I need to find the Equivalent Resistance across A and B. The problem is, I don't know which one's are in series and which ones are in parallel. How do I identify which ones are parallel or series?
[Physics] Determine which circuits are parallel and which are in series
electric-circuitselectrical-resistancehomework-and-exercises
Related Solutions
Alfred got in before me, but I have a diagram!
I've marked all continuous bits of wire in the same colour, and marked the corresponding colours on the ends of the resistors. A quick redraw later and I get:
which is a lot simpler!
When the switch is opened, the circuit is the equivalent of this, so I think you can clearly see that the resistors are in series and so are the capacitors.
It seems that you already understand how to calculate series resistance, so I'll show how to understand series capacitance. First, you probably already know that capacitance is defined as the ratio of stored charge to voltage. That is, $C=\frac{Q}{V}$. When both of these capacitors are charged, they must have the same charge. That is, $Q_{C1} = Q_{C2}$. Also, the voltages across them must add up to the total voltage, so $V_t = V_{C1} + V_{C2}$. Since we know that $V = \frac{Q}{C}$, we can say $$\begin{eqnarray}V_t &=& V_{C1} + V_{C2} \\ \frac{Q}{C_t} &=& \frac{Q_{C1}}{C1} + \frac{Q_{C2}}{C2} \\ Q\frac{1}{C_t} &=& Q\left(\frac{1}{C1} + \frac{1}{C2}\right)\\ \frac{1}{C_t} &=& \frac{1}{C1} + \frac{1}{C2}\\ C_t &=& \frac{1}{\frac{1}{C1} + \frac{1}{C2}} \end{eqnarray}$$
Best Answer
If all of the current leaving one resistor enters another resistor, the two resistors are in series.
The resistances of series connected resistors can be added together to find the equivalent resistance of a single resistor, e.g.,
$$R_{eq} = R_1 + R_2 $$
If all of the voltage across one resistor is across another resistor, the two resistors are in parallel.
The conductances of parallel connected resistors can be added together to find the equivalent conductance of a single resistor, e.g.,
$$G_{eq} = \frac{1}{R_{eq}} = G_1 + G_2 = \frac{1}{R_{1}} + \frac{1}{R_{2}}$$
or
$$R_{eq} = \frac{1}{\frac{1}{R_{1}} + \frac{1}{R_{2}}} $$