Homework on RC Circuits - 2000 version (solutions)

Feel free to work on the homework in groups. The work you hand in, however, should reflect your understanding of the material. You should show all of your calculations (neatly) and justify all of your answers for full credit.

Power supply hooked in series to light bulb and capacitor A capacitor of capacitance C = 1.0 F (Farad) is placed in a circuit with a light bulb, as shown here. The power supply provides a potential difference of V₀ = 10.0 V, and the light bulb has a resistance of R = 45.0 W (Ohms) when the potential difference across it is 10.0 V.
In answering some of these questions, you will need to remember that the potential difference around the complete circuit is zero, so
V₀ -V_R - V_C = 0 You should also remember the definitions of resistance and capacitance: R = V/i (i is current)
C = Q/V (Q is charge)

1. What is the maximum amount of charge Q₀ the capacitor will hold in this circuit?
The maximum voltage on the capacitor is V0. Using the definition of capacitance one finds Q₀ = C V₀ = (1 F)(10 V) = 10 C

2. How much current will pass through the light bulb when it has a potential difference of 10.0 V?
Since resistance and voltage are given, one can solve for the current using ohms law:

3. What is the time constant of the circuit?
Using the definition of the time constant, t, t = RC = (45.0 W)(1.0 F) = 45 s

4. The capacitor is completely discharged, then the charging circuit shown above is connected. What will the voltage on the capacitor be after one time constant has elapsed?
The expression for charging of a capacitor is V = V₀ (1 – e ^–t/t ) When t is equal to one time constant, t, the exponential reduces to e^-1 and one can write V = V₀ (1 – e^-1) = (10.0 V)(0.63) = 6.3 V

5. How long after the charging circuit is connected will it take for the voltage across the capacitor to reach 5.0 Volts? How long will it take for the voltage across the capacitor to reach 7.5 Volts?
Here, again, the expression for a charging capacitor is used. However, in this case the equation must be solved for t since V is known. Substituting in the appropriate values one finds t = 31 s and 62 s required for the capacitor to charge to 5.0 V and 7.5 V, respectively.

6. What will the voltage across the capacitor be after 93 seconds have elapsed? Is this consistent with your responses to question 5?
Substituting t = 93 s into the charging expression gives the result: V = V₀ (1 –e^-t/t) = 10.0 V (1 – e^-93/45) = 8.73 V. This is consistent with question 5, since we found 31 seconds to be the “half-life”, or the time for the capacitor to cut the difference between its current voltage and the maximum voltage in half. After 93 seconds this difference will be halved three times, leaving 1/8 remaining. The difference between V₀=10.0 V and V = 8.73 V is indeed about 10.0V/8 = 1.25 V.

7. How much energy is stored in the capacitor when it is fully charged?
The energy stored in a capacitor is given by U = ½ CV² Substituting appropriate values one has U = ½ (1.0 F)(10.0 V)² = 50 J

8. Through the magic of age-old technology known as imagination, the spacing between the plates of the capacitor is doubled while the capacitor is still hooked up to the power supply. What happens to (a) the capacitance, (b) the charge on the capacitor, (c) the energy stored in the fully-charged capacitor, and (d) the energy density of the fully-charged capacitor? (e) Does the light bulb light up when you separate the plates?
(a) The general expression for the capacitance of a parallel plate capacitor is C = eA/d In this case the distance between the plates, d, doubles while the area, A, remains constant. Thus the capacitance is halved.
(b) From the definition of capacitance one can write Q = CV In this case the voltage across the capacitor is held constant by the power supply. In order to maintain this voltage when the capacitance is halved the charge on the plates must be redistributed so that the charge on the plates is also halved.
(c) Going back to the expression for the energy stored in a capacitor's electric field, U = ½ CV², it is clear that if voltage is fixed, a decrease in the capacitance by half will reduce the stored energy by half as well.
(d) The energy density in the capacitor is given by the expression u = ½ eE² The electric field between two parallel capacitor plates is related to the potential between them by E = V/d so that for a fixed potential an increase in d by a factor of two will reduce E by one half. From the expression for energy density it may be determined that the energy density will then fall to one fourth of its previous value.
(e) Yes, the redistribution of charge between the plates produces a current which, if strong enough, could light the bulb. This would depend on the rate at which the plates were separated.

