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so shouldn't you use the formula V=Vs(e^(-t/RC)) instead of the formula V=Vs(1-e^(-t/RC)), which is for charging capacitors? Please correct me and tell me why if I'm wrong. Thank you!

so shouldn't you use the formula V=Vs(e^(-t/RC)) instead of the formula V=Vs(1-e^(-t/RC)), which is for charging capacitors? Please correct me and tell me why if I'm wrong. Thank you!

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For b of this question, I think it's using the wrong formula. The question itself specifically states that capacitors are discharging, so shouldn't you use the formula V=Vs(e^(-t/RC)) instead of the formula V=Vs(1-e^(-t/RC)), which is for charging capacitors? Please correct me and tell me why if I'm wrong. Thank you!

**Textbook Problem** At time \( t = 0 \), an \( RC \) circuit consists of a 12.0-V emf device, a 60.0-\(\Omega\) resistor, and a 150.0-\(\mu F\) capacitor that is fully charged. The switch is thrown so that the capacitor begins to discharge. a. What is the time constant \( \tau \) of this circuit? b. How much charge is stored by the capacitor at \( t = 0.5\tau \), \( 2\tau \), and \( 4\tau \)?
Transcribed Image Text:**Textbook Problem** At time \( t = 0 \), an \( RC \) circuit consists of a 12.0-V emf device, a 60.0-\(\Omega\) resistor, and a 150.0-\(\mu F\) capacitor that is fully charged. The switch is thrown so that the capacitor begins to discharge. a. What is the time constant \( \tau \) of this circuit? b. How much charge is stored by the capacitor at \( t = 0.5\tau \), \( 2\tau \), and \( 4\tau \)?
**Expression for Potential Across a Capacitor in an RC Series Circuit** To find the potential across the capacitor with respect to time constant, use the following formula: \[ V_c = V_s \left( 1 - e^{(-t/RC)} \right) \quad \text{(II)} \] **Where:** - \( V_c \) is the capacitor voltage. - \( V_s \) is the emf device voltage. - \( R \) is the circuit resistance. - \( C \) is the circuit capacitance. - \( t \) is the time in seconds. **Expression for Charge Stored in a Capacitor** The charge stored in the capacitor is given by the formula: \[ Q = CV_c \quad \text{(III)} \] **Where:** - \( Q \) is the charge stored in the capacitor. - \( C \) is the circuit capacitance. - \( V_c \) is the capacitor voltage. **Conclusion:** For a practical example, substitute the following values into equation (II): - \( V_s = 12.0 \, \text{V} \) - \( C = 150.0 \, \mu \text{F} \) - \( R = 60.0 \, \Omega \) - \( t = 0.5\tau \)
Transcribed Image Text:**Expression for Potential Across a Capacitor in an RC Series Circuit** To find the potential across the capacitor with respect to time constant, use the following formula: \[ V_c = V_s \left( 1 - e^{(-t/RC)} \right) \quad \text{(II)} \] **Where:** - \( V_c \) is the capacitor voltage. - \( V_s \) is the emf device voltage. - \( R \) is the circuit resistance. - \( C \) is the circuit capacitance. - \( t \) is the time in seconds. **Expression for Charge Stored in a Capacitor** The charge stored in the capacitor is given by the formula: \[ Q = CV_c \quad \text{(III)} \] **Where:** - \( Q \) is the charge stored in the capacitor. - \( C \) is the circuit capacitance. - \( V_c \) is the capacitor voltage. **Conclusion:** For a practical example, substitute the following values into equation (II): - \( V_s = 12.0 \, \text{V} \) - \( C = 150.0 \, \mu \text{F} \) - \( R = 60.0 \, \Omega \) - \( t = 0.5\tau \)
Expert Solution
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Yes. You are right , the equation used is wrong. 

The equation is used for charging and the value of potential difference will increase according to this equation.

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