Question

In a circuit with a capacitor (C = 30 × 10⁻⁶), resistance (R = 250 × 10³Ω), and voltage (E = 300V), when the switch is closed:

a) What is the charge on the capacitor?
b) What is the voltage at the resistor?
a) Q = 0.009C, V_R = 300V
b) Q = 9C, V_R = 0.12V
c) Q = 0.12C, V_R = 9V
d) Q = 300C, V_R = 0.009V

Answer 1

a) The charge on the capacitor is 0.009C. b) The voltage at the resistor is 0V.

Step-by-step explanation:

a) To determine the charge on the capacitor, we can use the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. Plugging in the given values, we get Q = (30 × 10-6) × 300 = 0.009C. a) The charge on the capacitor is 0.009C. b) The voltage at the resistor is 0V.

b) The voltage at the resistor can be found using Ohm's Law, V = IR, where V is the voltage, I is the current, and R is the resistance. Since the capacitor is fully charged, it acts as an open circuit and no current flows through the resistor. Therefore, the voltage at the resistor is 0V.