Problema 10-15    Source:   Problem elaborated by the author of the site.

In the circuit show in Figure 10-15.1 all the bulbs are 4 volts and have a power of 16 watts each. It is understood that when the lamp dissipates the nominal power (16 watts) it will be at maximum brightness. It is requested:

a) With S₁ closed and S₂ open, the lamps will light at maximum brightness?

b) With S₁ open and S₂ closed, which lamps will light at maximum brightness?

c) With both keys, S₁ and S₂ closed, which lamps will light at maximum brightness?

Solution of the problem 10-15

Item a

When only the key S₁ is closed, the lamps that are connected to the voltage source are the lamps L₁ , L₂ , L₃ and L₆. Therefore, for all lamps to light up at maximum brightness we must have a voltage source with V = 4 x 4 = 16 volts. Since the circuit supply provides only 12 volts, then no bulb will light up at maximum brightness because all of them are in series.

Item b

With the keys S₁ open and S₂ closed, the lamps that will be connected to the voltage source will be L₄ , L₅ e L₆. In this case we have three lamps connected in series and for terms maximum brightness we must have a voltage of V = 4 x 3 = 12 volts. As the voltage source provides 12 volts then the three bulbs will light at maximum brightness.

Item c

With the keys S₁ and S₂ closed, all the lamps will be in the circuit being that, L₁ , L₂ and L₃ form a parallel circuit with L₄ and L₅. In turn, this whole set is in series with the lamp L₆ . It is possible to calculate how much power each lamp will dissipate by calculating the electrical resistance of each lamp. As the working voltage and power of each lamp is known, the resistance can be calculated without problems. So.

Therefore as L₁, L₂ e L₃ form a series circuit, for this circuit there is a total of 3 ohms. L₄ and L₅ also form a series circuit. In this case, we have 2 ohms. And as was said earlier, these two sets are in parallel. Therefore, the equivalent resistance of this circuit is:

Adding this value to the lamp resistance L₆ is the total system resistance. Thus, it is possible to calculate the electric current that circulates in each resistor and, consequently, the power that each lamp will dissipate. So:

The electric current that flows through the circuit is:

However, from the problem data it is possible to calculate the maximum electric current that each lamp supports. Like this:

With this, it is concluded that the lamp L₆ will not support current I_total and its filament will be destroyed . As it is in series in the circuit will cause no lamp to light.