Question

What is the output voltage of a 3.0000-V lithium cell in a digital wristwatch that draws 0.300 mA, considering the cell’s internal resistance is 2.00Ω?

(a) 3.0000 V
(b) 2.908 V
(c) 2.678 V
(d) 2.450 V

Answer 1

The output voltage of a 3.0000-V lithium cell with an internal resistance of 2.00Ω drawing a current of 0.300 mA is 2.9994 V, which is closest to the given option (a) 3.0000 V.

Step-by-step explanation:

The student is asking about the output voltage of a 3.0000-V lithium cell with an internal resistance of 2.00Ω when a current of 0.300 mA is drawn by a digital wristwatch. To find the output voltage, we can use Ohm's law which relates voltage (V), current (I), and resistance (R).

The formula for the voltage drop across a resistor is V = IR, where I is the current and R is the resistance. The current must be converted from milliamperes to amperes by dividing by 1000, so the actual current is 0.300 mA = 0.000300 A.

Applying Ohm's law, the voltage drop across the internal resistance is:

V = IR
V = (0.000300 A)(2.00 Ω)
V = 0.0006 V

The output voltage is the difference between the initial voltage of the cell and the voltage drop across its internal resistance.

Output voltage = Initial voltage - Voltage drop
Output voltage = 3.0000 V - 0.0006 V
Output voltage = 2.9994 V

Considering the options given, the output voltage is 2.9994 V which is not listed, suggesting that there may be a typo in the provided options. The closest option to the calculated value would be: (a) 3.0000 V