electric-circuitselectrical-resistance
I know how to detect when resistors are arranged in parallel or series arrangement and
I can also find their equivalent resistance in simple circuits or when resistances are connected in form of
triangle but what happens when the arrangement is complex like this :
Which resistors are parallel and which are in series ? How can I find the equivalent resistance in such cases ? Is there rule or method for figuring this out ?
Resistances, in series, add:
$$R_{EQ} = R_1 + R_2$$
This follows from KVL and Ohm's Law: $V = IR$. Since series connected circuit elements have identical current $I$ through:
$$V_{EQ} = IR_1 + IR_2 = I(R_1 + R_2) = IR_{EQ}$$
Conductances, in parallel, add:
$$G_{EQ} = G_A + G_B$$
This follows from KCL and the dual of Ohm's Law: $I = VG$. Since parallel connected circuit elements have identical voltage $V$ across
$$I_{EQ} = VG_A + VG_B = V(G_A + G_B) = VG_{EQ}$$
Now, it is clear that conductance is the reciprocal of resistance, thus:
$$G_{EQ} = \frac{1}{R_{eq}} = \frac{1}{R_A} + \frac{1}{R_B} \Rightarrow R_{eq} = \dfrac{1}{\frac{1}{R_A} + \frac{1}{R_B}} $$
The physical interpretation is quite straightforward. Adding another path for current allows more current for a given voltage; putting a resistance in parallel is adding a conductance - an extra path for current. This is analogous to adding another path for water flow for a given pressure; allowing more flow for a given pressure.
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
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!