail.png)
![Step 2
Applying KVL in lower part of circuit.
−4+i₁ +6i - i₁ + 4i − i₁ − 5 + 10i-i₁ +0.1i = 0 - 4 +i₁ +6i − 6i₁ + 4i - 4i₁
-20 +10i - 10i₁ +0.1i = 020.1i — 19i₁ = 24.....
.eqn 1
Step 3
Applying KVL in upper part of circuit.
−4+i+i₁ + 2i₁ −5+2i₁ +0.1i = 0 − 4 + i +i₁ + 2i₁ − 10 + 2i₁ + 0.1i = 01.1
i + 5i₁ = 14.
eqn 2
Step 4
Solving equation 1 and equation 2.
i = 3.17 mA, ₁ = 2.10 mA](https://content.bartleby.com/qna-images/question/75150320-ecb1-43ba-9fdd-271ebc30937c/1bd75488-148b-42a8-9415-9ce4b2896b6a/5brgwib_thumbnail.png)
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