Question

Consider the circuit and assume the battery has no internal resistance. (i) Just after the switch is closed, what is the current in the battery? (a) 0 (b) e/2R (c) 2e/R (d) e/R (e) impossible to determine (ii) After a very long time, what is the current in the battery? Choose from the same choices.

Answer 1

Just after the switch is closed, the current in the battery is e/R. After a very long time, the current in the battery is also e/R.

Step-by-step explanation:

In order to determine the current in the battery just after the switch is closed, we need to analyze the circuit. Since the battery has no internal resistance, the current in the battery will be determined by the total resistance in the circuit. When the switch is closed, the equivalent resistance of the circuit will be R. According to Ohm's Law, the current in the battery will be e/R, where e is the electromotive force of the battery and R is the resistance. Therefore, the answer to the first part of the question is (d) e/R.

After a very long time, the capacitor in the circuit will be fully charged and act as an open circuit. This means that no current will flow through the capacitor, and the equivalent resistance of the circuit will be the sum of the resistances of R and R2. According to Ohm's Law, the current in the battery will be e/(R + R2). Therefore, the answer to the second part of the question is also (d) e/R.

Answer 2

Just after the switch is closed, the current in the battery is (e/R). After a very long time, the current in the battery will be zero.

Step-by-step explanation:

In the given circuit, just after the switch is closed, the current in the battery is (e/R).

To understand why, we need to apply Kirchhoff's loop rule to the circuit. Kirchhoff's loop rule states that the sum of the potential drops across each component in a closed loop is equal to the sum of the potential rises. When the switch is closed, the circuit forms a closed loop which only includes the battery and the resistor. Since the battery has no internal resistance, there is no potential drop across it. Therefore, the potential drop across the resistor, which represents the voltage across the resistor, is equal to the potential rise of the battery.

Using Ohm's Law, which states that V = IR, we can express the potential drop across the resistor in terms of current and resistance: V = IR. Since the potential drop across the resistor is equal to the potential rise of the battery, we can write: IR = e. Solving for I, we get: I = e/R. Therefore, the current in the battery just after the switch is closed is e/R.

After a very long time, the current in the battery will be (0). This is because the capacitor in the circuit will become fully charged and act as an open circuit, blocking the flow of current. Since there is no path for the current to flow, the current in the battery will be zero.