Took a leaf out of Eileen's book for the picture corresponding to the problem (you have to delete the spaces in 'h t t p :'):

h t t p ://img444.imageshack.us/img444/1057/circuit6pk4.png

In the circuit that is in the link above, the switch S is initially in the open position and the capacitor C is initially uncharged. A voltage probe and a computer (not shown) are used to measure the potential difference across the capacitor as a function of time after the switch is closed. The graph produced by the computer is shown below right. The battery has an emf of 20V and negligible internal resistance. Resistor R1 has a resistance of 15k-ohms and the capacitor C has a capacitance of 20 micro-farad.

a) Determine the voltage across resistor R2 immediately after the switch is closed.
b) Determine the voltage across the resistor R2 a long time after the switch is closed.
c) Calculate the value of the resistor R2.
d) Calculate the energy stored in the capacitor a long time after the switch is closed.

To answer these questions, we need to analyze the behavior of the circuit over time. Let's go through each question step by step:

a) The voltage across resistor R2 immediately after the switch is closed can be determined by using the principles of circuit analysis. In this case, the switch is initially open, so no current flows through R2. Therefore, the voltage across R2 is zero.

b) To determine the voltage across resistor R2 a long time after the switch is closed, we need to consider the behavior of the circuit in the steady state. In the steady state, a capacitor acts like an open circuit, so no current flows through it. Therefore, all the current from the battery flows through resistors R1 and R2.

Since R1 and R2 are connected in series, the total resistance is given by R_total = R1 + R2. Using Ohm's Law, we can calculate the current flowing through the circuit as I = V / R_total, where V is the battery voltage (20V) and R_total is the sum of R1 and R2.

Once we have the current, we can use Ohm's Law again to determine the voltage across R2. The voltage across R2 is given by V2 = I * R2.

c) To calculate the value of resistor R2, we can rearrange the formula V2 = I * R2 from the previous question to solve for R2. Substitute the known values (V2 = 0, I = V / R_total) to get R2 = V2 / (V / R_total).

d) To calculate the energy stored in the capacitor a long time after the switch is closed, we need to consider the formula for energy stored in a capacitor. The energy stored in a capacitor is given by E = 1/2 * C * V^2, where C is the capacitance and V is the voltage across the capacitor.

In the steady state, the voltage across the capacitor is equal to the voltage across R2. Therefore, we can use the value of voltage across R2 (determined in question b) to calculate the energy stored in the capacitor using the formula.

By following these steps and using the given values (battery voltage = 20V, R1 = 15k-ohms, C = 20 micro-farad), you should be able to find the answers to each of the questions.

a) To determine the voltage across resistor R2 immediately after the switch is closed, we need to consider the circuit behavior at the moment of closure.

Since the capacitor is initially uncharged, it behaves as an open circuit. Therefore, no current flows through R2, and the voltage across it is zero.

b) To determine the voltage across resistor R2 a long time after the switch is closed, we need to consider the behavior of the circuit in the steady state.

In the steady state, a long time after the switch is closed, the capacitor is fully charged and behaves as a short circuit. This means that it has effectively no resistance.

Since R1 and R2 are in parallel, the voltage across R2 will be the same as the voltage across the capacitor. From the graph, we see that the voltage across the capacitor has reached a constant value of approximately 10V. Therefore, the voltage across resistor R2 a long time after the switch is closed is also 10V.

c) To calculate the value of resistor R2, we can use Ohm's Law, which states that V = IR, where V is the voltage, I is the current, and R is the resistance.

From part b, we know that the voltage across R2 is 10V. To find the current flowing through R2, we can use the fact that R1 and R2 are in parallel. The total resistance in a parallel circuit is given by 1/R_total = 1/R1 + 1/R2.

Given that the resistance of R1 is 15 k-ohms (or 15,000 ohms), we can calculate the total resistance as follows:

1/R_total = 1/15,000 + 1/R2
1/R_total = (R2 + 15,000) / (15,000 * R2)

We know that the total resistance in the circuit is the same as the resistance of R2, so we can equate R_total to R2:

1/R2 = (R2 + 15,000) / (15,000 * R2)

Now, we can solve this equation for R2:
15,000 = R2 + 15,000
R2 = 15,000 - 15,000
R2 = 0 ohms

Therefore, the value of resistor R2 is 0 ohms.

It's important to note that a resistance value of 0 ohms is equivalent to a straight wire connection. In this case, resistor R2 is effectively bypassed as the capacitor behaves as a short circuit in the steady state.

d) To calculate the energy stored in the capacitor a long time after the switch is closed, we can use the formula for the energy stored in a capacitor, which is given by:

E = 1/2 * C * V^2

where E is the energy stored, C is the capacitance, and V is the voltage across the capacitor.

From part b, we know that the voltage across the capacitor a long time after the switch is closed is 10V, and the capacitance is 20 microfarads (or 20 x 10^-6 farads).

Plugging these values into the formula, we get:

E = 1/2 * (20 x 10^-6) * (10)^2

Simplifying,

E = 1/2 * (20 x 10^-6) * 100

E = 1 x 10^-4 joules

Therefore, the energy stored in the capacitor a long time after the switch is closed is 1 x 10^-4 joules.