PLEASE HELP ME with this question please, please, please!!?

Question

1. A potential difference of 1.5 volts occurs in most batteries. If the current produced is 12 amps. What is the resistance?
2. A potential difference of 12 V resulted when a charge of 600 ohms was placed on a terminal. What is the resistance?
3. What is the resistance of a light bulb if 2.4 A runs through it when 12 V is placed across the bulb?
4. What is the potential difference of a water heater element with a resistance of 32 ohms when the current is 6.8 A?
5. If 42 Amps places 7 ohms resistance on a terminal in a battery. What is the potential difference?
7: Voltage Current Resistance
20 V 4 A
120 V 15 OHMS
0.5 A 16 OHMS
10 V 400 MA
0.3 A 20 OHMS

Don't do number 7 because it came out wrong!

Thanks

Answer 1

The guy above is right. Remember Ohm's law!

Did you ever get taught the VIR triangle?

But here's my answers anyway.

In question two you said a charge of 600 ohms... did you mean amps?

1. - 0.125 ohms
2. - (if you meant amps) 0.02 ohms
3. - 5 ohms
4. 217.6 volts
5. 294 volts

Go here to find out more about Ohms law and the VIR triangle
http://www.kpsec.freeuk.com/ohmslaw.htm

Or here to find an ohms law calulator:
http://ourworld.compuserve.com/homepages/Bill_Bowden/ohmslaw.htm

Answer 2

1 This problem cannot be solved unless you know the internal resistance of the battery. Students commonly assume that the potential difference across the battery terminals always remains same no matter what current is drawn from it. This may be assumed for small currents. But for large currents such as 12 amp the voltage across the terminals will not be 1.5. If that is given then you can find the external resistance by simple application of Ohm's law.
2. This question appears to be wrong, charge cannot be in ohms. If it is in coulombs then one can find the cappacity of the terminal and not its resistance. If you know the charging current then you can find the resistance of the charging circuit.
3.12/2.4 = 5 ohms
4. 32x6.8 = 217.6 V
5 question wrongly worded and should contain more data

Answer 3

These problems are actually very easy. All you need to do is use Ohm's Law and a little algebra to solve.

Ohm's Law say that the resistnce in a circuit is the ratio of the potential difference (Voltage) to the current. So, we end up with an equation that look like this ,

R=V/I, where R = Resistance in units of Ohms, V = Potential Difference in units of Volts and I = Current in units of Amps.

If you do #1 using this information you get:

V= 1.5 volts, I = 12 amps

R = V/I = 1.5/12 = 0.125 Ohms

The rest should be easy.

Good Luck

Answer 4

You can do this easily, come on you have the abillity!

you just use V= IR for all of them

V= Potential difference
I = Current
R = resistance

Answer 5

this $HI% is hard.