I have two questions specifically.

  • Can the usage of Ohm's law be utilized to predict current, voltage in circuits such as parallel, complex and series? How would I explain this?
  • How does the power dissipated by light by a light bulb affect its brightness?
  • 4
    $\begingroup$ One time, one question. $\endgroup$ – user36790 May 3 '16 at 2:47
  • $\begingroup$ Ohm's law doesn't apply to circuits but to linear circuit elements. You probably mean Kirchhoff's laws and, yes, there is sophisticated software to do that for you for circuits of virtually any complexity. $\endgroup$ – CuriousOne May 3 '16 at 3:17
  • $\begingroup$ $I = \eta \frac{P}{A}$. Where $\eta$ is efficiency of bulb. I is intensity of light and P is power used by bulb and A is area of light coverage. $\endgroup$ – Anubhav Goel May 3 '16 at 9:41
  • $\begingroup$ Presumably the efficiency, $\eta$, depends on the operating point of the bulb? $\endgroup$ – jim May 3 '16 at 12:52
  1. Along the lines of what @CuriousOne said you can use Kirchhoff's laws to understand the current and voltage drop at each circuit element with any complexity you desire. Kirchhoff's laws deal with the fact of the total sum of currents and voltages in any circuit junction and loop, respectively, must equal zero. The second law essentially uses Ohm's law to sum up the voltage drops at each element. For more details and an example you can refer to the Wikipedia page: https://en.m.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws.

  2. The power dissipated in something like a light bulb will go as $P=I^{2}R $ which in a light bulb the power it dissipates is the conversion of the electrical energy into visible light by some means. For example, in an incandescent bulb, current is used to heat up a tungsten wire to point in which it glows. The current that was used to heat up the wire and make it glow is the power dissipated. So we can see that the amount of power dissipated would be related to the resistance of the light bulb (for example the thickness of the tungsten would change the resistance and therefore the power dissipation) and the current we input (larger current, greater brightness).

Hope that helps!


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