Assume that E = 59.5 V . The battery has negligible internal resistance. Part A

ID: 1659476 • Letter: A

Question

Assume that E = 59.5 V . The battery has negligible internal resistance.

Part A

Compute the equivalent resistance of the network in (Figure 1) .

Express your answer in ohms to three significant figures.

Part B

Find the current in the 3.00 resistor.

Express your answer in amperes to three significant figures.

Part C

Find the current in the 6.00 resistor.

Express your answer in amperes to three significant figures.

Part D

Find the current in the 12.0 resistor.

Express your answer in amperes to three significant figures.

Part E

Find the current in the 4.00 resistor.

Express your answer in amperes to three significant figures.

3.00 12.0 6.00 4.00

E = 59.5 V

Let
R1 = 3 ohms
R2 = 6 ohms
R3 = 12 ohms
R4 = 4 ohms

Part A)
Let the equivalent resistance of R1 and R2 be R12
Let the equivalent resistance of R3 and R4 be R34

1/R12 = 1/R1 + 1/R2
R12 = 2 ohms

1/R34 = 1/R3 + 1/R4
1/R34 = 16/(12*4)
R34 = 3 ohms

R12 and R34 are in series hence,
Rnet = R12 + R34
Rnet = 2 + 3 = 5 ohms

Part B)

Inet = V/Rnet
Inet = 59.5/5
Inet = 11.9 A

Now,
Voltage drop across R12 = 11.9*2 = 23.8 V
V12 = 23.8 V

I1 = V12/R1
I1 = 23.8/3 = 7.93 A

Part C)

I2 = V12/R2
I2 = 23.8/6 = 3.97 A

Part D)
Voltage drop across R34 = 11.9*3 = 35.7 V
V34 = 35.7 V

I3 = V34/R3
I3 = 35.7/12 = 2.975 A

Part E)
I4 = V34/R4
I4 = 35.7/4 = 8.925 A

Please upvote if the answers are correct.

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