Math Problem Statement

Considere el siguiente circuito RC donde la FEM tiene 9 V. (a) ¿Cuál es la corriente inicial en el circuito una vez que se cierra el interruptor? (b) ¿Cuál es la corriente final a tiempos muy largos en el circuito una vez que se cierra el interruptor? Y en este caso, ¿con qué voltaje queda cargado cada capacitor?

Solution

The problem states:

"Consider the following RC circuit where the EMF (electromotive force) is 9 V.

(a) What is the initial current in the circuit once the switch is closed?

(b) What is the final current at very long times in the circuit after the switch is closed?

(c) In this case, what voltage remains on each capacitor after the switch is closed?"


To solve this, let's break down each part:

Given Information

  1. The EMF (voltage source) is 9 V.
  2. Each resistor has a resistance of 1kΩ1 \, k\Omega.
  3. Each capacitor has a capacitance of 10μF10 \, \mu F.

Part (a): Initial Current

When the switch is first closed, the capacitors act like short circuits (since they are uncharged and do not resist the flow of current initially). Therefore, the initial current can be calculated as if the capacitors were not there.

  1. Total resistance in series: Rtotal=1kΩR_{\text{total}} = 1 \, k\Omega

  2. Using Ohm's Law: Iinitial=VRtotal=9V1kΩ=9mAI_{\text{initial}} = \frac{V}{R_{\text{total}}} = \frac{9 \, V}{1 \, k\Omega} = 9 \, mA

Part (b): Final Current at Long Times

At very long times, the capacitors become fully charged, which means no current flows through them. Therefore, the final current in the circuit will be zero because the capacitors will act as open circuits, blocking any current flow.

Part (c): Voltage on Each Capacitor at Long Times

At steady state, each capacitor will have a voltage across it that depends on the voltage division in the circuit. Since the resistors and capacitors are in parallel branches, the EMF will be divided across them.

For this specific configuration, each capacitor will end up with a voltage equal to the EMF, since they are effectively in parallel once charged.

Therefore: Vcapacitor=9VV_{\text{capacitor}} = 9 \, V


Would you like a detailed derivation of any part, or do you have questions?

Related Questions

  1. What happens to the current in an RC circuit as time progresses after the switch is closed?
  2. How do you calculate the time constant in an RC circuit?
  3. What role do capacitors play in limiting current flow in an RC circuit?
  4. How can you experimentally measure the initial current in an RC circuit?
  5. What would change if the EMF were increased in this circuit?

Tip

In an RC circuit, the time constant τ=R×C\tau = R \times C determines how quickly the capacitor charges or discharges.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Electric Circuits
RC Circuits
Current and Voltage Analysis
Ohm's Law

Formulas

I = V / R (Ohm's Law)
Voltage across capacitor in steady-state = EMF of the source in parallel

Theorems

Ohm's Law
Steady-State Analysis in RC Circuits

Suitable Grade Level

Undergraduate Electrical Engineering or Physics