Series Parallel circuit table problem?

tokaygeckoguytokaygeckoguy
in Science & Mathematics

I am having problems with this circuit. Go down to question 8 on this site. I know the answers but I am not getting how to get the answers on the table. Can you explain how to get the answers please.

http://www.allaboutcircuits.com/worksheets/dc_sp.h...

Comments

  • billrussell42billrussell42

    This is really easy. Calculate the total R, which is the two on the left in series, in parallel with the one on the right. Rt = 6.6•4.7/(6.6+4.7) = 2.75k

    the 10 mA current produces a voltage across the two sides of E = IR = 27.5 volts

    And that causes current in each side, just apply ohms law.

  • fred-davis_60c2e09fbe599fred-davis_60c2e09fbe599

    You're probably getting confused by the source being in the middle. Redraw the diagram so that all the resistors are on one side.

    Add R2 and R3 together. They are equivalent to a single 6.6kOhm resistor. I'll call it Rnew. Redraw the diagram with Rnew replacing R2 and R3. You now have R1 and Rnew in parallel with each other. If you run the calculation, they are equivalent to one resistor that is 2.745 kOhm. (If you don't know how to do that part, then you're too lost, and you should ask your teacher for a tutor.)

    V=IR

    I = 0.010 amps (given)

    R = 2745 ohms (calculated)

    V = 27.45 volts

    That's the voltage across the source. Now look at R1. It has a voltage of 27.45 across it. Use V=IR to calculate the current through that resistor.

    Next, you have 27.45 volts across Rnew (the combination of R2 and R3). Use V=IR to calculate the current. That's the current going through R2 and R3.

    Use P=IV to calculate the powers.

  • anonymousanonymous

    i'm no longer sparkling on the way they're related. Are you announcing that C2 and C3 are in sequence and that C1 is expounded for the time of the two certainly one of them so as that it relatively is in parallel with the sequence pair.? bear in mind that for capacitors in parallel Ct = C1 + C2 + ..... the place Ct = finished capacitance For capacitors in sequence a million/Ct = a million/C1 + a million/C2 + ...... So for the two in sequence we've a million/Ct = a million/C2 + a million/C3 = (C2 + C3)/ C2C3 Ct = C2C3 / (C1 + C2) [it particularly is only the Ct for the sequence pair] upload this to C1 and Ct = C1 + C2C3/(C2+C3) [ it particularly is the main suitable Ct] Your blunders replaced into in no longer including the two fractions [a million/C2 + a million/C2] then turning the consequence the different way as much as get Ct for the sequence pair.

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