series connection! help me with this! could you? please?

Question

help me please!

question: two resistors are connected in series across an ideal 12.0-V battery. Resistor A has a value of 24.0-ohms and the potential difference across resistor B is 3.60 V. (a) What is the potential difference across A? (b) What is the current in the resistors? (c) What is the resistance of B? (d) What is the total power dissipated in the circuit?

please help me solve this! i really dont have an idea! thanx.

Answer 1

1 If resistors are in SERIES, the total potential diff. is across BOTH. So, if resistor B has a 3.60 V potential diff, resistor A has a 8.40 V potential diff.

2. If resistors are in SERIES, the potential diff across EACH resistor can be used to find the current, which is the same for BOTH resistors.
8.4 = 24x I (Va = Ra I )
I = 8.4/24 = 0.35

3. Since we know amps and Vb, we can find Rb
Vb= Rb I
3.6 = Rb 0.35, Rb = 10.3 ohm

4. Power = V I = 12 x 0.35 = 4.2 watts.

Better get this stuff down, there is a lot more coming.

Answer 2

The lamps on your place are related in parallel. The wiring materials an equivalent voltage to all of them, and anybody could be became on or off without affecting the others. some holiday easy strings (quite those with the smallest bulbs) are related in sequence; removing anybody bulb will end the flow of modern and all of them will flow out. The bulbs in such strings oftentimes have an association so as that if the filament burns out, the bulb will become a quick circuit, the only bulb is dark however the the remainder of them stay lit, slightly extra brightly. Such strings are geared up with a fuse interior the plug so as that if many bulbs short out, the present flow is interrupted till now an overcurrent situation might reason a fireplace.

Answer 3

E=IR U can souffle it for all solutions .
E=Voltage
I=current
R=resistance