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Hello Electrical Engineering stack exchange! I was taking a physics practice test and I ran into the following circuit

enter image description here

"What is the current across the 2 olhm resistor when the switch is open?"

Now I know that V=IR but I don't really have any idea on how to tackle this question because of the opposing batteries. I've figured that once the capacitor is filled that no current should flow there so the switch doesn't matter right?, but I'm just stuck. Since this is a practice problem I'm looking for an explanation on how to tackle these circuits with opposing batteries. Thanks!

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You simply add the battery voltages, observing polarity. The resulting voltage to use when calculating currents is therefore 6 volts.

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  • \$\begingroup\$ Quick response! Anyways if I add the batteries I should remove the 4V battery and replace the 10V battery with one of 6V in the same place? \$\endgroup\$ – John Jul 23 '17 at 3:06
  • \$\begingroup\$ @John, Have you learned Kirchoff's Voltage Law (aka KVL)? If so, just apply that. Otherwise, the short answer is, it won't end up mattering how you rearrange the two batteries and two resistors, since there's only one loop in the circuit (for the open switch case). \$\endgroup\$ – The Photon Jul 23 '17 at 3:47
  • \$\begingroup\$ You could re-draw the circuit with a 6v battery, and applying the same logic you could replace the two resistors with a 6ohm resistor which gives you the answer instantly!!! \$\endgroup\$ – Solar Mike Jul 23 '17 at 10:23

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