I already know the answer of question 1, 2 and 3.
1) 448 ohm
2) 1.6 V
3) 0.13 V
About the question 4, why 6V/300 = 0.02A = 20mA is wrong? Can I use current divider like 0.01339 * (3.3 * 10^3 / (300 + 3.3 * 10^3))? But, the answer still wrong.
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2\$\begingroup\$ Start by showing how you might begin to address question 3. Nobody's going to help you without some means of seeing where you go wrong. \$\endgroup\$– Andy akaJan 31, 2020 at 9:11
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1\$\begingroup\$ Break it up into smaller problems like you did for questions 1, 2 and 3. If you’ve gotten as far as answer #3 I’m sure you can apply the same techniques to manage the other two as well. \$\endgroup\$– StarCatJan 31, 2020 at 9:28
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1\$\begingroup\$ What is the voltage across the 300 ohm resistor? \$\endgroup\$– user253751Jan 31, 2020 at 10:56
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\$\begingroup\$ To further nudge you in the right direction, the 6V isn't dropped entirely over the 300Ω resistor; if it was, that's saying the resistor is connected between 6V and ground. So you can either (a) Find the drop across the 300Ω, so I = (6-Vx) / 300, or (b) Find the current over that entire resistor branch. That current is the very same that passes through 300Ω, so I = 6 / (the equivalent resistance of that branch) \$\endgroup\$– OrotaviaJan 31, 2020 at 13:31
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1 Answer
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By superposition, the current thru R=300 splits into two paths that results in the sum for the answer in 4).
R1 = 300+100+100 = 500
R2 = 300+3k+1k+100 = 4k4
I(R=300)=6/500 +6/4k4
