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I'm having trouble with this circuit. I'm trying to find the current through and voltage drop across the resistors when the switch is open and closed.

When the switch is open, I've found that the voltage drop across the upper resistor and lower resistor is 10V each. I'm not sure how to deal with the second current source when the switch is closed.

Any help is appreciated!

enter image description here

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  • \$\begingroup\$ Maybe start with the resistor that is common for both current sources, find the current through that. \$\endgroup\$ – Dampmaskin Apr 15 '17 at 23:47
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    \$\begingroup\$ Just use KCL at the node between the resistors. 10 mA comes in one branch. 5 mA comes in the other branch. So you know right away how much is flowing out the third branch. \$\endgroup\$ – The Photon Apr 16 '17 at 0:08
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    \$\begingroup\$ When you close the switch, where does the extra 5 mA go? Can it go through the other current source? If not, what is left? So what is the current through the lower resistor? What is the current through the upper? What are the voltages across each? \$\endgroup\$ – WhatRoughBeast Apr 16 '17 at 1:56
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    \$\begingroup\$ I don't want to be the one opening that switch, there's no telling how high the voltage could go. \$\endgroup\$ – Spehro Pefhany Apr 16 '17 at 9:49
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Q) What do you know about current sources?

A) That the current at their terminals is always the same as the number written beside them.

Q) So what is the current through the top 1k resistor, whether the switch is open or closed?

A) As it's connected only to the 10mA source, it's always 10mA.

Q) So where does the extra 5mA flow when you close the switch?

A) The only place it can flow ...

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  • \$\begingroup\$ Other useful knowledge: a two-pin current source has the same current out pin #1 as comes in pin #2. \$\endgroup\$ – Whit3rd Apr 16 '17 at 8:58
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When the switch is turned on. According to Kirchhoff's law. In the node which connects both resistors and switches:

IR lower + Iswitch - IR upper = 0;

Iswitch = I5mA source = 5mA;

IR upper = I10mA source = 10mA;

IR lower =IR upper - Iswitch = 10-5 = 5 mA;

UR upper= 10mA * 1 kOhm = 10 V;

UR lower= 5mA * 1 kOhm = 5 V.

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