Recent content by jawhnay

@cwatters I'm from the united states.

how would I have known that the voltage rating of the lightbulb is 120 V if it isn't stated in the question? Is there no other way to solve part c given what is in the question?

So, I get I = 12/(2.4+35) = 0.321 amps. P = I^2R = (.321)^2(2.4) = 0.247 watts. I got the wrong answer. The answer is actually 0.46 watts.

the resistance would be R = (12)^2/60 = 2.4 ohms?

The current would increase to .188 amps to make up for the decrease in resistance if the voltage of the battery is the same.

The voltage of the battery and the internal resistance of it doesn't change right? So wouldn't the current be the same regardless if the 125 ohm resistor is gone? I tried to use P = IV to calculate the power output for the lightbulb

oh my gosh, I cannot believe I didn't see it. Thank you for the help!

Homework Statement How do I find the answer for part C? 3) A battery with an Emf of 12 volts is hooked up to a 125 ohm resistor and an ammeter. The ammeter measures a current of 75 mA. a) Calculate the internal resistance of the battery and the actual voltage across the battery. b) Calculate...

yes, I know how to write loop equations. However, I thought the equivalent resistance was just adding up the resistances so I didn't think I would need to show my loop equations. I have edited my first post with the loop equations and edited picture.

Homework Statement Please take a look at the attachment to see the circuit. My question is how do I find the equivalent resistance of this circuit? I have done the other two parts but this part has got me stumped. I also drew my currents on the picture. Homework Equations The...

How do you know that r(0) = .45 x 10-3 and r(.38) = 9.67 x 10-3 ? I'm kind of confused... Can anyone make this clearer for me?

where are you getting r(0 m) = 0.45e-3 m and r(38e-2 m) = 9.67e-3 m ?

Can anyone else offer any insight on how to do this problem?

so you're saying R = pL/A ... dR = pdr/A what would my area be? Would it be A = ∏r2 or A = 2∏rL How exactly do I go about doing this?

Homework Statement A Copper wire has a shape given by a radius that increases as R(x)= aex+ b. Its initial radius is .45 mm and final radius is 9.67 mm and its horizontal length is 38 cm. Find its resistance. Homework Equations R = pL/A where p = resistivity of copper A = ∏r2 L = length of...