Asked by Amy
A 20-Ω resistor and a 30-Ω resistor are wired in parallel and connected to a 9.0-V battery.
a. How much current is drawn from the battery?
b. How much power is dissipated in the circuit?
a. How much current is drawn from the battery?
b. How much power is dissipated in the circuit?
Answers
Answered by
Anonymous
i = V/R
for 20-Ω
i = 9/20
for 30-Ω
i = 9/30
so
total current from battery = 9/20 + 9/30 = 27/60 + 18/60 = 45/60 = 3/4 amp
====================
or use parallel resistor formula
1/R = 1/R1 + 1/R2
R = R1 R2 /(R1+R2) = 600 / 50 = 12 Ohms
i = V/R = 8/12 = 3/4 amp again
for 20-Ω
i = 9/20
for 30-Ω
i = 9/30
so
total current from battery = 9/20 + 9/30 = 27/60 + 18/60 = 45/60 = 3/4 amp
====================
or use parallel resistor formula
1/R = 1/R1 + 1/R2
R = R1 R2 /(R1+R2) = 600 / 50 = 12 Ohms
i = V/R = 8/12 = 3/4 amp again
Answered by
Anonymous
Power = volts * amps = 9 * 3/4 = 27/4 watts
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.