Assume there is a 100V to 1000V transformer and the capacitor connected in the second circuit.(You know, transformers have two circuits)

The second circuit's voltage is 1000V, but the amount current is lower than the first circuit.(because of the power conservation)

So my question is that the formula $Q=CV$ still stands in the case of decreased current?


If Q is the charge on the capacitance C and V is its voltage, yes, that relation holds.

  • $\begingroup$ OK. I've understood now. The problem was easy one. If the current goes lower, then the charging time will increases. $\endgroup$ – user28936 Aug 29 '13 at 4:36

The relationship $Q = C\,V$ always holds when $V$ refers to the charge on the capacitor and $V$ the potential difference between the capacitor's terminals and as long as the conditions that let it be well modeled by an ideal capacitor hold (mostly these will be an upper limit on the frequency).

But you seem to asking what the capacitor "looks like" from a circuit standpoint if it is connected across the secondary winding of an ideal transformer and we do measurements on the primary winding. The answer is that it still looks like a capacitor, but now it has a reflected capacitance given by $n^2 C$, where $n$ is the voltage ratio of the secondary to the primary. You get the squared factor because $n$ times as much current must flow in the primary as in the secondary to charge the capacitor AND the voltage we see at the primary is $n^{-1}$ times that on the secondary.

The idea of reflected impedance is quite general: see the my answer to a related mechanical problem, where I show how it applies to the ideal electrical transformer in the second half of the answer. If you put an impedance $Z$ across the transformer's output, the reflected impedance is $n^{-2} \, Z$, and a capacitor has in impedance $I(t) \mapsto V(t) = C^{-1} \int_0^t I(u)\,\mathrm{d} u$, which is inversely proportional to $C$, whence the $n^2$ factor. So, from the primary's terminals, the capacitance looks one hundred times bigger. You can also do this problem from first principles: We have $V_s(t) = n V_p(t)$, $I_p = n I_s(t)$ and $V_s(t) = C\,\int_0^t I_s(u)\,\mathrm{d} u$, where naturally $V_p$, $V_s$, $I_p$ and $I_s$ are respectively the primary and secondary voltage, primary and secondary current. Eliminating $I_s$ and $V_s$ yields $V_p = n^{-2} C^{-2} \,\int_0^t I_p(u)\,\mathrm{d} u$.

  • $\begingroup$ Oops, got the reflected impedance ratio around the wrong way: I've just fixed things. $\endgroup$ – WetSavannaAnimal Aug 29 '13 at 4:54

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