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Chapter 26: Problem 3

A circuit consists of a source of emf, a resistor, and a capacitor, all connected in series. The capacitor is fully charged. How much current is flowing through it? a) \(i=V / R\) b) zero c) neither (a) nor (b)

Short Answer

Expert verified
a) i=V/R b) zero c) neither (a) nor (b) Answer: b) zero

Step by step solution

01

Understand the behavior of a charged capacitor

A fully charged capacitor means that the potential difference across its plates is equal to the source of emf (V). In other words, it can no longer store any more charge; hence no more current will flow through it.
02

Analyze the current in the circuit

Once the capacitor gets fully charged, it acts like an open circuit, disconnecting the circuit, and no current flows through it. The current through a capacitor is equal to the rate of change of charge with respect to time (\(i= dq/dt\)). If the capacitor is not charging or discharging, this rate becomes zero and no current flows through it.
03

Choose the correct answer

Since no current flows through a fully charged capacitor, the correct answer is: b) zero

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Most popular questions from this chapter

A circuit consists of two \(100 .-\mathrm{k} \Omega\) resistors in series with an ideal \(12.0-\mathrm{V}\) battery. a) Calculate the potential drop across one of the resistors. b) A voltmeter with internal resistance \(10.0 \mathrm{M} \Omega\) is connected in parallel with one of the two resistors in order to measure the potential drop across the resistor. By what percentage will the voltmeter reading deviate from the value you determined in part (a)? (Hint: The difference is rather small so it is helpful to solve algebraically first to avoid a rounding error.)

A capacitor bank is designed to discharge 5.0 J of energy through a \(10.0-\mathrm{k} \Omega\) resistor array in under \(2.0 \mathrm{~ms}\) To what potential difference must the bank be charged, and what must the capacitance of the bank be?

An ammeter with an internal resistance of \(53 \Omega\) measures a current of \(5.25 \mathrm{~mA}\) in a circuit containing a battery and a total resistance of \(1130 \Omega\). The insertion of the ammeter alters the resistance of the circuit, and thus the measurement does not give the actual value of the current in the circuit without the ammeter. Determine the actual value of the current.

A battery has \(V_{\text {emf }}=12.0 \mathrm{~V}\) and internal resistance \(r=1.00 \Omega\). What resistance, \(R,\) can be put across the battery to extract \(10.0 \mathrm{~W}\) of power from it?

Two light bulbs for use at \(110 \mathrm{~V}\) are rated at \(60 \mathrm{~W}\) and \(100 \mathrm{~W}\), respectively. Which has the filament with lower resistance?
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