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Homework Help: [Electricity & Magnetism] Finding the voltage of a battery.

  1. Oct 26, 2014 #1
    1. The problem statement, all variables and given/known data
    Hello. I just need someone to check if I'm approaching the problem correctly (the steps I've taken are given below).

    I drew a quick sketch of the circuit here: http://i.imgur.com/mClttXn.png?1

    In the problem, I am given the capacitances of A, B, and C. I am also given the voltage across capacitor B. My goal is to find the voltage of the battery.

    2. Relevant equations
    Equivalent capacitance (series) = (cap1*cap2)/(cap2+cap1)
    Equivalent capacitance (parallel) = cap1 + cap2 + ... + capn
    Voltage = Q/C
    Total voltage (series) = V1 + V2
    Charge = CV

    3. The attempt at a solution
    Here are the steps I would take to find the voltage of the battery:

    I first find the equivalent capacitance across A and B, the two caps in parallel. Since capacitors in parallel have equal voltages, the voltage across the combined capacitor is the same as the one across capacitor B (let's pretend it was 9V. So the voltage across the combined cap is 9V).

    Now, I have the combined cap in series with capacitor C. Since the voltage across the battery is the sum of the voltages of the combined cap and cap C (please correct me if I'm wrong here), I need to find the voltage across C. To do this, I'll need the charge on C. Since the two caps are in series, I can find the charge Q of cap C by finding the charge Q on the combined cap (since the two charges should be equal). Once I find Q, I just divide that by the capacitance of C to obtain the voltage across C.

    The voltage of the battery = voltage across combined cap + voltage across C.

    Am I doing this correctly?
  2. jcsd
  3. Oct 27, 2014 #2


    User Avatar

    Staff: Mentor

    Your method looks okay.
  4. Oct 27, 2014 #3
    Thank you!
  5. Oct 27, 2014 #4
    Yep you are on track, you only made it much harder than it is. Think of the caps as Impedance loads (Z), Just happen to be capacitance this time.

    you got Series and || down for them.

    Think to use a Voltage diver.... Xc is 1/wc and Xl is wL and Xr is just R

    If you think Impedance be easier if one of the Caps is a R or an L. I think Z ......
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