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 current through the circuit is $I = \frac{10.0 \, \text{V}}{1430.0 \, \Omega} = 0.006993 \, \text{A}$. The potential drop across the $590.0 \, \Omega$ resistor is $V = I \times 590.0 \, \Omega = 4.1 \, \text{V}$.

An Ideal 10 590.0−Ω590.0 - \Omega 590.0−Ω Resistor in Series with An

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