Question

A 10.-ohm resistor and a 5.0-ohm resistor are connected. If the current through the 10.-ohm resistor is 1.0 ampere, then the current through the 5.0-ohm resistor is?

And it's a parallel circuit.

Answer 1

The two resistors are in parallel so they will have the same voltage across them. Therefore, if one has half the resistance of the other one it must have twice the current going through it.
\[V=IR\]

Answer 2

With a little algebra, you can work out that the effective resistance of a pair of resistors in parallel is
\[R = \frac{1}{\frac{1}{R_1}+\frac{1}{R_2}} = \frac{1}{\frac{R_2}{R_1R_2}+\frac{R_1}{R_1R_2}} = \frac{R_1R_2}{R_1+R_2}\]

Here we've got 1 amp through a 10 ohm resistor, so by Ohm's law, \(V = IR = 1A *10 \Omega = 10 V \) across resistor \(R_1\). That means \(R_2\) also has 10 V across it, so rearranging Ohm's law, we get \(I = V/R = 10 V/5 \Omega = 2A\). The total current through the two resistors is 3A, so using Ohm's law yet again, \(R = V/I = 10V/3A = 3 \frac{1}{3}\Omega\) which is just what our formula predicts:
\[R = \frac{10\Omega*5\Omega}{10\Omega + 5\Omega} = 3 \frac{1}{3} \Omega \]

Answer 3

The formula has a nice intuitive result if you have two identical resistors: the resistance is then
\[\frac{R*R}{R+R} = \frac{R^2}{2R} = \frac{R}{2}\]which makes sense if you think of the current as water in a pipe. If you double the number of pipes, you can move twice as much water per unit time.

Answer 4

Since the problem states the resistors are connected in parallel then you know the voltage drop across both resistors are equal. The 10 Ohm resistor having a current of 1 Amp would produce a 10 volt drop, therefore, the 5 Ohm resistor must have 2 Amps flowing through it.