electric-circuitselectrical-resistance
Some circuits are easier to see which resistors are in parallel or series. However, when faced with a more complex circuit, I can't figure out which ones are in series or parallel.
I am currently facing the following problem:
I am instructed to find the equivalent resistance; but I do not know how to determine which are in parallel or series. Is there a particular method to do so?
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
The trick is to look at the nodes in the circuit. A node is a junction in the circuit. Two resistor are in parallel if the nodes at both ends of the resistors are the same. If only one node is the same, they are in series. So, R1 and R2 are in parallel and R3 is in series with R1||R2.