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A \(4.60 \mu F\) capacitor that is initially uncharged is connected in series with a \(7.50 \mathrm{k} \Omega\) resistor and an \(\mathrm{emf}\) source with \(\mathcal{E}=245 \mathrm{~V}\) and negligible internal resistance. Just after the circuit is completed, what are (a) the voltage drop across the capacitor; (b) the voltage drop across the resistor; (c) the charge on the capacitor; (d) the current through the resistor? (e) A long time after the circuit is completed (after many time constants) what are the values of the quantities in parts (a)-(d)?

Short Answer

Expert verified
The initial values just after the circuit is completed are \(V_c = 0\), \(V_r = 245 V\), \(Q = 0 C\), and \(I = 245/7.5k = 32.67 mA\). After a long time, the values become \(V_c = 245 V\), \(V_r = 0 V\), \(Q = 4.6μF * 245V = 1.127 C\), and \(I = 0 A\).

Step by step solution

01

Calculate initial and final values

Just after the circuit is completed, the voltage across the capacitor \(V_c\) is 0 because the capacitor has not had time to charge. Thus, the voltage drop across the resistor \(V_r\) is equal to the emf source \(\mathcal{E}\). Since there is no charge on the capacitor initially, \(Q = 0 C\). The current \(I\) in the circuit can be calculated using Ohm's law, \(I = V/R\), where \(V = \mathcal{E}\) is the voltage supplied by the emf source and \(R\) is the resistance.
02

Calculation for a long time after the circuit is complete

A long time after the circuit is completed, the capacitor behaves like an open circuit (it's fully charged) so there's no current flowing in the circuit. Therefore, voltage drop across the resistor \(V_r\) is 0. However, voltage drop across the capacitor \(V_c\) will be equal to emf source \(\mathcal{E}\). Since the capacitor is fully charged, the charge \(Q\) on the capacitor can be calculated using the formula for the charge on a fully charged capacitor, which is \(Q = CV\), where \(C\) is the capacitance and \(V\) is the voltage.
03

Plugging in the given values

Now we substitute the given values into the equations derived in steps 1 and 2 to get the individual quantities asked for.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Capacitor Charging
When a capacitor is connected to a power source, it starts charging up. Initially, a capacitor holds no charge, so the voltage across it is zero. This is because the electric field inside it has not yet been established.
As time progresses, the capacitor charges up, storing energy in the form of an electric field between its plates. The rate at which it charges depends on the resistance and capacitance in the circuit; these factors describe how quickly or slowly the capacitor can fill up with charge.
During this charging phase, the current flows from the power source, through the resistor, and into the capacitor. Eventually, the capacitor reaches a point where it is fully charged, meaning the voltage across its terminals equals the voltage from the source. At this stage, no more current flows in the circuit.
Ohm's Law
Ohm's Law is critical in analyzing circuits like the one described. It establishes a fundamental relationship between voltage (V), current (I), and resistance (R): \[I = \frac{V}{R} \]Where
  • \(I\) is the current in amperes (A),
  • \(V\) is the voltage in volts (V), and
  • \(R\) is the resistance in ohms (Ω).
For a new circuit, such as one where the capacitor just starts to charge, the entire supplied voltage appears across the resistor. Since resistors are designed to limit current flow, Ohm's Law helps predict the rate of current flow when a given voltage is applied.
In our exercise, immediately after connection, all voltage from the source appears across the resistor as the capacitor has not charged up yet. This allows us to determine the initial current flowing through the circuit.
Time Constant
The time constant \(\tau\), a key concept in RC circuits, determines how fast the capacitor charges or discharges. It is defined as the product of the resistance (R) and the capacitance (C): \[\tau = R \cdot C \]
  • For our circuit, \(R = 7.50 \text{k}\Omega\) and \(C = 4.60 \mu F\) (microfarads).
  • The time constant gives us a measure: essentially, it is the time required for the voltage across the capacitor to reach approximately 63% of its maximum value, or in reverse, to drop to 37% during discharge.
After several time constants (about five times \(\tau\)), the capacitor is considered fully charged, and almost no current flows through the circuit anymore. This allows engineers to forecast and manage the behavior of the circuit over time.
Voltage Across Resistor
The voltage across a resistor is significant in understanding the behavior of an RC circuit. Initially, just after the circuit is connected, the entire voltage from the power source is across the resistor since the capacitor has not started charging yet. Hence, we have: \[ V_r = \mathcal{E} \]Where
  • \(V_r\) is the voltage drop across the resistor.
  • \(\mathcal{E}\) is the electromotive force provided by the source.
As the capacitor charges over time, this voltage gradually decreases. Eventually, when the capacitor becomes fully charged, the voltage across the resistor drops to zero, because the capacitor assumes the full supply voltage. Understanding this transition helps in predicting how the circuit consumes energy and operates under steady-state conditions.

