A 10.0 V battery is connected across two resistors in series.If the resistors have resistances of 840.0 Ω and 590.0 Ω ,what is the voltage drop across the 590.0 Ω resistor?
A)4.1 V
B)5.9 V
C)14 V
D)7.0 V

Question

Quizplus · Accepted Answer

The total resistance in series is 1430 Ω (840 Ω + 590 Ω). The current through the circuit is I = V/R = 10 V / 1430 Ω = 0.006993 A. The voltage drop across the 590 Ω resistor is V = IR = 0.006993 A * 590 Ω = 4.1 V.

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