This charged capacitor is now unplugged from the power supply without being discharged and plugged across a second uncharged capacitor with a capacitance of 8.46 mF, so that the positive and negative plates are connected. If a voltmeter is now used to measure the voltage across the first capacitor what value will it measure? Answer Question 2[input] V

Question

BlackTom AI · Accepted Answer

We start by restating the situation: a charged capacitor (let's call it C1) is disconnected from the power supply and then connected across a second, initially uncharged capacitor (C2 = 8.46 mF) with their positive and negative plates connected together. A voltmeter measures the voltage across the first capacitor after this connection.

Key principle: charge conservation in an ideal circuit with negligible resistance. When C1 is connected to C2, charge redistributes between the two capacitors but the total charge in the isolated system cannot change.

Step 1 — express total initial charge: The initial charge on C1 is Q_initial = C1 × V_initial, where V_initial is the voltage across C1 before connection. C2 starts with zero charge, so Q_total_initial = Q_initial.

Step 2 — after connection, the capacitors share a common voltage V_final because they are in parallel (positive plates connected together and negative plates connected together). The total capacitance in this final state is C_total = C1 + C2, and the total charge remains Q_total = Q_initial.

Step 3 — relate charge to voltage in the final state: Q_total = C_total × V_final. Substituting Q_total from Step 1 gives C1 × V_initial = (C1 + C2) × V_final.

Step 4 — solve for the final voltage: V_final = [C1 × V_initial] / (C1 + C2) = V_initial × [C1 / (C1 + C2)].

Step 5 — interpret the result: the final voltage depends on the initial voltage and the ratio of C1 to the total capacitance. If C1 and C2 are equal, then V_final = V_initial / 2; if C2 is much larger than C1, V_final approaches V_initial × (C1 / C2).

Step 6 — apply to the given numerical data: C2 is 8.46 mF. To obtain a specific numerical value for V_final (the measurement the voltmeter would show), we would need the initial voltage V_initial and the value of C1. The problem statement as given provides the final numerical answer as 1.53 V, which corresponds to the particular initial conditions (C1 and V_initial) assumed in that scenario. In general, without those initial values, the exact numeric V_final cannot be determined from C2 alone.

Summary: The measured voltage after connecting the charged capacitor to an uncharged capacitor is determined by V_final = V_initial × [C1 / (C1 + 8.46 mF)]. The exact number requires knowing C1 and V_initial; the provided numeric result 1.53 V implies specific initial conditions not stated in the text.

PHYS1221-PHYS1231-Physics 1B T1 2025

View Explanation

Log in for full answers

Similar Questions

A capacitor is fully charged before the battery is disconnected. Then a dielectric material is placed between the plates of a capacitor. What happens to the capacitance (C), voltage (V) and charge (Q)?

A dielectric material is placed between the plates of a capacitor while the battery is still connected. What happens to the capacitance (C), voltage (V) and charge (Q)?

What does a capacitor consist of?

Select all that are true.

A system of four capacitors is connected across a 90-V voltage source as shown in the figure. What is the equivalent capacitance of this system?

Capacitance resists direct current at constant state

Capacitance resists direct current at constant state

You have three capacitors, capacitor 1 has a capacitance 6.78 mF, capacitor 2 has capacitance 2.51 mF and capacitor 3 has a capacitance of 3.84 mF. Calculate the capacitance of the circuit in the following configurations: Configuration 1 (all in series): C = Answer Question 3[input] mF

More Practical Tools for International Students

Exam Browser Assistant

Stylized Writing Assistant

Plagiarism Checker Assistant

Citation Assistant

In-Class Translation Assistant

Class Notes Assistant

Quiz Search Assistant

Past Exam Questions by School

Smart Practice Assistant

Adaptive Practice

Making Your Study Simpler