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

Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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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
Transcribed Image Text: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
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Follow-up Question

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
Transcribed Image Text: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
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
Transcribed Image Text: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
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