electric-circuitselectrical-resistance
So I was told in the physics class that the resistance in a parallel circuit is smaller than the resistance in a series circuit. Why does that happen?
Is this statement also true for circuits which have no resistors or resistance-offering devices connected to them?
And also from my calculations the total resistance in a parallel circuit is smaller than the resistances of any of the devices connected in the circuit. Now I really don't understand how this can happen.
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!
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.
Best Answer
Question: Which of two pipes of equal length offers less resistance to the flow of water, one of which has twice the cross sectional area of the other?
Answer: The one of twice the cross sectional area.
But the one of twice the cross sectional area can be thought of as two of the smaller cross sectional area pipes in parallel.
This analogy gives an idea of the smaller resistance of a parallel arrangement of resistors although not completely.
A length $l$ of wire of cross sectional area $A$ has a resistance $R$ given by the equation $R = \dfrac {\rho l}{A}$ where $\rho$ is the resitivity of the wire.
Doubling the cross sectional area of a fixed length of wire decreases the resistance by a factor of two which is equivalent to having two wires of area $A$ in parallel.
However for a pipe the "resistance" to fluid flow is proportional to $\dfrac {1}{\text{area}^2}$.
So in the case of a pipe having the area increase by a factor of two decreases the resistance to fluid flow by a factor of four.