# Homework Help: Basic Elecricity

1. Oct 28, 2006

### Energize

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.

#### Attached Files:

• ###### homework.JPG
File size:
10.1 KB
Views:
112
2. Oct 29, 2006

### Staff: Mentor

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.

3. Oct 29, 2006

### andrevdh

The total resistance of resistors in parallel can be calculated with

$$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$$

this can be changed to

$$R_p=\frac{R_1R_2}{R_1 + R_2}$$

4. Oct 29, 2006

### Energize

When I do that I get 6x2 = 12ohm 12ohm/8 ohms = 1.5 ohms, is that right?

5. Oct 29, 2006

### andrevdh

Yes. That seems to be it.

6. Oct 29, 2006

### Energize

Am I right in thinking that if you have a 12v battery and short it the current would be very high?

Last edited: Oct 29, 2006
7. Oct 29, 2006

### andrevdh

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!).

8. Oct 29, 2006

### Energize

Ok I think I'm not sure If I've got this right. I worked out I1 and I2 using fractions.

#### Attached Files:

• ###### homework2.JPG
File size:
10.8 KB
Views:
93
9. Oct 30, 2006

### Energize

Anyone?.........

10. Oct 30, 2006

### Staff: Mentor

11. Oct 30, 2006

### andrevdh

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

$$I = I_1 + I_2$$

So $$I$$ will be be determined by the total resistance of the circuit and the voltage of the power supply.

12. Oct 31, 2006

### Energize

Ok we went through that question today and I got A-F right but G was 0.9A.

13. Oct 31, 2006

### CPL.Luke

12/5 does not equal 1.2, maybe you sent us the wrong version of the problem, or the teacher did it wrong

14. Oct 31, 2006

### Energize

Ah sorry I put 12v instead of 6, doh.