Talk:A-level Physics (Advancing Physics)/Potential Dividers/Worked Solutions

Potential Dividers/Worked Solutions
I think the last question is wrong...

2.102.61.95 (discuss) 13:39, 31 December 2011 (UTC)

Oh, good, it wasn't just me who thought that. xD


 * Just by thinking about the ratios between the resistances, I made it $$V_{3\mho, 2\mho} = 9.0 \times \frac{\frac{1}{3} + \frac{1}{2}}{0.8 + \frac{1}{3} + \frac{1}{2} + 1} \approx 2.85\mbox{ V}$$
 * Also, the circuit current would only be 2A if the total series resistance was 4.5 ohms, and it is actually 2.63333 ohms. Recent Runes (discuss • contribs) 20:56, 12 January 2012 (UTC)