When two resistors are connected in series across a 12-V battery,1.1A flows.When they are connected in parallel across the same battery,6.3A flows.What is the smaller resistor's value?

Question

Quizplus · Accepted Answer

When the resistors are in series, the total resistance can be found using Ohm's law, $V = IR$, where $V = 12V$ and $I = 1.1A$. Thus, the total resistance in series is $R_{series} = \frac{V}{I} = \frac{12V}{1.1A} = 10.91\Omega$. When in parallel, the total current is $6.3A$, implying the total resistance in parallel is $R_{parallel} = \frac{V}{I} = \frac{12V}{6.3A} = 1.905\Omega$. For two resistors in parallel, $R_{parallel} = \frac{R_1R_2}{R_1 + R_2}$. Given $R_{series} = R_1 + R_2 = 10.91\Omega$ and $R_{parallel} = 1.905\Omega$, we can solve for $R_1$ and $R_2$. Assuming $R_1 < R_2$, and using the relationship between series and parallel resistances, we find that one of the resistors must be $2.5\Omega$, which is the smaller value.

When Two Resistors Are Connected in Series Across a 12-V Ω\varOmegaΩ

When Two Resistors Are Connected in Series Across a 12-V $\varOmega$