Finding Current in Resistors & Charge of Capacitor

In summary: I just calculated the voltage and the charge using that voltage and I got the correct result. Thank you so much for the help!
  • #1
ananonanunes
18
6
Homework Statement
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Relevant Equations
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Suppose the switch has been closed for a long time so that the capacitor is fully charged and current is constant.
a)Find the current in each resistor and charge Q of the capacitor.
b)The switch is now opened at t=0s. Write the equation for the current for the resistor of 15kΩ as a function of time and find the time interval needed for the charge to drop to 1/5 of its original value.
1682093015932.png

I was able to find the current in the resistors. I assumed the current would be 0A in the resistor of ##3k\Omega## and since there is no current in this "branch", the other two resistors are in series so ##I_{R_1}=I_{R_2}## and ##R_{eq}=12+15=27k \Omega##. ##I=\frac{\mathcal{E}}{R_{eq}}=333mA##.
I don't know how to find the charge in the capacitor. I thought that, because it was fully charged, ##Q=C\mathcal{E}## but that is not the correct answer.
 
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  • #2
##Q=CV.##
What do you think the voltage across the capacitor is? Hint: There is no current through the 3 kΩ resistor.
 
  • #3
kuruman said:
##Q=CV.##
What do you think the voltage across the capacitor is? Hint: There is no current through the 3 kΩ resistor.
Since there is no current through the ##3k\Omega## there is also no voltage across that resitor. So I thought the voltage across the capacitor would be the same as the electromotive force, because the voltage wouldn't need to be "split" between the capacitor and the resistor. But this is incorrect and I don't really understand why.
 
  • #4
ananonanunes said:
Since there is no current through the ##3k\Omega## there is also no voltage across that resitor. So I thought the voltage across the capacitor would be the same as the electromotive force, because the voltage wouldn't need to be "split" between the capacitor and the resistor. But this is incorrect and I don't really understand why.
Look at the circuit again. The voltage across the capacitor is the same as the voltage across the 15kΩ resistor. What is that voltage?
 
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  • #5
kuruman said:
Look at the circuit again. The voltage across the capacitor is the same as the voltage across the 15kΩ resistor. What is that voltage?
I just calculated the voltage and the charge using that voltage and I got the correct result. Thank you so much for the help!
 
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Related to Finding Current in Resistors & Charge of Capacitor

1. How do you calculate the current through a resistor in a series circuit?

In a series circuit, the current through each resistor is the same. You can calculate the current by using Ohm's Law, which states I = V/R, where I is the current, V is the total voltage across the series circuit, and R is the total resistance. The total resistance in a series circuit is the sum of the individual resistances.

2. How do you find the equivalent resistance in a parallel circuit?

To find the equivalent resistance (R_eq) in a parallel circuit, you use the formula 1/R_eq = 1/R1 + 1/R2 + 1/R3 + ... for all resistors in the circuit. After calculating the reciprocal of the equivalent resistance, take the inverse to find R_eq.

3. How do you determine the charge on a capacitor in a DC circuit?

The charge (Q) on a capacitor in a DC circuit can be determined using the formula Q = C * V, where C is the capacitance of the capacitor, and V is the voltage across the capacitor. This formula assumes the capacitor is fully charged.

4. What happens to the current through a resistor when additional resistors are added in parallel?

When additional resistors are added in parallel, the total resistance of the circuit decreases. According to Ohm's Law, if the voltage remains constant, a decrease in resistance will result in an increase in the total current supplied by the source. However, the current through each individual resistor will depend on its own resistance and the applied voltage.

5. How does the time constant affect the charging and discharging of a capacitor?

The time constant (τ) of an RC circuit, which is the product of resistance (R) and capacitance (C), affects the rate at which a capacitor charges or discharges. A larger time constant means the capacitor charges or discharges more slowly. The voltage across the capacitor during charging or discharging can be described by exponential functions involving the time constant.

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