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I want to find the current through branch AB, CD, EF and GH and find out which batteries are being charged

However, the potential at A,C and E are the same(because B,D,F are equipotential and because of the way the batteries are arranged) and hence by ohm's law no current should flow through AC or CE. Consequently, the circuit across AB and CD is not a closed loop i.e it is an open circuit.

Why should any current flow through AB or DC at all?

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  • \$\begingroup\$ Voltage sources are never considered open circuits, even in equipotential states. \$\endgroup\$ – Ignacio Vazquez-Abrams Aug 17 '17 at 17:17
  • \$\begingroup\$ It's not an open circuit. Otherwise, I think you already have the answer. \$\endgroup\$ – Brian Drummond Aug 17 '17 at 17:23
  • \$\begingroup\$ The correct answer (given in the solution) is: 3A through AB, 4A through DC, 9A through EF and 8A though HG. Batteries w/20V and 15V(in EF) are being charged. \$\endgroup\$ – xasthor Aug 17 '17 at 17:32
  • \$\begingroup\$ I think your circuit doesn't match the circuit you were given. Looking at the rightmost loop (EFGH) you have two 15 volt batteries with opposite polarities in series so there should be no current in that loop. \$\endgroup\$ – Peter Bennett Aug 17 '17 at 17:45
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It sounds like the solution you're looking at is wrong, because your point is correct. There is no current through AB and DC. As far as the solution you gave in the comments:

3A through AB, 4A through DC, 9A through EF and 8A though HG. Batteries w/20V and 15V(in EF) are being charged

Either the solution is wrong or your schematic is wrong. The quickest way to check that is to realize that EF and HG represent essentially identical paths, but your solution has them with different currents.

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