8 AA cells are connected in series. These make a true battery of 49 volts. This

Question

8 AA cells are connected in series. These make a true battery of 49 volts. This voltage is then connected to three resistors. R1=60 R2=550 R3-360 R1 and R2 are in parallel with each other, and that parallel combination is in series with R3. Start by drawing a schematic diagram of the circuit, then proceed to the calculations. Round all answers to 2 sigfigs precision. The combined resistance of the three resistors is 49. What is the current flowing through the 6002 resistor?4 What is the current flowing through the 550 resistor? What is the current flowing through the 360 resistor? What is the voltage drop across the 600 resistor? What is the voltage drop across the 5502 resistor?40 What is the voltage drop across the 3602 resistor?40^

Explanation / Answer

each AA produces 1.5 V ,So net Voltage of 8 AA Cells is

V=8*1.5=12 V

a)

equivalent resistance

Req =360+(1/550+1/60)-1=414.1 ohms

b)

Total Current flowing in the circuit is

I=12/414.1=0.02898 A

Current through each resistor is

I60=0.02898*(550/550+60)=0.02613 A

I550=0.02898-0.02613=0.00285 A

I360=0.02898 A

C)

Voltage drop across each resistor is

V60=0.02613*60=1.5678 Volts

V550=0.00285*550=1.5678 Volts

V360=0.02898*360=10.4322 Volts

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