Assume that E = 55.0 V and R = 2.95 Part A Compute the equivalent resistance of

Question

Assume that E = 55.0 V and R = 2.95

Part A

Compute the equivalent resistance of the network in the figure. The battery has negligible internal resistance.

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Part B

Find the current in the 2.95 resistor.

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Part C

Find the current in the 6.00 resistor.

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Part D

Find the current in the 12.0 resistor.

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Part E

Find the current in the 4.00 resistor.

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Explanation / Answer

Solve 2.95 and 6 ohm, both in parallel

Req1 = 1.977 ohm

Now for 12 ohm and 4 ohm

Req2 = 3 ohm

Total Req = 4.977 ohm

Total I = 11.05 A

The current splits into 2.95 and 6 initially then combines and then splits into 12 and 4 ohm and then combines again.

For 2.95||6 ohm resisters

I in 6ohm resister = 3.64 A and remaining 7.40 A in 2.95 Ohm

similarly since 12ohm is 3 times 4 ohm, 11.05A splits into 2.76A and 8.28A

8.28A flows into 4ohm and 2.76A flows into 12ohm

PART A

Req = 4.977 ohm

PART B

I = 7.40 A

PART C

I = 3.64 A

PART D

I = 2.76 A

PART E

I = 8.28 A

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