Ravster
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#1
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#1
hey, so I'm stuck on part b.

I know for part a) the voltage is: (0.24A)(15) = 3.6V

and for part b) I did 6 / 4 = 1.5 A but this is wrong??
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atsruser
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(Original post by Ravster)
hey, so I'm stuck on part b.

I know for part a) the voltage is: (0.24A)(15) = 3.6V

and for part b) I did 6 / 4 = 1.5 A but this is wrong??
Note that since the 4 ohm and parallel pair are in series, then:

6 = V_{4 \Omega} + V_{\parallel}

and you already know V_{\parallel}
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wew.lad
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Try to do a Kirchhoff's second law loop. Something useful for me when dealing with electricity questions was learning the correct definition of Kirchhoff's second law, where the some of all p.ds in any loop = 0.

There's three (or four if you could consider the parallel component as one part) potential loops for this circuit. The equation for one going through the 4 ohm resistor and resistor R would be:

6 - 4I1 - RI2 = 0

Basically, find the single loop or path in the circuit that leaves I1 as the only remaining variable.
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