Two resistors of 5.0 and 9.0 ohms are connected inparallel. A 4.0 Ohm resistor is then connected in series withthe parallel combination. A 6.0V battery is then connected tothe series-parallel combination. What is the current throughthe 9.0 ohm resistor?

sjeikdom0

sjeikdom0

Answered question

2020-12-02

Two resistors of 5.0 and 9.0 ohms are connected inparallel. A 4.0 Ohm resistor is then connected in series withthe parallel combination. A 6.0V battery is then connected tothe series-parallel combination. What is the current throughthe 9.0 ohm resistor?

Answer & Explanation

Jozlyn

Jozlyn

Skilled2020-12-03Added 85 answers

R1|| R2 so equivalant R12= 5×9(5+9)=3.21429W
Further ; R12& R3 are in series so equivalant R = R12+R3 or R = 7.21429W
emf of the battery (E) = 6.0 V
Circute current ( I ) = E / R = 0.831683 A
Voltage drop across R12 is givenas V12= I × R12 = 2.67327 V
The current through the 9.0 ohm resistor (I9)=V12/R2=0.297 A

Jeffrey Jordon

Jeffrey Jordon

Expert2021-09-30Added 2605 answers

The current flows through the 9.0 ohm resistor is

current I=VR=2.7 ohm9.0 ohm = 0.297 ohm

Answer is 0.297 A.

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