If R = 4.0 kΩ, C = 3.0 mF, ε = 15 V, Q = 12 mC, and I = 2.0 mA, what is the potential difference V_b − V_a? A) +3.0 V B) −19 V C) −3.0 V D) +27 V E) +21 V

Question

Quizplus · Accepted Answer

We can use Kirchhoff's Voltage Law (KVL) to find the potential difference. Along the loop that includes VS1U1B1bS1U1B0 and VS1U1B1aS1U1B0, the potential difference can be expressed as:

VS1U1B1bS1U1B0 − U1B1bC − I(R+R) − ε + U1B1aC + U1B1aS1U1B0 = 0

Simplifying and solving for the potential difference gives:

VS1U1B1bS1U1B0 − VS1U1B1aS1U1B0 = ε - 2IR - 2UC

Substituting the given values:

VS1U1B1bS1U1B0 − VS1U1B1aS1U1B0 = 15V - 2(4kΩ)(2mA) - 2(3mF)(12mC)/(3mF+3mF)

VS1U1B1bS1U1B0 − VS1U1B1aS1U1B0 = -3V

Therefore, the answer is C.

If R = 40 KΩ, C = 3