ohms law - Finding Source Voltage of a DC series circuit

K

KLL

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Hello,

I'm trying to figure out how to determine source voltage of a DC series circuit. The only information given is

Resistor 1 = 12ohm
Resistor 2 = 9ohm
Resistor 3 = 3ohm

And the voltage drop across R1 is 6V.

The answer is 12V for the DC battery. But I can't figure out how to get to 12.

If I put 12V down as the battery size, and combine all three resistors 24ohm and use ohms law I can't come up with 12V.

R total = 24ohm
12V source battery

24 divided by 12 = 2Amps

Now if I go to R1 and knowing that the circuit amperage is 2, I get a voltage drop of 2A * 12ohm = 24V not 6V.

Where am I going wrong?
 
KLL said:
Hello,

I'm trying to figure out how to determine source voltage of a DC series circuit. The only information given is

Resistor 1 = 12ohm
Resistor 2 = 9ohm
Resistor 3 = 3ohm

And the voltage drop across R1 is 6V.

The answer is 12V for the DC battery. But I can't figure out how to get to 12.

If I put 12V down as the battery size, and combine all three resistors 24ohm and use ohms law I can't come up with 12V.

R total = 24ohm
12V source battery

24 divided by 12 = 2Amps

Now if I go to R1 and knowing that the circuit amperage is 2, I get a voltage drop of 2A * 12ohm = 24V not 6V.

Where am I going wrong?
Click to expand...
With the information you have given, the problem is impossible. All we can do is find the current through $R_1$:
[imath]V_{R1} = I R_1[/imath]

[imath]6 = I \cdot 12[/imath]

[imath]I = 0.5[/imath] A.

We don't know how the other resistors are connected, what the total potential drop across the whole circuit is, etc.

Is this the whole problem statement?

-Dan
 
topsquark said:
We don't know how the other resistors are connected,
Click to expand...
khansaheb said:
You did not mention anything about series/parallel connection!
Click to expand...
I'm confused. It clearly says it's a series circuit:
KLL said:
I'm trying to figure out how to determine source voltage of a DC series circuit.
Click to expand...
So I assume all three resistors are in series. That's also why it's valid to add the resistances to get 24 ohms. (We would prefer, though, that you show the actual problem, with any included picture, so we could be sure how it is worded.)

So here's the error:
KLL said:
R total = 24ohm
12V source battery

24 divided by 12 = 2Amps
Click to expand...
Why do you divide the resistance by the voltage? If V = IR, then I = V/R, not R/V.

Fix that, and everything will check out.

But you didn't say how you tried to get the answer of 12V; your work appears to be just a check of the book's answer. Do you need help with the actual problem, too? Post #2 gives you the start.
 
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KLL said:
Hello,

I'm trying to figure out how to determine source voltage of a DC series circuit. The only information given is

Resistor 1 = 12ohm
Resistor 2 = 9ohm
Resistor 3 = 3ohm

And the voltage drop across R1 is 6V.

The answer is 12V for the DC battery. But I can't figure out how to get to 12.

If I put 12V down as the battery size, and combine all three resistors 24ohm and use ohms law I can't come up with 12V.

R total = 24ohm
12V source battery

24 divided by 12 = 2Amps

Now if I go to R1 and knowing that the circuit amperage is 2, I get a voltage drop of 2A * 12ohm = 24V not 6V.

Where am I going wrong?
Click to expand...
Okay, so. Back to the start.

Earlier I showed you how to get the current in [imath]R_1[/imath]. The current in a series circuit is the same in all resistors. So we know that the current in the circuit as a whole is 0.5 A.

Can you finish?

-Dan
 
KLL said:
Hello,

I'm trying to figure out how to determine source voltage of a DC series circuit. The only information given is

Resistor 1 = 12ohm
Resistor 2 = 9ohm
Resistor 3 = 3ohm

And the voltage drop across R1 is 6V.

The answer is 12V for the DC battery. But I can't figure out how to get to 12.

If I put 12V down as the battery size, and combine all three resistors 24ohm and use ohms law I can't come up with 12V.

R total = 24ohm
12V source battery

24 divided by 12 = 2Amps

Now if I go to R1 and knowing that the circuit amperage is 2, I get a voltage drop of 2A * 12ohm = 24V not 6V.

Where am I going wrong?
Click to expand...
What a kerfuffle! (Mainly due to peeps not reading things properly! ?)

There's no need to involve Ohm's Law to calculate circuit current, it's just a simple matter of some additions if you draw the circuit diagram (always a good idea ?)

As I keep repeating: "A picture's worth a thousand words!" (FR Barnard?)

We (should) know that the total voltage drop around the circuit is equal to the source voltage and so (ignoring the connecting wires' resistance as we invariably do in these problems) the battery's voltage must be...?


Battery Voltage.jpg​
 
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@KLL,

Always use the old triangle (or Trapezium ?) "trick" for these three-part formulae that crop up so often in Physics.

That way you only need to memorize one form of the formula (eg: V = I × R) and when you jot that down (in the triangle, as shown) you can easily get the other forms just by putting your finger over the term you want to calculate.

Then you are much less likely to make the kind of mistake you made above (when you got the formula wrong) ?

3 Part Formula.jpg
​
PS: Is equally handy for the Trigonometric Ratios too: \(\displaystyle \sf\left(\frac{OPP}{SIN | HYP}\right)\) ?
 
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