Question
information that is known already
An uncharged capacitor and a resistor are connected in series to a battery as shown in the figure, where epsilon = 12.0 V, C = 5.40 mu F, and R = 8.70 times 105 ohm. The switch is thrown to position a. Find the time constant of the circuit, the maximum charge on the capacitor, the maximum current in the circuit, and the charge and current as functions of time. Now, a second resistor R2 of 3.10 x 105 ft is connected in parallel to the existing resistor in the circuit, and a second capacitor C2 = 3.50 nF is connected to the existing capacitor in parallel. What will be the new time constant? Treat the system as a simple RCcircuit with an equivalent capacitance for C and an equivalent resistance for R to calculate the time constant for each, What will be the maximum current in the circuit (leaving the battery terminal)? Now, we connect the second resistor R2 of 3.10 x 105 ft and the second capacitor C2 = 3.50 nF both in series to the other elements of the circuit. What will be the new time constant? as a simple RC circuit with an equivalent capacitance for C and an equivalent resistance for R to calculate the time constant for each, What will be the maximum current in the circuit (leaving the battery terminal)?
Explanation / Answer
solution:
Part a:
New resistance in the circuit, R_new = R1 || R2 = (8.7*10^5) || (3.1*10^5) = 2.285*10^5 ohm
New capacitance in circuit, C_new = C1 + C2 = (5.4*10^-6) + (3.5*10^-6) = 8.9*10^-6 F
New time constant , = R_new * C_new = (2.285*10^5) * (8.9*10^-6) = 2.034 s
Max current, I_max = E / R_new = 12V / (2.285*10^5 ohm) = 52.516 *10^-6 A
Part b:
New resistance in the circuit, R_new = R1 + R2 = (8.7*10^5) + (3.1*10^5) = 11.8*10^5 ohm
New capacitance in circuit, C_new = C1 || C2 = (5.4*10^-6) || (3.5*10^-6) = 2.12*10^-6 F
New time constant , = R_new * C_new = (11.8*10^5) * (2.12*10^-6) = 2.51 s
Max current, I_max = E / R_new = 12V / (11.8*10^5 ohm) = 10.17 *10^-6 A
