An ideal 10.0-V dc is connected across a $590.0 - \Omega ,$ resistor in series with an $840.0 - \Omega$ resistor.What is the potential drop across the $590.0 \Omega$ resistor?
A)4.1 V
B)5.9 V
C)14 V
D)7.0 V

Question

Quizplus · Accepted Answer

The total resistance in the series circuit is $590.0 \Omega + 840.0 \Omega = 1430.0 \Omega$. The total current ($I$) flowing through the circuit can be calculated using Ohm's law, $I = V/R$, where $V = 10.0 \text{ V}$ and $R = 1430.0 \Omega$. Thus, $I = 10.0 \text{ V} / 1430.0 \Omega$. The potential drop ($V_{590}$) across the $590.0 \Omega$ resistor can then be found using $V = IR$, where $I$ is the current calculated earlier and $R = 590.0 \Omega$. Substituting the values, $V_{590} = I \times 590.0 \Omega$, which gives approximately $4.1 \text{ V}$.

An Ideal 10 590.0−Ω,590.0 - \Omega ,590.0−Ω, Resistor in Series with An

An Ideal 10 $590.0 - \Omega ,$ Resistor in Series with An