If a 9-v radio requires 200 mA to operate and in order to operate the radio from a 12-v battery at the same current what size resistor must be connected in series with it?

Answer 1

Kirchoff's current law states that the current at every point in a series circuit is the same. If the radio is drawing 200 mA, then so will the resistor.

Kirchoff's voltage law states that the sum of the signed voltage drops in a series circuit all add up to zero, or that the voltage across parallel elements is the same. If the battery is providing 12 V, and the radio is using 9 V, then the resistor must be using 3 V.

Ohm's law states that resistance is voltage divided by current. If the voltage across the resistor is 3 V, and the current is 200 mA, then the resistor must be 15 ohms.

A consequence of the definition of voltage, current, and power is that power (joules per second) is voltage (joules per coulomb) times current (coulombs per second). The power dissipated by the resistor must then be 600 mW, so it should be rated at least one watt.

If the current drawn by the radio is not constant, a better solution might be to place a reverse biased 3 V zener diode in series with the radio, instead of a resistor. This will ensure that the voltage drop is always 3 V. Power across the diode will still be about 600 mW, for the 200 mA case.