Engineering Board Exam Reviewer · Series Circuits · medium

In the series circuit shown, V=9 V, R1=3.3 k , and R2=5.7 k . Find the voltage at point A with respect to the negative terminal of the source.

  1. 3.3 V
  2. 4.5 V
  3. 5.7 V
  4. 6.75 V

Answer

C. 5.7 V

Solution

I=V/RT=1 mA. The voltage at A equals the total drop across the resistors between A and the negative terminal: VA=I(R2)=(1 mA)(5.7 k )=5.7 V.

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