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Series Circuits - Engineering

Q1:

Two resistors are in series: a 5.6 k resistor and a 4.7 k resistor. The voltage drop across the 5.6 k resistor is 10 V. The voltage across the 4.7 k resistor is

A 8.39 V

B 10 V

C 2.32 V

D 0 V

ANS:A - 8.39 V

Voltage in 4.7 kv = 4.7 k*i.

i = Voltage in 5.6k/5.6k.

i = 10/5.6 k A.

v in 4.7 k = 4.7 k*10/5.6 k.

= 8.39 V.