A 12 V supply drives a total current of 4 A through two resistors connected in parallel. One resistor is R1 = 6 ohms. Find the value of R2.

Step-by-step solution

1. 1

   Use Ohm's law on the circuit as a whole to get the equivalent resistance. The supply voltage and the total current both refer to the whole network, so they combine directly:

   $R_{\text{eq}}=\frac{V}{I}=\frac{12\ \text{V}}{4\ \text{A}}=3\ \Omega$

   This single number is what the two resistors must jointly imitate.

2. 2

   Confirm the topology from the numbers before choosing a formula. The equivalent resistance $3\ \Omega$ is smaller than the known resistor $R_1=6\ \Omega$. Adding resistors in series can only increase resistance, so a series arrangement is impossible here; parallel is the only consistent reading. This check is worth doing every time, because the series and parallel formulas give wildly different answers and the wrong one produces a negative resistance.

3. 3

   Write the parallel-resistance relation. For two resistors side by side, conductances (not resistances) add:

   $\frac{1}{R_{\text{eq}}}=\frac{1}{R_1}+\frac{1}{R_2}$

   Substituting the known values:

   $\frac13=\frac16+\frac{1}{R_2}$

4. 4

   Solve for $R_2$. Isolate the unknown reciprocal and use a common denominator of $6$:

   $\frac{1}{R_2}=\frac13-\frac16=\frac26-\frac16=\frac16\quad\Longrightarrow\quad R_2=6\ \Omega$

   Do not forget the final reciprocal — stopping at $\tfrac1{R_2}=\tfrac16$ and writing $R_2=\tfrac16$ is the classic slip in this calculation.

5. 5

   Check with the branch currents. Both resistors sit across the full $12\ \text{V}$, so

   $I_1=\frac{12}{6}=2\ \text{A},\qquad I_2=\frac{12}{6}=2\ \text{A},\qquad I_1+I_2=4\ \text{A}$

   which reproduces the given total current exactly. ✓ The result also matches the shortcut for two equal parallel resistors, $R_{\text{eq}}=R/2$: since $6/2=3$, equal branches were the only way to reach $3\ \Omega$ with one branch fixed at $6\ \Omega$.