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Question regarding voltage and current across a short circuit

  1. May 31, 2012 #1
    When a short circuit occurs, why won't the light bulb light up as well? Since the short circuit and the main branch where the lightbulb is are connected in parallel so won't the two have equal voltages and hence the bulb will light up (P=V^2/R)? Some people told me that its because the resistance of the short circuit is alot lesser than the lightbulb. But it still has a value R right? If not wont the same thing happen if one branch has a R of 1ohm and the other with 10ohm, but we know something like this doesn't happen right? So I'm pretty confused here. Thanks for the help!
     
  2. jcsd
  3. May 31, 2012 #2

    K^2

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    Science Advisor

    It's because there is resistance in your power supply and some resistance in the wires. Solve the circuit with battery in parallel with a short and a light bulb. Now, add a small resistor to each wire segment, and one in series with battery. The later simulates internal resistance of a battery. Say, light bulb is 100 Ohm, each wire segment is 1 Ohm, and 5V battery has 10 Ohm in series with it. What's the voltage across the light bulb now?
     
  4. May 31, 2012 #3
    Oh that makes the voltage across the bulb close to zero or even 0 if negligible? But in these questions we do not take that in consideration though. Like in questions we just take the short to take all V and all I. But is this the actual reason behind this? Thanks for the help!
     
  5. May 31, 2012 #4
    BTW in your circuit, is there a resistor on the short and also on the lightbulb and in series with the battery? Is it a bulb with a 10 ohm resistance, with a 1 ohm resistor in series then a parallel branch which is the fuse and the bulb, in each of them there is a 1 ohm resistor while the bulb has 100ohms? If so, then the V of the bulb is 0.415V but then if this is the correct then the sport circuit has a voltage of 0.42V so current is 0.42V which isn't high. Is there anything wrong with my working? Thanks K^2!
     
    Last edited: May 31, 2012
  6. May 31, 2012 #5
    In your situation since you are begining to study a short circuit, you would think of the short as connecting the two ends of the light bulb together. You would then have 0 volts across the light bulb and all the current from the battery would flow from one terminal of the battery to the short and to the other terminal.
    And situations such as this do happen in real life with circuitry where bare wires may rub against each other to erode the insulation down exposing bare metal.

    In other cases of short circuit, as as you learn more about short circuits, internal resistance of a battery, and the resistance inherent in wiring, as K2 stated, you will find ( as you did in your calculation if correct ) that yes, there would be a current throught the light bulb which would glow with less exhuberence because of the faulty circuit.
     
  7. Jun 1, 2012 #6
    Hi! But I was wondering if my calculations were correct since it didn't really fit the idea where the current is extremely high. I'm taking the light bulb and resistor branch to have 101ohm, the short circuit to have 1ohm and in series to the parallel branches and battery to be 11ohm. The values which I got didn't seen very correct though. Thanks!
     
  8. Jun 1, 2012 #7

    NascentOxygen

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    Staff: Mentor

    An ideal short circuit is precisely 0 ohms. The voltage across it is always 0 volts, regardless of the current that flows through it. Practical short circuits may only approximate this.
     
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