A tablet draws 10.0 A when plugged into a 150-V line. If another tablet has twice as much resistance, what would the current of the second tablet be if plugged into the same line?
0.60 A
1.5 A
5 A
15 A
The Correct Answer and Explanation is:
To solve this problem, we need to use Ohm’s Law:I=VRI = \frac{V}{R}I=RV
Where:
- III is the current (in amperes, A)
- VVV is the voltage (in volts, V)
- RRR is the resistance (in ohms, Ω\OmegaΩ)
Step 1: Determine the resistance of the first tablet
The first tablet draws 10.0 A of current when connected to a 150-V line. Using Ohm’s Law:R1=VI=150 V10.0 A=15 ΩR_1 = \frac{V}{I} = \frac{150 \, \text{V}}{10.0 \, \text{A}} = 15 \, \OmegaR1=IV=10.0A150V=15Ω
Step 2: Find the resistance of the second tablet
The second tablet has twice as much resistance as the first tablet. Therefore:R2=2×R1=2×15 Ω=30 ΩR_2 = 2 \times R_1 = 2 \times 15 \, \Omega = 30 \, \OmegaR2=2×R1=2×15Ω=30Ω
Step 3: Calculate the current for the second tablet
The voltage remains the same (150 V), so the current I2I_2I2 for the second tablet is:I2=VR2=150 V30 Ω=5.0 AI_2 = \frac{V}{R_2} = \frac{150 \, \text{V}}{30 \, \Omega} = 5.0 \, \text{A}I2=R2V=30Ω150V=5.0A
Final Answer:
The current for the second tablet is 5 A.
Explanation:
This question demonstrates the relationship between resistance and current for a fixed voltage. When resistance increases, current decreases proportionally because they are inversely related in Ohm’s Law. Doubling the resistance (2×15 Ω=30 Ω2 \times 15 \, \Omega = 30 \, \Omega2×15Ω=30Ω) reduces the current by half since the voltage (150 V150 \, \text{V}150V) stays constant.
In the first case, a resistance of 15 Ω15 \, \Omega15Ω allowed 10 A10 \, \text{A}10A of current to flow. With twice the resistance (30 Ω30 \, \Omega30Ω), the current decreases to 5 A5 \, \text{A}5A. This reflects the principle that electrical devices with higher resistance draw less current when connected to the same voltage source.
The correct answer is 5 A.