For her electronics class, Gabriella configures a circuit as shown in the figure. Find the following. (Assume C, = 32.0 pF and C, = 3.63 pF.) 6.00 pF Cz pF =C, pF 9.00 V (a) the equivalent capacitance (in uF) 4.185 Remember that parallel capacitors add like resistors in series, and series capacitors add like resistors in parallel. pF (b) the charge on each capacitor (in uC) C, (left) c, (right) C2 6.00 µF capacitor (c) the potential difference across each capacitor (in V) c, (left) 1.18 Subtract the value of the potential difference across the parallel combination you determined in part (b) from the total potential difference. Because the two capacitors C, are equal, the potential difference across each is half of the difference. V C, (right) C2 3.91 Apply Q = CAV, and solve for AV. V 3.91 6.00 µF capacitor Apply Q= CAV, and solve for AV. V

For her electronics class, Gabriella configures a circuit as shown in the figure. Find the following. (Assume C, = 32.0 pF and C, = 3.63 pF.) 6.00 pF Cz pF =C, pF 9.00 V (a) the equivalent capacitance (in uF) 4.185 Remember that parallel capacitors add like resistors in series, and series capacitors add like resistors in parallel. pF (b) the charge on each capacitor (in uC) C, (left) c, (right) C2 6.00 µF capacitor (c) the potential difference across each capacitor (in V) c, (left) 1.18 Subtract the value of the potential difference across the parallel combination you determined in part (b) from the total potential difference. Because the two capacitors C, are equal, the potential difference across each is half of the difference. V C, (right) C2 3.91 Apply Q = CAV, and solve for AV. V 3.91 6.00 µF capacitor Apply Q= CAV, and solve for AV. V

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter20: Electric Potential And Capacitance

Section: Chapter Questions

Problem 46P

See similar textbooks

Similar questions

Is a circuit breaker wired (a) in series with the device it is protecting, (b) in parallel, or (c) neither in series or in parallel, or (d) is it impossible to tell?
The power dissipated in a resistor is given by P = V2/R, which means power decreases if resistance increases. Yet this power is also given by P = I2R, which means power increases if resistance increases. Explain why there is no contradiction here.
Figure P18.26 shows a voltage divider, a circuit used to obtain a desired voltage Vout from a source voltage . Determine the required value of R2 if = 5.00 V, Vout = 1.50 V and R1 = 1.00 103 (Hint: Use Kirchhoff's loop rule, substituting Vout = IR2, to find the current. Then solve Ohms law for R2. Figure P18.26
Consider the circuit shown in Figure P28.21 on page 860. (a) Find the voltage across the 3.00-0 resistor, (b) Find the current in the 3.00-12 resistor.
The values of the components in a simple series RC circuit containing a switch (Fig. P21.53) are C = 1.00 F, R = 2.00 106 , and = 10.0 V. At the instant 10.0 s after the switch is closed, calculate (a) the charge on the capacitor, (b) the current in the resistor, (c) the rate at which energy is being stored in the capacitor, and (d) the rate at which energy is being delivered by the battery.
Figure P18.26 shows a voltage divider, a circuit used to obtain a desired voltage Vout from a source voltage . Determine the required value of R2 if = 5.00 V, Vout = 1.50 V and R1 = 1.00 103 (Hint: Use Kirchhoff's loop rule, substituting Vout = IR2, to find the current. Then solve Ohms law for R2. Figure P18.26
A child's electronic toy is supplied by three 1.58-V alkaline cells having internal resistances of 0.0200 inseries with a 1.53-V carbon-zinc dry cell having a 0.100- internal resistance. The load resistance is 10.0 . (a) Draw a circuit diagram of the toy and itsbatteries, (b) What current flows? (c) How much power is supplied to the load? (d) What is the internal resistance of the dry cell if it goes bad, resulting in only 0.500 W being supplied to the load?
A battery with an internal resistance of 10.0 produces an open circuit voltage of 12.0 V. A variable load resistance with a range from 0 to 30.0 is connected across the battery. (Note: A battery has a resistance that depends on the condition of its chemicals and that increases as the battery ages. This internal resistance can be represented in a simple circuit diagram as a resistor in series with the battery.) (a) Graph the power dissipated in the load resistor as a function of the load resistance. (b) With your graph, demonstrate the following important theorem: The power delivered to a load is a maximum if the load resistance equals the internal resistance of the source.
A capacitor with initial charge Q0 is connected across a resistor R at time t = 0. The separation between the plates of the capacitor changes as d = d0/(1 + t) for 0 t 1 s. Find an expression for the voltage drop across the capacitor as a function of time.
Consider a circuit with three capacitors in parallel, where C1 = 1.50 μF, C2 = 8.50 μF, and C3 = 4.80 μF, find the equivalent capacitance.
A circuit with three resistors connected in series and parallel connection with a supply DC voltage is shown in the figure. The supply voltage of, Vsupply=120 V, Resistances R1=100, R2 = 200, R3=300 and R4= 400, R6=190 (R5 is not connected). What will be the magnitude of voltage, Vab between a and b?
Using the exact exponential treatment, find how much time is required to charge an initially uncharged 50-uF (micro-Farad) capacitor through a 75 M-Ohm (mega-ohm) resistor to 60.0% of its final voltage. Group of answer choices a)320 days b)57 minutes c)3.2x10^-2 sec d)4.5 minutes