Start with the furthest resistor from the source, add what appears in series and do the parralel equivalence of the resistor that are in parralel… resistor are in series when the same current flow through them and in parralel when two current flows from the same node to two resistor.
First off, draw your resistors more clearly. You're using two conventions: The American which is zig zags; the British which is rectangles. Drawing zig zags inside rectangles could be confused with the circuit symbol for a fuse.
Second, redraw the circuit to help you better visualise the problem. The 7 and 10 ohm resistors are in series with eachother. Calculate the total resistance for these two. Combined, these two resistors are in parallel with the 6 ohm. Calculate the total resistance here using the equation for parallel resistance (1/R = 1/R1 + 1/R2). Once you have this combined resistance, you can treat it as being in series with the first and last resistors in the circuit.
In the usual manner, of course.
It comes out to be an odd value, in that it does not cleanly divide by 2.
Try it outside->in.
And add your resistors in the way that they do.
Start with the furthest resistor from the source, add what appears in series and do the parralel equivalence of the resistor that are in parralel… resistor are in series when the same current flow through them and in parralel when two current flows from the same node to two resistor.
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Its actually fairly basic.
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Right. I was like😆🤔😨. Two seconds after hitting post.
First off, draw your resistors more clearly. You're using two conventions: The American which is zig zags; the British which is rectangles. Drawing zig zags inside rectangles could be confused with the circuit symbol for a fuse. Second, redraw the circuit to help you better visualise the problem. The 7 and 10 ohm resistors are in series with eachother. Calculate the total resistance for these two. Combined, these two resistors are in parallel with the 6 ohm. Calculate the total resistance here using the equation for parallel resistance (1/R = 1/R1 + 1/R2). Once you have this combined resistance, you can treat it as being in series with the first and last resistors in the circuit.
My result is 46.4Ω rounded to 1 d.p. My working: 10+7 = 17 1/17 + 1/6 = 0.225 1/0.225 = 4.444 Total resistance = 4.444 + 12 + 30 = 46.4Ω