12.1 Resistors in series and parallel  (Page 6/17)

Now the current $I_2$ through resistance $R_2$ is found using Ohm’s law:

Discussion for (c)

The current is less than the 2.00 A that flowed through $R_2$ when it was connected in parallel to the battery in the previous parallel circuit example.

Strategy and Solution for (d)

The power dissipated by $R_2$ is given by

Discussion for (d)

The power is less than the 24.0 W this resistor dissipated when connected in parallel to the 12.0-V source.

Practical implications

One implication of this last example is that resistance in wires reduces the current and power delivered to a resistor. If wire resistance is relatively large, as in a worn (or a very long) extension cord, then this loss can be significant. If a large current is drawn, the $\text{IR}$ drop in the wires can also be significant.

For example, when you are rummaging in the refrigerator and the motor comes on, the refrigerator light dims momentarily. Similarly, you can see the passenger compartment light dim when you start the engine of your car (although this may be due to resistance inside the battery itself).

What is happening in these high-current situations is illustrated in [link] . The device represented by $R_3$ has a very low resistance, and so when it is switched on, a large current flows. This increased current causes a larger $\text{IR}$ drop in the wires represented by $R_1$ , reducing the voltage across the light bulb (which is $R_2$ ), which then dims noticeably.