Engineering Board Exam Reviewer · Series Circuits · medium
In the series circuit shown, V=60 V, R1=19, , and R2=5, . How much power is dissipated by R1?
- 118.75 W
- 189.47 W
- 150 W
- 237.5 W
Answer
A. 118.75 W
Solution
I=V/RT= 60 V24, =2.5 A. Power in R1: P=I2R1=(2.5 A)2(19, )=118.75 W.
More Series Circuits questions
- 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.
- In the series circuit shown, V=48 V, R1=44, , R2=12, , and R3=24, . Find the voltage across R2.
- According to Kirchhoff's voltage law, in a series circuit the sum of the voltage drops across the resistors equals
- In the series circuit shown, V=48 V, R1=24, , R2=22, , and R3=50, . Find the current in the circuit.
- In the series circuit shown, V=24 V, R1=36, , and R2=12, . Find the current in the circuit.
- In the series circuit shown, V=9 V, R1=48, , R2=15, , and R3=27, . Find the total resistance.
- In the series circuit shown, V=30 V, R1=800, , and R2=2.2 k . Find the total resistance.
- Three resistors R1=3 k , R2=2 k , and R3 are in series across a 12 V battery. A voltmeter across R2 reads 4 V. Find R3.