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.
When a resistor is added the current goes down, that is expressed in the equation current= voltage/ resistance
The voltage of a battery goes as the current times the resistance (V=IR). Because the voltage is being held constant, the resistor that draws the most current will have the lower resistance.
12 milliamps
Internal resistance is approximately equal to 94.667
The 12V battery connected to the 2.4 Ohm combination will supply 12/2.4 or 5A. The individual currents will be 12/3 or 4A for the 3 Ohm resistor, and 12/12 or 1A for the 12 Ohm resistor. The 2.4 Ohm parallel combination is obtained from a simple product-over-the-sum calculation.
If a 9.0 volt battery is connected to a 4.0-ohm and 5.0-ohm resistor connected in series, the current in the circuit is 1.0 amperes. If a 9.0 volt battery is connected to a 4.0-ohm and 5.0-ohm resistor connected in parallel, the current in the circuit is 0.5 amperes.
When a resistor is added the current goes down, that is expressed in the equation current= voltage/ resistance
The voltage of a battery goes as the current times the resistance (V=IR). Because the voltage is being held constant, the resistor that draws the most current will have the lower resistance.
All the way along when the crocodile clip is connected to a resistor, when the other end of the resistor is connected to the other side of the battery.
12 milliamps
No, a battery is not a resistor. A battery provides electrical energy, while a resistor is a component that restricts the flow of current in a circuit.
Internal resistance is approximately equal to 94.667
The 12V battery connected to the 2.4 Ohm combination will supply 12/2.4 or 5A. The individual currents will be 12/3 or 4A for the 3 Ohm resistor, and 12/12 or 1A for the 12 Ohm resistor. The 2.4 Ohm parallel combination is obtained from a simple product-over-the-sum calculation.
Voltage = Current * ResistanceVoltage = 12VResistance = 10 ohmsCurrent = Voltage/ResistanceCurrent = 12V/10 ohmsCurrent = 1.2 Amps
That will depend on the sum of the load resistance and the internal resistance of the battery (this is true for all power sources, not just 6 volt batteries). Small compact batteries tend to have higher internal resistance and therefore are more limited in the current they can deliver to a given load than larger batteries.
4.8 ohms
yes... this is possible if a diode i connected in reverse bias with a battery and a resistor for example. A diode in reverse bias means its anode will be connected to positive terminal of the battery and its cathode to the negative terminal of the battery. In such a case, minimal current flows through the circuit which can be neglected.