Anyone know aobut basic electronic circuits?

Question

Here is a question on my homework that is very puzzling:
So you have a 9V battery that is connected to 4 light bulbs. First is a parallel circuit with a 50w and 25w bulb. Next is a single 40w bulb, and last is a 25w bulb. What is the current that flows through these, and how does the voltage change?

Answer 1

I'm not sure there is sufficient information given to answer this question. How are the wattages of the bulbs determined?(Are you sure it is wattage and not Ohmage?) If they have these wattages when connected alone to a single 9V source, then their actual wattages in your circuit will be different from this. If their wattages are exactly as described in the arrangement described, then that makes the problem a little easier.
Let us assume the latter is the case. Therefore the total wattage of the system is simply additive, so the total wattage is 50+25+40+25 which equals 140 watts Since wattage equals volts times amps, that means the amps are 140/9 or 15.555..... amps. According to my calculations, the first two bulbs in parallel both see a voltage of 4.82 Volts, while the single 40 watt bulb sees a voltage of 2.57 Volts, and the last 25 watt bulb sees a voltage of 1.61 Volts, so 4.82+2.57+1.61=9.00Volts with a total amperage through the system of 15.556 Amps. I hope this helps.
P.S. Devilsadvocate's calculations are wrong.

Answer 2

Before this one can be answered, you must specify at what voltages these wattage ratings are specified. You must also assume that, unlike real light bulbs, the resistance of the bulbs does not depend on the current through them or the voltage across them.

That said, if it is assumed that the wattage ratings of all of the bulbs are taken at the same voltage and the bulbs can be treated as ideal resistors - a questionable assumption that will be discussed in detail later - the voltages across each of the bulbs can still be calculated, though the currents cannot unless the voltage rating is specified. The resistances of the lights will then be inversely proportional to their wattage ratings. Taking a 100w bulb as being 1 unit of resistance, the 50w bulb would then be 2 units; the 40w; 2.5 units; and the two 25w bulbs, 4 units each.

The parallel combination would then have 4/3 units of resistance across it, in series with two other bulbs, one with 2.5 units and the other with 4 units of resistance. This gives a total resistance in the circuit of 47/6 units. The voltage drop across each bulb or (parallel combination of bulbs) would then be equal to

Vbulb = Vtotal * Rbulb / Rtotal.

For the parallel combination, the voltage drop would be

9 volts * (4/3) / (47/6) = 9 volts * (4/3) * (6/47) = 72/47 or 1.53 volts,

For the 40w bulb,

9 volts * (5/2) / (47/6) = 135/47 or 2.87 volts,

for the 25w bulb in series,

9 volts * 4 / (47/6) = 216/47 or 4.60 volts.

Full current, whatever that is, flows through the two series bulbs, with 1/3 of the current flowing through the parallel 25w bulb and 2/3 flowing through the 50w bulb.

Real light bulbs don't have a constant resistance. The brighter the bulb glows, the hotter its filament is and the higher its resistance. If you measure a household light bulb with an ohm-meter, you will get a resistance reading lower by a factor of 20 or so than you would expect from its voltage and wattage ratings. The difference is that you are measuring the resistance when the filament is cold and the resistance low. The resistance will indeed increase to the expected value when operated at the rated voltage.

Answer 3

P = power, v = voltage, I = current

Using P = V*I then we can find the voltage distriution in the circuit. The voltage distribution is propotional to the power distribution since the total current is a constant value. Therefore we have 3 series power loads

1. 50w + 25w = 75w
2. 40w
3. 25w

total power is the sum which is 140w. Note the power at each element is proportional to its voltage. Therefore,

The voltage acros the 75w combined element:

v1 = 9v * 75/140 = 4.8 volts (75w element)
v2 = 9v * 40/140 = 2.6 volts (40w element)
v3 = 9v * 25/140 = 1.6 volts (25w element)

I total = total power/applied voltage

Itotal = 140/9 = 15.6 amps

current through the 50w element is the power divided by the voltage which is 4.8 volts, then that current = 10.4 amps

Likewise the current through the first 25w elemnet is 25 divided by 4.8 = 5.2 amp.

The current through the other two elements is the total current which is 15.6 amps

Answer 4

The voltage remains the same on the parallel 50W and 25W bulbs. I assume the last two bulbs are in series?
The total current is derived by using the 9V reference on three parallel circuits - the last circuit being the series 40W and 25W combined to form the third parallel circuit.
The power values across the circuits are 50W, 25W and 65 W.
The current is (50W)= 5.6A (25W)=2.8A (65W)=7.2A
The only voltage change is at the series circuit of the 40W and 25W bulbs. The voltage drop across both is 9V, the drop across the 40W is 5.6V with the remaining 3.4V across the 25W.

Answer 5

Do you mean watts or ohms? If ohms, calculate the equivalent resistance of the circuit:R= 25 ohms (last bulb)+40 ohms(2nd to last bulb)+50/3 ohms (parallel 50 and 25 ohm bulbs). Then the current is V/R by ohms law. You can calculate the voltage drop across each bulb as V=I*R