Three resistors of 4 ohms, 4 ohms and 2 ohms are connected in series and carry a current of 12 A. Find the supply voltage.

Step-by-step solution

1. 1

   Identify the configuration, because it decides the formula. In a series circuit there is only one path, so the same current flows through every component and the resistances add. In a parallel circuit the voltage is shared instead and the reciprocals add. With a single current value of $12\text{ A}$ quoted for all three resistors, the series reading is the intended one.

2. 2

   Add the resistances. For resistors in series,

   $R_{\text{total}}=R_{1}+R_{2}+R_{3}=4+4+2=10\,\Omega.$

   Note the total exceeds the largest individual resistance — always true in series, and a quick check that the addition was done correctly.

3. 3

   Apply Ohm's law. With $V=IR$ and the same $12\text{ A}$ through the whole loop,

   $V=12\text{ A}\times10\,\Omega=120\text{ V}.$

   The units multiply to give volts, since $1\text{ A}\times1\,\Omega=1\text{ V}$ by definition.

4. 4

   Check by summing the individual voltage drops. Each resistor drops $V_{i}=IR_{i}$:

   $V_{1}=12\times4=48\text{ V},\quad V_{2}=12\times4=48\text{ V},\quad V_{3}=12\times2=24\text{ V},$

   and $48+48+24=120\text{ V}$ ✓. Kirchhoff's voltage law requires exactly this agreement, so it is a genuine independent check rather than a restatement.

5. 5

   Compare with the parallel case. Had the three been in parallel, $\frac{1}{R}=\frac14+\frac14+\frac12=1$, so $R=1\,\Omega$ and the supply would only need $V=12\times1=12\text{ V}$ — a tenfold difference. The total power is also worth noting: $P=VI=120\times12=1440\text{ W}$ in the series case.