1.	What is the maximum amount of charge Q₀ the capacitor will hold in this circuit? The maximum voltage on the capacitor is V0. Using the definition of capacitance one finds Q₀ = C V₀ = (1 F)(10 V) = 10 C
2.	How much current will pass through the light bulb when it has a potential difference of 10.0 V? Since resistance and voltage are given, one can solve for the current using ohms law:
3.	What is the time constant of the circuit? Using the definition of the time constant, t, t = RC = (45.0 W)(1.0 F) = 45 s
4.	The capacitor is completely discharged, then the charging circuit shown above is connected. What will the voltage on the capacitor be after one time constant has elapsed? The expression for charging of a capacitor is V = V₀ (1 – e ^–t/t ) When t is equal to one time constant, t, the exponential reduces to e^-1 and one can write V = V₀ (1 – e^-1) = (10.0 V)(0.63) = 6.3 V
5.	How long after the charging circuit is connected will it take for the voltage across the capacitor to reach 5.0 Volts? How long will it take for the voltage across the capacitor to reach 7.5 Volts? Here, again, the expression for a charging capacitor is used. However, in this case the equation must be solved for t since V is known. Substituting in the appropriate values one finds t = 31 s* and 62 s required for the capacitor to charge to 5.0 V and 7.5 V, respectively.*
6.	What will the voltage across the capacitor be after 93 seconds have elapsed? Is this consistent with your responses to question 5? Substituting t = 93 s into the charging expression gives the result: V = V₀ (1 –e^-t/t) = 10.0 V (1 – e^-93/45) = 8.73 V. This is consistent with question 5, since we found 31 seconds to be the “half-life”, or the time for the capacitor to cut the difference between its current voltage and the maximum voltage in half. After 93 seconds this difference will be halved three times, leaving 1/8 remaining. The difference between V₀=10.0 V and V = 8.73 V is indeed about 10.0V/8 = 1.25 V.
7.	How much energy is stored in the capacitor when it is fully charged? The energy stored in a capacitor is given by U = ½ CV² Substituting appropriate values one has U = ½ (1.0 F)(10.0 V)² = 50 J
8.	Through the magic of age-old technology known as imagination, the spacing between the plates of the capacitor is doubled while the capacitor is still hooked up to the power supply. What happens to (a) the capacitance, (b) the charge on the capacitor, (c) the energy stored in the fully-charged capacitor, and (d) the energy density of the fully-charged capacitor? (e) Does the light bulb light up when you separate the plates? (a) The general expression for the capacitance of a parallel plate capacitor is C = eA/d In this case the distance between the plates, d, doubles while the area, A, remains constant. Thus the capacitance is halved. (b) From the definition of capacitance one can write Q = CV In this case the voltage across the capacitor is held constant by the power supply. In order to maintain this voltage when the capacitance is halved the charge on the plates must be redistributed so that the charge on the plates is also halved. (c) Going back to the expression for the energy stored in a capacitor's electric field, U = ½ CV², it is clear that if voltage is fixed, a decrease in the capacitance by half will reduce the stored energy by half* as well.* (d) The energy density in the capacitor is given by the expression u = ½ eE² The electric field between two parallel capacitor plates is related to the potential between them by E = V/d so that for a fixed potential an increase in d by a factor of two will reduce E by one half. From the expression for energy density it may be determined that the energy density will then fall to one fourth of its previous value. (e) Yes, the redistribution of charge between the plates produces a current which, if strong enough, could light the bulb. This would depend on the rate at which the plates were separated.