Question

A 4.60-\(\mu\)F capacitor that is initially uncharged is connected in series with a 7.50-k\(\Omega\) resistor and an emf source with \(\varepsilon =\) 245 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

(a) 0 V initially, 245 V long-term; (b) 245 V initially, 0 V long-term; (c) 0 C initially, 1.127 mC long-term; (d) 32.7 mA initially, 0 A long-term.

Step-by-step solution

Understanding Initial Conditions

When the circuit is first completed, the capacitor is uncharged, meaning no voltage is across it. Hence, the initial voltage drop across the capacitor, \( V_C = 0 \) V.

Voltage Drop Across Resistor Initially

Just after the circuit is completed, since the capacitor acts as a short circuit (uncharged), all the source voltage appears across the resistor. Therefore, \( V_R = \varepsilon = 245 \) V.

Initial Charge on the Capacitor

Charge on the capacitor is given by \( Q = C \times V \). Initially, \( V=0 \), so \( Q = 4.60 \times 10^{-6} \times 0 = 0 \) C.

Initial Current Through the Resistor

The current through the resistor is given by Ohm's law \( I = \frac{V}{R} \). Initially, \( I = \frac{245}{7500} = 0.0327 \) A or 32.7 mA.

Long-term Voltage Drop Across Capacitor

After many time constants, the capacitor is fully charged and acts like an open circuit. Thus, the voltage across it equals the emf source, \( V_C = 245 \) V.

Long-term Voltage Drop Across Resistor

With the capacitor acting as an open circuit, no current passes through the resistor, so \( V_R = 0 \) V.

Long-term Charge on the Capacitor

Charge on the capacitor after a long time is given by \( Q = C \times V = 4.60 \times 10^{-6} \times 245 = 1.127 \times 10^{-3} \) C.

Long-term Current Through the Resistor

With the circuit open, the current through the resistor is \( I = 0 \) A.

Key concepts

Capacitor Charging

In an RC circuit, when a capacitor begins to charge, its voltage starts at zero and gradually increases as it accumulates charge. Initially, the capacitor acts like a short circuit because its voltage is zero. As a result, the entire voltage from the emf source is applied across the resistor, leaving the capacitor uncharged at the very beginning. Over time, however, the capacitor charges up and the voltage across it rises until it matches the emf source's voltage. The rate at which this charging occurs is an important aspect of circuit analysis, commonly described by the time constant, a key characteristic that determines how quickly the capacitor charges over time.

Voltage Drop

Voltage drop refers to the reduction of electric potential across elements in a circuit. Just after the circuit is closed, the voltage drop across the capacitor is zero because there is no initial charge on it, allowing the full voltage to drop across the resistor instead. As the capacitor charges, the voltage drop across it increases until it equals the battery voltage, effectively reducing the voltage drop across the resistor to zero. In analyzing circuits, it’s important to calculate these voltage drops to understand how the electrical potential is distributed.

Ohm's Law

Ohm's Law is a principle used to calculate current, voltage, and resistance in electrical circuits. It states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points, represented by the formula \( I = \frac{V}{R} \). In the initial state of the RC circuit, the entire emf source voltage appears across the resistor because the capacitor starts uncharged, resulting in a current through the resistor easily calculated using Ohm’s Law. As the capacitor becomes fully charged, the voltage across the resistor becomes zero, and subsequently, the current also drops to zero.

Time Constant

The time constant of an RC circuit is denoted by \( \tau = RC \). It represents the time taken for the voltage across the capacitor to reach approximately 63.2% of its maximum possible value during charging. Similarly, it also dictates how long it takes for the voltage to decay slightly below 36.8% of its initial value if discharging occurs. This time constant serves as a measure of how quickly the circuit responds to changes, whether charging or discharging, allowing for predictions of capacitor behavior over time. After a period equivalent to several time constants, the capacitor is considered to be fully charged, acting as an open circuit in the analysis.