Answer :
Let's go through the problem step by step, analyzing the behavior of the circuit both immediately after it is completed and after it has been connected for a long time.
### Just After the Circuit is Completed
When a capacitor is initially uncharged and connected in a circuit, its behavior is determined by the instantaneous response, as follows:
(a) Voltage drop across the capacitor:
- An uncharged capacitor initially has no voltage drop across it, so the voltage drop across the capacitor is:
[tex]\[ V_{C_{initial}} = 0 \text{ V} \][/tex]
(b) Voltage drop across the resistor:
- Immediately after completing the circuit, the full potential of the emf source appears across the resistor, assuming negligible internal resistance. The voltage drop across the resistor is equal to the emf:
[tex]\[ V_{R_{initial}} = 125 \text{ V} \][/tex]
(c) Charge on the capacitor:
- Since the capacitor is initially uncharged, the charge on it is:
[tex]\[ Q_{C_{initial}} = 0 \text{ C} \][/tex]
(d) Current through the resistor:
- Using Ohm's Law [tex]\( V = I \cdot R \)[/tex], the initial current in the resistor can be computed by:
[tex]\[ I_{R_{initial}} = \frac{E}{R} = \frac{125 \text{ V}}{7.50 \times 10^3 \text{ Ω}} = 0.0167 \text{ A} \, (16.67 \, \text{mA}) \][/tex]
### After a Long Time (Many Time Constants)
After a long period—typically several time constants—the capacitor will be fully charged and the current in the circuit will have reached zero.
(a) Voltage drop across the capacitor:
- As the capacitor becomes fully charged, it takes on the entire voltage of the emf source. So the voltage drop across the capacitor is:
[tex]\[ V_{C_{final}} = 125 \text{ V} \][/tex]
(b) Voltage drop across the resistor:
- With no current flowing in the circuit after the capacitor is fully charged, the voltage drop across the resistor is zero:
[tex]\[ V_{R_{final}} = 0 \text{ V} \][/tex]
(c) Charge on the capacitor:
- The charge on the capacitor in the fully charged state can be determined using the capacitance formula [tex]\( Q = C \cdot V \)[/tex]:
[tex]\[ Q_{C_{final}} = 4.60 \times 10^{-12} \text{ F} \times 125 \text{ V} = 5.75 \times 10^{-10} \text{ C} \][/tex]
(d) Current through the resistor:
- As mentioned, after the capacitor is fully charged, no current flows through the circuit, so the current through the resistor is:
[tex]\[ I_{R_{final}} = 0 \text{ A} \][/tex]
### Summary
- Immediately after the circuit is completed:
- Voltage across the capacitor: [tex]\(0 \text{ V}\)[/tex]
- Voltage across the resistor: [tex]\(125 \text{ V}\)[/tex]
- Charge on the capacitor: [tex]\(0 \text{ C}\)[/tex]
- Current through the resistor: [tex]\(0.0167 \text{ A}\)[/tex]
- After a long time:
- Voltage across the capacitor: [tex]\(125 \text{ V}\)[/tex]
- Voltage across the resistor: [tex]\(0 \text{ V}\)[/tex]
- Charge on the capacitor: [tex]\(5.75 \times 10^{-10} \text{ C}\)[/tex]
- Current through the resistor: [tex]\(0 \text{ A}\)[/tex]
These values should give you a thorough understanding of the behavior of the RC circuit both at the moment it is completed and after it has been connected for a significant period.
### Just After the Circuit is Completed
When a capacitor is initially uncharged and connected in a circuit, its behavior is determined by the instantaneous response, as follows:
(a) Voltage drop across the capacitor:
- An uncharged capacitor initially has no voltage drop across it, so the voltage drop across the capacitor is:
[tex]\[ V_{C_{initial}} = 0 \text{ V} \][/tex]
(b) Voltage drop across the resistor:
- Immediately after completing the circuit, the full potential of the emf source appears across the resistor, assuming negligible internal resistance. The voltage drop across the resistor is equal to the emf:
[tex]\[ V_{R_{initial}} = 125 \text{ V} \][/tex]
(c) Charge on the capacitor:
- Since the capacitor is initially uncharged, the charge on it is:
[tex]\[ Q_{C_{initial}} = 0 \text{ C} \][/tex]
(d) Current through the resistor:
- Using Ohm's Law [tex]\( V = I \cdot R \)[/tex], the initial current in the resistor can be computed by:
[tex]\[ I_{R_{initial}} = \frac{E}{R} = \frac{125 \text{ V}}{7.50 \times 10^3 \text{ Ω}} = 0.0167 \text{ A} \, (16.67 \, \text{mA}) \][/tex]
### After a Long Time (Many Time Constants)
After a long period—typically several time constants—the capacitor will be fully charged and the current in the circuit will have reached zero.
(a) Voltage drop across the capacitor:
- As the capacitor becomes fully charged, it takes on the entire voltage of the emf source. So the voltage drop across the capacitor is:
[tex]\[ V_{C_{final}} = 125 \text{ V} \][/tex]
(b) Voltage drop across the resistor:
- With no current flowing in the circuit after the capacitor is fully charged, the voltage drop across the resistor is zero:
[tex]\[ V_{R_{final}} = 0 \text{ V} \][/tex]
(c) Charge on the capacitor:
- The charge on the capacitor in the fully charged state can be determined using the capacitance formula [tex]\( Q = C \cdot V \)[/tex]:
[tex]\[ Q_{C_{final}} = 4.60 \times 10^{-12} \text{ F} \times 125 \text{ V} = 5.75 \times 10^{-10} \text{ C} \][/tex]
(d) Current through the resistor:
- As mentioned, after the capacitor is fully charged, no current flows through the circuit, so the current through the resistor is:
[tex]\[ I_{R_{final}} = 0 \text{ A} \][/tex]
### Summary
- Immediately after the circuit is completed:
- Voltage across the capacitor: [tex]\(0 \text{ V}\)[/tex]
- Voltage across the resistor: [tex]\(125 \text{ V}\)[/tex]
- Charge on the capacitor: [tex]\(0 \text{ C}\)[/tex]
- Current through the resistor: [tex]\(0.0167 \text{ A}\)[/tex]
- After a long time:
- Voltage across the capacitor: [tex]\(125 \text{ V}\)[/tex]
- Voltage across the resistor: [tex]\(0 \text{ V}\)[/tex]
- Charge on the capacitor: [tex]\(5.75 \times 10^{-10} \text{ C}\)[/tex]
- Current through the resistor: [tex]\(0 \text{ A}\)[/tex]
These values should give you a thorough understanding of the behavior of the RC circuit both at the moment it is completed and after it has been connected for a significant period.
