Imagine a parallel circuit with 3 lamps in each branch. Connected to the circuit is an ammeter and a constant-voltage d.c. supply. Each bulb operates at normal brightness and the ammeter (of negligible resistance) registers a steady current.
Then, the filament of one of the bulbs breaks. What happens to the ammeter reading and to the brightness of the remaining bulbs?
The answer is that the ammeter reading decreases and the bulb brightness remains unchanged.
I don't understand why the ammeter reading decreases and why the bulbs' brightness remain the same. I thought the bulbs' brightness will be increased cos one of the filament of the bulb breaks so more current will flow through the other 2 bulbs, isn't it? With more current, shouldn't the other two bulbs shine more brightly? Please answer my 2 questions stated above, thanks.
2007-11-07 11:48:33 · 8 answers · asked by ♪£yricảl♪ 4 in Science & Mathematics ➔ Physics
To Nomadd: So if it is not a constant voltage supply, the voltage in the circuit would change and thus the ammeter reading and the brightness of the lamps would be affected?
2007-11-07 12:29:12 · update #1
There is a constant voltage across each lamp. So when one burns out it does not effect the other lamps. The voltage is still constant. That's the nice thing about parallel circuits. as for the ammeter reading decreasing, it has to because one of the bulbs burned out. That means one of the current paths has stopped.
2007-11-07 11:53:48 · answer #1 · answered by Anonymous · 0⤊ 0⤋
I believe you meant to say" imigine a parallel circuit with three branches and with one lamp in each branch". Never the less it was given that the source Voltage is constant regardless of how many bulbs you connect in parallel accross the Voltage . This means that the current in any branch is determined by deviding the source voltage by the resistance of the lamp in that branch and is independent of the current in the other two branches. The sum of the currents in all the branches is equal to total current the source Voltage is supplying and this total current is what the ammeter is reading. Now when the filament of one bulb breaks and opens the circuit in that branch then the resistance of that branch goes to infinity and the current in that branch goes to zero. Since the ammeter is reading the total current of all three branches the ammeter reading will decrease by one third if we assume all three lamps had equal resistance. The reason the brightness of the other two bulbs remains the same is because the Voltage applied to them is still the same and the internal resistace of them is still the same and thus the current through them is still the same. The current is what determines the brightness therefore the brightness will remain the same. Another way of explaining why the ammeter reading decreases when one of the filements breaks open is the fact that resisters [ in this case filaments] inserted in parallel with other resistors lowers the total resistance of the circuit and increases the total current that the source voltage supplys. Conversely removing a branch resister [ in this case a filament] increases the total resistance of the circuit and lowers the current that the source Voltage supplys. Hope I haven`t baffled you even more with my ill worded explanation. P. S. If the circuit was driven by a constant current source rather than a constant voltage source then your line of reasoning would be valid.
2007-11-07 21:40:35 · answer #2 · answered by Mr. Un-couth 7 · 0⤊ 0⤋
While in parallel, assume that the resistance of each bulb is R. A voltage of V is applied across all the bulbs in parallel. Thus the current drawn by each bulb is V/R. The ammeter in series reads V/R + V/R + V/R = 3V/R before. When one circuit is cut, it reads 2V/R. Note that neither V nor R has changed to cause different currents to flow in the circuit.
2007-11-07 19:54:23 · answer #3 · answered by noitall 5 · 0⤊ 0⤋
The key here is the voltage source: constant voltage! With constant voltage and constant resistance in the bulbs (assuming pure resistive load, no inductance or capacitance), then Ohm's law predicts the current through each bulb: V/R = I. Changing the number of bulbs won't affect the current through each one. If you had an infinitely powerful constant voltage source, you could add bulbs in parallel forever and the current (and, therefore brightness) of each would be the same.
2007-11-07 19:55:33 · answer #4 · answered by Berry K 4 · 0⤊ 0⤋
The voltage across the remaining lamps doesn't change just because the one bulb breaks. That's what constant voltage supply means. So why would the current through the bulbs change?
2007-11-07 20:20:59 · answer #5 · answered by Nomadd 7 · 0⤊ 0⤋
It has been a while, but I'll do my best.
Since one light burns out, the resistance decreases. This will change all readings, since all readings are integral to what exists in the functioning circuit.
As far as the lights, if you are on a parallel, there is no increase or decrease to the individual lights. Each light gets the same amount of current (one light does not increase or decrease the current flowing to the other lights). When in parallel, it's as if each light were separately attached to the power source. This is sometimes hard to envision. If you work with it for a while, you may find it becomes easier to see.
Hope this helps.
2007-11-07 20:30:21 · answer #6 · answered by Militia-Angel 3 · 0⤊ 1⤋
Resistance goes up because you lose one of the legs. Say each lamp was 1 ohm. Then it would go from 1.5 ohms to 3 ohms (total resistance).
Voltage is constant, so do arrow analysis
V = IR (R goes up x2, so I goes down x2)
Now you have half the total current.
W = IV (I goes down x2 and V is constant, so W goes down x2)
Double the resistance, contant voltage, half the current, half the power.
2007-11-07 19:57:08 · answer #7 · answered by Anonymous · 0⤊ 0⤋
no
2007-11-07 19:52:31 · answer #8 · answered by M 3 · 0⤊ 1⤋