(II) A battery with an emf of 12.0 V shows a terminal voltage of 11.8 V when operating in a circ

(II) A battery with an emf of 12.0 V shows a terminal...

(II) A battery with an emf of 12.0 $\mathrm{V}$ shows a terminal voltage of 11.8 $\mathrm{V}$ when operating in a circuit with two lightbulbs rated at 3.0 $\mathrm{W}(\text { at } 12.0 \mathrm{V})$ which are connected in parallel. What is the battery's internal resistance?

Key Concepts

Electromotive Force (emf)

This is the maximum potential difference that a battery or power source can provide when no current flows. It represents the ideal voltage generated by the source and serves as a benchmark for comparing the actual performance under load.

Terminal Voltage

Terminal voltage is the voltage measured across the output terminals of a battery when it is delivering current to a circuit. Due to the battery’s internal resistance, the terminal voltage is lower than the emf when current is flowing, reflecting voltage drops internal to the battery.

Internal Resistance

Internal resistance is an inherent property of all real batteries and power sources. It causes a voltage drop inside the battery when current flows, reducing the output voltage available to the external circuit. This concept is crucial for analyzing real-world performance in electric circuits.

Parallel Circuits

In parallel circuits, components are connected such that the same voltage is applied across each component. Understanding how parallel configurations distribute current and affect overall resistance is key to analyzing how devices rated for specific power or voltage operate when connected to a common voltage source.

Transcript

00:01 Okay, in this question, two light bulbs are connected in parallel with the battery.

00:08 The electromagnetic force of the battery is given to be 12 volts.

00:14 The internal resistance of the battery, lowercase r, is not given and is the one that we are going to find.

00:27 And the resistance of the light bulbs also is not given.

00:32 But instead the power of the light bulbs are given.

00:37 So the power of the light bulbs are 3 watt at 12 volts.

00:47 So that means if the terminals of the light bulbs are connected to a 12 volt, the power dissipated at the light bulbs would be 3 watt.

01:02 So having these two information like it means, me redraw that here.

01:08 For example, if this is our light bulb, if the terminals are connected to 12 volts exactly, the power dissipation would be 3 watt.

01:23 So in this situation, we have this formula for power.

01:28 Power equals v squared over r.

01:37 So and from the parameters here, we have v and p and r is unknown so using this equation we can simply find the resistance of these light bulbs so the resistance would be voltage squared over the power so simply plug in in voltage 12 volt squared divided by 3 so this is volt and this is watt so the result would be in and it will give us 48 oh so now we have the resistance of light bulbs to be 48 on each 48 now these two light bulbs are connected in parallel so we can find an equivalent resistance for this combination to solve the circuit.

02:52 So the equivalent resistance of two parallel resistances would be, we can find it using this equation...

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