How could the potential difference be constant across all the resistors of parallel connecting resistors although each resistor has a specific resistance?
[Physics] potential difference in parallel connecting resistors
electric-circuitselectrical-resistancevoltage
Related Solutions
The diagram shows first four resistors in series then four resistors in parallel.
For the resistors in series the current flowing into the wire, $I_{in}$ must be the same as the current flowing out, $I_{out}$ because the current can't escape from the wire. There is only one route for the current to flow through the wire so the current has to pass through all the resistors in turn. That's why the current passing through every resistor must be the same.
Now look at the resistors in parallel. The point here is that the top ends of the resistors are all connected together so they must all be at the same voltage $V_{in}$. Likewise the bottom ends are all connected together so they must be at the same voltage $V_{out}$. That means all the resistors have the same voltage drop across them of $V_{in} - V_{out}$.
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.
Best Answer
Kirchoff's laws tell us that the potential drop across any closed loop in a circuit must be equal to the voltage sources in the loop, from which we conclude that the voltage drop across resistors in parallel must be equal.
Ohm's law states:
$$V=IR$$
From which we conclude that, since $V$ is fixed, if the different resistors have different $R$'s, then the current ($I$) through each must also be different (and obey Ohm's law).