**In a laboratory demo, a 12 V car battery powers two 12 V, 24 W lamps in parallel.**
The current is measured using an ammeter (of resistance 10 ohms) which is connected in series with each of the bulbs in turn. What will the ammeter read? Comment on your answer.
**Possible Answers:**
1) **1.2 A**
Explanation: Using Ohm's Law, [tex]I = \frac{V}{R}[/tex], so [tex]\frac{12 \, \text{V}}{10 \, \text{Ohms}} = 1.2 \, \text{A}[/tex]
2) **0.75 A**
Explanation:
- The current of the bulbs normally is [tex]I = \frac{P}{V}[/tex], so [tex]I = \frac{24 \, \text{W}}{12 \, \text{V}} = 2 \, \text{A}[/tex].
- Resistance of each bulb: [tex]R = \frac{V}{I}[/tex], so [tex]R = \frac{12 \, \text{V}}{2 \, \text{A}} = 6 \, \text{Ohms}[/tex].
- Total resistance with the ammeter: [tex]6 \, \text{Ohms} + 10 \, \text{Ohms} = 16 \, \text{Ohms}[/tex].
- Current: [tex]I = \frac{V}{R}[/tex], so [tex]\frac{12 \, \text{V}}{16 \, \text{Ohms}} = 0.75 \, \text{A}[/tex].
Answer (2)
Your second solution is excellent, and well done. Ammeters are usually assumed to have zero resistance, and that would be your first solution. Your second one shows how much difference it can make when the meter itself has substAntial resistance. / / / Notice also that the poor meter has to dissipate a little more than 5 watts. That's a lot.
The ammeter will read 0.75 A when connected in series with one of the 12 V, 24 W lamps due to the inclusion of its own resistance of 10 Ohms. This effectively increases the total resistance in the circuit resulting in lower current than the bulbs would normally draw. The power dissipated by the ammeter would be about 5.625 W.
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