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I'm confused on a homework question in my intro to electrical systems class. I have a simple dc circuit with a voltage source of 4.7 volts in series with a 40 ohm resistor, that then connects to a 50 ohm resistor paralleled with a diode in series with a 20 ohm resistor that then connects back to the negative terminal of the voltage source. The problem is that I've calculated the current through the first resistor as 117.5 mA, and I'm trying to find the current through the diode, which after getting the question wrong the homework program tells me the answer is 45.3 mA. Yet that would mean the 50 ohm resistor has 72.2 mA flowing across, which is greater than the current flowing across the 20 ohm resistor.

Why is it that a paralleled resistor does not have more current flowing across the path with less resistance?

what affect does the diode have on current?

assuming the diode is ideal and has voltage drop of .6 V

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    \$\begingroup\$ 1 schematic worth 10000 words. Sometimes more. \$\endgroup\$ – Eugene Sh. Oct 22 '15 at 20:46
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    \$\begingroup\$ Also, it would show you're ready to put some effort to solve the problem, which is a big step towards getting an answer. \$\endgroup\$ – Dmitry Grigoryev Oct 22 '15 at 20:51
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    \$\begingroup\$ Current is never 'across', it's 'through'. You really need to get that sorted. \$\endgroup\$ – user1844 Oct 22 '15 at 21:35
  • \$\begingroup\$ Can someone just walk me through how you get the current I3? \$\endgroup\$ – zack Oct 22 '15 at 21:39
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KCL for the junction is $$I_1=I_2+I_3$$ The voltage on the junction is \$V_1\$ Then $$I_1=(4.7-V_1)/40$$ $$I_2=V_1/50$$ $$I_3 = (V_1-0.6)/20$$

Solving: $$ (4.7-V_1)/40 = V_1/50 + (V_1-0.6)/20 $$ giving us $$V_1=1.55V$$ Hence $$I_3=(1.55-0.6)/20=0.0475A=47.5mA$$ Close enough to \$45.3\$..`.

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  • \$\begingroup\$ okay thank you! ya I guess I just misunderstood the use for I=V/R, and will use Ohm's law from now on. because now that means that the majority of the current is flowing through the 20 ohm resistor which makes sense! \$\endgroup\$ – zack Oct 22 '15 at 21:57
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The problem is that I've calculated the current across the first resistor as 117.5 mA,

This is not correct. It applies only if the whole 4.7 V is applied across the first resistor. From your description it sounds like the 4.7 V is shared between several resistors and a diode, so you can't assume that the voltage across the first resistor is 4.7 V.

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  • \$\begingroup\$ maybe I wasn't clear. the first resistor of 40 ohms is in series with the voltage source. the current flows directly from the voltage source through only the 40 ohm resistor, and then the circuit splits after this resistor. \$\endgroup\$ – zack Oct 22 '15 at 21:28
  • \$\begingroup\$ That doesn't match what is in your schematic. In your schematic it shows that the current flows through the 40-ohm resistor, and then it flows through the other resistors and the diode. Therefore the voltage across the 40-ohm will not be the full 4.7 V, and the current wil be less than 117.5 mA. \$\endgroup\$ – The Photon Oct 22 '15 at 22:17
  • \$\begingroup\$ ya i just sort of misunderstood how V=IR can be applied, my mistake. I feel like my EL systems teacher doesnt actually teach me anything... lol. Thank you for the help! I've got it sorted out now. \$\endgroup\$ – zack Oct 22 '15 at 22:18

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