Ohm's Law Practice?

Question

I'm looking for some help on these questions - I don't need answers, just a step in the right direction would be helpful.

Three resistors, 25, 45 and 75 Ω are connected in series, and a .51 amp current passes through them. What is the a) equivalent resistance and b) the potential difference across the 3 resistors?

So for a), I think you just add up all the resistance, since it's a series circuit, but for b), I have no idea what to do...any hints?

Three resistors (9, 5, and 1 Ω) are connected in series across a 24-V battery. Find a) the current in, b) the voltage across, and c) the power dissipated in each resistor.

To find the current, I think you use V=IR, which leads me to 24/9 amps, 24/5 amps, and 24 amps for question A, but that doesn't seem right.

For B, I think the same voltage flows throughout the entire circuit, so I think 24 is the answer for that one, but I'm not sure.

And I have no idea how to do C.

Answer 1

for b

across all three total resistors?

V = IR (current times resistance)

so .51A times 145ohm equivalent resistance = 73.95

OR ARE YOU trying to get the voltage drop across EACH INDIVIDUAL resistor?

which would just be V1 = IR1,  V2 = IR2,  and V3 = IR3

------

in series current will be the same in each resistor.

so to find I:

V = I R(the equivalent resistance)

in series the voltage drop across each resistor will be proportional to each resistance.

so V = V1 + V2 + V3

V = IR1 + IR2 + IR3



pretty simple.

once you have those values you can find the power dissipation of each resistor pretty easily with P = IV