Amy needs 5.0 V for some integrated circuit experiments. She uses a 7.5 V battery and two resistors to make a voltage divider. One resistor is 310 . She decides to make the other resistor smaller. What value should it have?
She wants 2/3 the voltage across one resistor (the larger), and 1/3 across the smaller.
So, the smaller must be 1/2 the larger, since V=IR and the same current flows through both resistors.
To determine the value of the second resistor in the voltage divider, we need to use the voltage divider formula:
V_out = (R2 / (R1 + R2)) * V_in
In this case, Amy wants an output voltage (V_out) of 5.0 V, and she is using a 7.5 V battery (V_in). The value of one resistor is given as 310 Ω (R1).
Let's plug in these values and solve for R2:
5.0 V = (R2 / (310 Ω + R2)) * 7.5 V
To simplify this equation, we can cross-multiply:
5.0 V * (310 Ω + R2) = R2 * 7.5 V
Now, expand the left side of the equation:
1550 Ω + 5.0 V * R2 = 7.5 V * R2
Next, isolate R2 by moving the terms involving R2 to one side of the equation:
7.5 V * R2 - 5.0 V * R2 = 1550 Ω
Now, combine and factor out R2:
2.5 V * R2 = 1550 Ω
Divide both sides of the equation by 2.5 V:
R2 = 1550 Ω / 2.5 V
R2 = 620 Ω
Therefore, to achieve an output voltage of 5.0 V with a 7.5 V battery and a 310 Ω resistor, Amy should make the other resistor 620 Ω.