Basic Electricity: Learn Basics of Electricity

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Can someone please check my answers here, there are loads of formulaes been give that I haven't used so I'm not sure if they are right or not, thanks.
 

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The first problem (resistors in series) is done correctly, but the second (resistors in parallel) is not. Recalculate the total resistance in that second circuit. Hint: When resistors are in parallel, the total resistance will be less than that of any single resistor.
 
The total resistance of resistors in parallel can be calculated with

[tex]\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}[/tex]

this can be changed to

[tex]R_p=\frac{R_1R_2}{R_1 + R_2}[/tex]
 
andrevdh said:
The total resistance of resistors in parallel can be calculated with

[tex]\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}[/tex]

this can be changed to

[tex]R_p=\frac{R_1R_2}{R_1 + R_2}[/tex]

When I do that I get 6x2 = 12ohm 12ohm/8 ohms = 1.5 ohms, is that right?
 
Yes. That seems to be it.
 
Am I right in thinking that if you have a 12v battery and short it the current would be very high?
 
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Not neccessarily. The amp-hours rating is a complicated affair. It is the amount of current it can supply (chosen for "normal" [designed] operating conditions) for a certain set period (I think it is normally for 20 hours, but I might be wrong) before the battery is completely drained and dies out. So you might get more out of it for a shorter period but only to a certain limit (everyone has his limitations!).
 
Ok I think I'm not sure If I've got this right. I worked out I1 and I2 using fractions.
 

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The current through the 2 ohm resistor is the current through all of the circuit since

[tex]I = I_1 + I_2[/tex]

So [tex]I[/tex] will be be determined by the total resistance of the circuit and the voltage of the power supply.
 
Ok we went through that question today and I got A-F right but G was 0.9A.
 
12/5 does not equal 1.2, maybe you sent us the wrong version of the problem, or the teacher did it wrong
 
Ah sorry I put 12v instead of 6, doh.