Circuit: Battery With Internal Resistance

[Destro96]

---

Members don't see this ad.

Physics gurus, please riddle me this.

Let's suppose that we have a simple circuit.

The battery produces a potential difference of 12V. It has an internal resistance of 20 ohms. That battery is then connected to a circuit with two 100 ohm headlights (wired in series for a TTL resistance of 200 ohms).

What is the current and voltage of the circuit?

The internal resistance thing is the part that is killing me...

 

[Chowdder]

I'm no physics guru, but I will try and help you understand this problem (at least see how I would go about to solve it). Internal resistance inside of a battery is basically a microscopic resistor attached to the macroscopic circuit. It adds in series to the effective resistance of the macroscopic circuit. The internal resistance tacks onto total resistance of the circuit dictated by ohm's law following kirchoff's law.

If R_internal is internal resistance, V is voltage output by the battery, R_eff is the effective resistance, and I is the total current.

Then ohm's law of V=IR becomes V=I*(R_internal+R_eff)

So to answer your question.

I=12V/(20Ohm+200Ohm)
I=0.06A

Voltage is 12 but I think you're referring to the e.m.f., which is lower than the stated voltage because of the internal resistance. Remember, whenever there is a resistor, there is a voltage drop.

Since this is a circuit in series, current flowing through R_internal is the same as current flowing through R_eff.
Therefore, the voltage drop at the internal reistance is

V_drop = 0.06A*20Ohm
V_drop = 1.2V

So the e.m.f. is simply the voltage of the battery minus the voltage drop at the internal resistance.

e.m.f = 12V-1.2V
e.m.f = 10.8 V