Assume that E = 49.0 V . The battery has negligible internal resistance. Assume

Question

Assume that E = 49.0 V . The battery has negligible internal resistance.

Assume that epsilon = 49.0 V. The battery has negligible internal resistance. Compute the equivalent resistance of the network in (Figure 1). Express your answer in ohms to three significant figures. Find the current in the 1.00 ohm resistor. Express your answer in amperes to three significant figures. Find the current in the 3.00 ohm resistor. Express your answer in amperes to three significant figures. Find the current in the 7.00o ohm resistor. Express your answer in amperes to three significant figures. Find the current in the 5.00 ohm resistor. Express your answer in amperes to three significant figures.

Explanation / Answer

a) 1 and 3 ohm are in series ==R13 = 4 ohm

7 and 5 ohm are in series ==R75 = 12 ohm

Now R13 and R75 are in parallel ====> R = 4*12/16 = 3 ohm

b) Current in circuit = 49/3 = 16.33 A

Let I1 current in R13 branch and (16.33 -I1) in R75 branch

then I1(1+3) = 49 =====> I1 = 12.25 A

Then Current in other brach = 16.33 - 12.25 = 4.08 A

b) Current Through 1.0 ohm resistor = I1 = 12.25 A

c) Current Through 3.0 ohm resistor = I1 = 12.25 A

d) Current Through 7.0 ohm resistor = 4.08 A

e) Current Through 5.0 ohm resistor = 4.08 A

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