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 is $590 \Omega + 840 \Omega = 1430 \Omega$. The current is $I = \frac{10.0 V}{1430 \Omega} = 0.006993 A$. The potential drop across the $590 \Omega$ resistor is $V = I \times 590 \Omega = 4.1 V$.

An Ideal 10 590.0−Ω590.0 - \Omega590.0−Ω Resistor in Series with An 840,0−Ω840,0 - \Omega840,0−Ω

An Ideal 10 $590.0 - \Omega$ Resistor in Series with An $840,0 - \Omega$