Engineering Board Exam Reviewer · Series Circuits · medium
Resistors of 10 k , 22 k , and 47 k are connected in series across a 12 V source. Find the voltage across the 22 k resistor.
- 4 V
- 5.62 V
- 6.68 V
- 3.34 V
Answer
D. 3.34 V
Solution
RT=10+22+47=79 k . Voltage divider: V=12·22/79=3.34 V.
More Series Circuits questions
- A 12 V battery and a 6 V battery are connected in series so that they aid each other, across a 9, resistor. Find the current.
- Two resistors in the ratio R1:R2=2:3 are connected in series across a 50 V source. What is the voltage across R2?
- The circuit shown has V=100 V and R2=24 k . The ammeter reads 4 mA. Find the unknown resistance Rx.
- In the series circuit shown, V=100 V, R1=24, , R2=22, , and R3=4, . Find the voltage at point B with respect to the negative terminal of…
- 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…
- In the series circuit shown, V=30 V, R1=1 k , R2=23.4 k , and R3=5.6 k . Find the total power delivered by the source.
- In the series circuit shown, V=60 V, R1=19, , and R2=5, . How much power is dissipated by R1?
- In the series circuit shown, V=24 V, R1=33, , R2=27, , and R3=60, . How much power is dissipated by R2?
- In the series circuit shown, V=48 V, R1=36, , R2=80, , and R3=76, . Find the voltage across R1.
- In the series circuit shown, V=24 V, R1=10, , R2=2, , R3=36, , and R4=12, . Find the voltage across R3.