ohms law can calculate the voltage drop across each resistance?

Answer 1

Yes. V = I * R

If you know the resistance of and the current through any given resistor, the voltage is determined.

Answer 2

It depends if the resistors are connected in series or parallel.
In a series circuit of two resistors of 10 ohms each the total resistance is 20 ohms. Assume the battery voltage is 10 volts, then, the current in the circuit will then be I=E/R..... I=10/20 ... then I equals .5 amps through each 10 ohm resistance. For the voltage drop across each resistance rewrite the formula as E=IR. Substitute and you get E = .5 times 20 which equals 10 volts. Since both resistors = 10 ohms then the voltage across each resistance is 10 volts.
The sum of the voltage drops across the resistors in a circuit must equal the supply voltage. (Kirchoff's Law). There are two resistors in the circuit and the voltage drop across each one is ten volts. 10 volts plus 10 volts equals the 20 volt supply...so then...the answer is correct. The calculation for a parallel circuit is different.

Answer 3

The emf source in the model battery represents the potential difference the battery can create in the absence of internal dissipation. Based on what you say you get, it should be clear that a battery driving a circuit will get warmer as it discharges. The only source that can provide the power to create that heat is the battery itself. As a result, not all the power the battery can produce goes to driving the circuit. Using the model of the emf source in series with a resistor, the internal resistance of the battery can be treated as just another resistor in the circuit and then you can solve (as usual) for the current through each branch and the voltage drop across each component. Once you know the current being supplied by the battery, you can compute the voltage drop across the internal resistor. Since that internal resistor lies inside the boundary of the battery terminal, that voltage drop reduces the effective voltage being supplied by the emf source to the rest of the circuit -- the voltage being supplied by the battery that the rest of the circuit sees is the voltage of the emf source minus the voltage drop across the internal resistor. In the limit that the internal resistance goes to zero, the power dissipation in the battery goes to zero, the voltage drop across the internal resistor goes to zero, and the voltage the rest of the circuit sees as a result of the battery goes to the potential difference supplied by the emf source.

Answer 4

Yes but you have to know the value of the resistance and the value of the current to find the voltage drop.

If you have multiple resistors in a circuit...
if they are in a series then the voltage drop across each is split

say you have 3 resistors in a series
R1=10 ohms R2=5 ohms and R3 =20 ohms

you apply a 120v source to the circuit

find the current first total the resistors
10+5+20=35
V=I*R in this case you are looking for current voltage divided by resistance
120/35=3.42 ohms now go back and find voltage drop across each resistor
3.42*10=34.2 volts
3.42*5=17.1 volts
3.42*20=68.4 volts
when you add them up will total 119.7 but my round off error accounts for being .3 short

Answer 5

Yes. if there are many resistors (and or other electrical components in the circuit such as inductors, capacitors, etc...)
then u can calculate the current through each component for a given voltage and vice-versa by using the following equations -
1. V=I*R
2. R={r1+r2+...} for resistors in series
3. R={r1*r2/[r1+r2]} for resistors in parallel.

and therefore determine the Voltage drop and current across each resistor.

Answer 6

"... each resistance?"

By that I assume that their more than one resistance. Since all of the 'resistors' are linear, simple linear math can be used;
I * R = V
where I and V are vectors and R is matrix describing the circuit. From the above a series of linear equations a made and be solved, at least in mathematically. By hand thing could be painful but should be doable.