A 12 volt source is connected across a 6 ohm resistor. Find the current, and check whether a reading of 4 amps is possible.

Step-by-step solution

1. 1

   Write Ohm's law and pick the form you need. The three quantities are linked by $V = IR$. Two of them are given ($V$ and $R$) and the third is wanted, so rearrange to

   $I = \frac{V}{R}$

2. 2

   Substitute the given values.

   $I = \frac{12\ \text{V}}{6\ \Omega} = 2\ \text{A}$

3. 3

   Test the claimed 4 A against the same law. If the current really were $4\ \text{A}$ at $12\ \text{V}$, the resistance would have to be $R = \tfrac{12}{4} = 3\ \Omega$; and if it were $4\ \text{A}$ through $6\ \Omega$, the source would have to be $V = 4 \times 6 = 24\ \text{V}$. So the trio $(12\ \text{V},\ 6\ \Omega,\ 4\ \text{A})$ is over-determined and inconsistent — at most two of the three can be correct.

4. 4

   Cross-check with power. At $2\ \text{A}$ the resistor dissipates $P = I^2R = 4 \times 6 = 24\ \text{W}$, and also $P = VI = 12 \times 2 = 24\ \text{W}$. The two independent power formulas agree, confirming $2\ \text{A}$. Assuming $4\ \text{A}$ instead would give $I^2R = 96\ \text{W}$ but $VI = 48\ \text{W}$ — a contradiction.

5. 5

   State the answer. With $12\ \text{V}$ across $6\ \Omega$ the current is $2\ \text{A}$, dissipating $24\ \text{W}$.