# If all you have is a 120-V line, would it be possible to light several 6-V lamps without burning them out? How?

Feb 18, 2017

This question is about series and parallel connections. Here, you must connect 20 of the bulbs in series to make it work.

#### Explanation:

When we connect a number of devices in series across a voltage source, the voltage is divided among those devices, in a proportion that is based on the resistance of each device.

As an example, suppose we have two resistances, $1 \Omega$ and $2 \Omega$ that we connect in series across a 12 volt power supply. Since the total resistance of the two in series is $3 \Omega$, the current that will run is

$I = \frac{V}{R} = \frac{12}{3} = 4 A$

When a $4 A$ current passes through a $1 \Omega$ resistor, the voltage drop is

$V = I \cdot R = \left(4 A\right) \cdot \left(1 \Omega\right) = 4 V$

The same current then passes through the $2 \Omega$ resistor, and the voltage drop is $8 V$.

In this way, the two series resistors have divided the source voltage, with the larger resistance causing the greatest voltage drop.

Now, if you must use a $120 V$ source, but have only $6 V$ across each bulb, it will be necessary to divide the 120 volts into twenty $6 V$ drops, which is exactly what happens if you connect twenty identical resistors (the bulbs) in series.
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