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Mathematics 6 Online
OpenStudy (moongazer):

What is the total resistance ? (sorry if this question is not allowed here) I'll just remove it if someone say that I should remove this. Thanks :)

OpenStudy (moongazer):

OpenStudy (moongazer):

I made R1 R2 R3 parallel. then get its resistance. then add it to R3 and R5 but it said I got the wrong answer I think I made a mistake to in choosing which resistors are parallel or in series to each other

OpenStudy (anonymous):

r3,r4, and r5 are series therefore : r345 = r3 + r4 + r5 r1, r2, and r345 are parallel therefore: rt = 1/((1/r1)+(1/r2)+(1/r345))

OpenStudy (anonymous):

i agree with @Hidden_Twilight

OpenStudy (anonymous):

Observe that\(\large R_3\), \(\large R_4\), and \(\large R_5\) are in series, so define \(\large R_6 = R_3+R_4+R_5\). Then we're left with \(\large R_1\), \(\large R_2\), and \(\large R_6\), which are parallel, so \(\large \dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_6}\).

OpenStudy (anonymous):

Ah, got ninja'd by @Hidden_Twilight. XD

OpenStudy (anonymous):

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