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Most popular questions from this chapter

A \(12.4 \mu \mathrm{F}\) capacitor is connected through a \(0.895 \mathrm{M} \Omega\) resistor to a constant potential difference of \(60.0 \mathrm{~V}\). (a) Compute the charge on the capacitor at the following times after the connections are made: \(0,5.0 \mathrm{~s}, 10.0 \mathrm{~s}, 20.0 \mathrm{~s},\) and \(100.0 \mathrm{~s} .\) (b) Compute the charging currents at the same instants. (c) Graph the results of parts (a) and (b) for \(t\) between 0 and \(20 \mathrm{~s}\).

The three-terminal device circled in Fig. \(\mathbf{P} 26.81\) is an NPN transistor. It works using three simple rules: I. The net current into the device is the same as the net current out of the device, so \(I_{\mathrm{C}}+I_{\mathrm{B}}=I_{\mathrm{E}} .\) II. The potential at point \(e\) is always \(0.60 \mathrm{~V}\) less than the potential at point \(b .\) Thus, \(V_{\text {in }}-V_{e}=0.60 \mathrm{~V}\). III. The current into point \(c\) is always a fixed multiple of the current into point \(b,\) so \(I_{\mathrm{C}}=\beta I_{\mathrm{B}},\) where \(\beta \gg 1\) is a parameter characteristic of the transistor. Use these rules to answer the questions. (a) What is the potential \(V_{\text {out }}\) in terms of \(V_{\text {in }},\) in the limit \(\beta \rightarrow \infty\) ? (Hint: Determine \(I_{E}\) in terms of \(V_{\text {in }}\), and use this to determine \(I_{\mathrm{C}}\) in terms of \(V_{\mathrm{in}}\) and \(\beta .\) The potential difference across the \(1.0 \mathrm{k} \Omega\) resistor can then be used to determine \(V_{\text {out }}\) Take the limit \(\beta \rightarrow \infty\) in this result. \((b)\) What value of \(V_{\text {in }}\) is needed so that \(V_{\text {out }}=7.5 \mathrm{~V} ?\) (c) If the input potential includes a small time-dependent "signal," so that \(V_{\text {in }}=15.0 \mathrm{~V}+v_{\text {in }}(t),\) then the output potential is \(V_{\text {out }}=7.5\) \(\mathrm{V}+G v_{\mathrm{in}}(t),\) where \(G\) is a "gain" factor. What is \(G\) for this circuit?

A capacitor with \(C=6.00 \mu \mathrm{F}\) is fully charged by connecting it to a battery that has emf \(50.0 \mathrm{~V}\). The capacitor is disconnected from the battery. A resistor of resistance \(R=185 \Omega\) is connected across the capacitor, and the capacitor discharges through the resistor. (a) What is the charge \(q\) on the capacitor when the current in the resistor is \(0.180 \mathrm{~A} ?\) (b) If the connection to the resistor is completed at time \(t=0,\) what is the value of \(t\) when the current has the value specified in part (a)?

A \(42 \Omega\) resistor and a \(20 \Omega\) resistor are connected in parallel, and the combination is connected across a \(240 \mathrm{~V}\) de line. (a) What is the resistance of the parallel combination? (b) What is the total current through the parallel combination? (c) What is the current through each resistor?

A capacitor that is initially uncharged is connected in series with a resistor and an emf source with \(\mathcal{E}=110 \mathrm{~V}\) and negligible internal resistance. Just after the circuit is completed, the current through the resistor is \(6.5 \times 10^{-5}\) A. The time constant for the circuit is \(5.2 \mathrm{~s}\). What are the resistance of the resistor and the capacitance of the capacitor?

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