Capacitor problem, easy but just can't seem to get it

  • Thread starter Thread starter supermenscher
  • Start date Start date
  • Tags Tags
    Capacitor
Click For Summary
SUMMARY

The discussion centers on calculating the potential difference and charge on two capacitors, specifically a 0.40E-6F and a 0.70E-6F capacitor, connected in series to a 12 V battery. The total capacitance (C_t) is calculated using the formula C_t = (1/(1/C_1 + 1/C_2))^-1. The total charge (Q_t) is determined by Q_t = V_t * C_t, where V_t is the total voltage. Each capacitor's voltage (V_1 and V_2) can then be found using V_1 = Q_1/C_1 and V_2 = Q_2/C_2, confirming that the charge on both capacitors is equal (Q_1 = Q_2 = Q_t).

PREREQUISITES
  • Understanding of capacitor basics, including capacitance and charge.
  • Familiarity with series circuits and how capacitance behaves in series.
  • Knowledge of the formulas V = Q/C and C_t = (1/(1/C_1 + 1/C_2))^-1.
  • Ability to manipulate algebraic equations to solve for unknowns.
NEXT STEPS
  • Study the derivation and application of the formula for total capacitance in series circuits.
  • Learn how to calculate voltage across capacitors in series using individual capacitance values.
  • Explore practical applications of capacitors in electronic circuits.
  • Review problems involving capacitors to reinforce understanding of charge and voltage relationships.
USEFUL FOR

Students in electrical engineering, hobbyists working with circuits, and anyone needing to understand capacitor behavior in series configurations.

supermenscher
Messages
47
Reaction score
0
A 0.40E-6F and 0.70E-6F capacitor are connected in series to a 12 V battery. Calculate the potential difference and charge on each capacitor.

I know this problem is really easy, but I just can't seem to get it started. How do you find the potential difference. Can someone help me out?
 
Physics news on Phys.org
supermenscher said:
A 0.40E-6F and 0.70E-6F capacitor are connected in series to a 12 V battery. Calculate the potential difference and charge on each capacitor.

I know this problem is really easy, but I just can't seem to get it started. How do you find the potential difference. Can someone help me out?
V = Q/C. You know the total voltage and total capacitance (I believe capacitors in series add up like resistors in parallel), and so you can calculate the total charge. Since they are in series, I would think that the charge on both capacitors would be the same, so you've got the charge values you need. Now, apply the V = Q/C equation again on each capacitor individually to get the voltage across each one, and you're done.
 
Thank you very much for your explanation, but it is still confusing me...do you think you could show me the first couple steps of what you mean to get me started...that would really help me out.
 
supermenscher said:
A 0.40E-6F and 0.70E-6F capacitor are connected in series to a 12 V battery. Calculate the potential difference and charge on each capacitor.

I know this problem is really easy, but I just can't seem to get it started. How do you find the potential difference. Can someone help me out?
Let C_1 and C_2 represent the capacitances of the capacitors. Let C_t represent the capacitance of both capacitors in series.

C_t = (\frac{1}{C_1} + \frac{1}{C_2})^{-1}

V_t = \frac{Q_t}{C_t}

Q_t = V_tC_t

Q_1 = Q_2 = Q_t

V_1 = \frac{Q_1}{C_1}

V_2 = \frac{Q_2}{C_2}

The above is exactly what I said before, but in equation form. As long as I'm not mistaken, lines 1 and 4 are correct, and so you'll get the right answers. If I am mistaken, and one of them are wrong, you'll probably want to find the right equations somewhere for capacitance and charge in series.
 

Similar threads

Replies
4
Views
2K
Replies
5
Views
2K
Replies
3
Views
2K
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
9K
Replies
5
Views
675
  • · Replies 18 ·
Replies
18
Views
3K