Assume that E = 53.0 V . The battery has negligible internal resistance. PART A-

Question

Assume that E = 53.0 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 1.00 resistor. Express your answer in amperes to three significant figures.

PART C- Find the current in the 3.00 resistor. Express your answer in amperes to three significant figures.

PART D- Find the current in the 7.00 resistor. Express your answer in amperes to three significant figures.

PART E- Find the current in the 5.00 resistor. Express your answer in amperes to three significant figures.

Explanation / Answer

A.

Remember:
For series combination
Req = R1 + R2 + R3 +...............
for parallel combination
1/Req = 1/R1 + 1/R2 + 1/R3 + ............
for 2 resistors in parallel it will be
Req = R1*R2/(R1+R2)
Using this Information:

1 and 3 ohm are in series, So

Rs1 = 1 + 3 = 4 ohm

7 and 5 ohm are in series, So

Rs2 = 7 + 5 = 12 ohm

Now 12 and 4 ohm are in parallel, So

Req = 12*4/(12 + 4) = 3 ohm

B.

Veq = ieq*Req

ieq = Veq/Req = 53/3 = 17.67 Amp.

Now remember in resistors parallel combination voltage distribution in each part will be same and in the series combination, the current distribution in each resistor will be same.

Voltage in Rs1 = Voltage in Vs2 = Veq = 53 V

Now Current in Rs1 will be

is1 = Vs1/Rs1 = 53/4 = 13.25 Amp.

Since current in series remains same, So

Current in 1 ohm = 13.25 Amp.

Current in 3 ohm = 13.25 Amp.

Now Current in Rs2 will be

is2 = Vs2/Rs2 = 53/12 = 4.42 Amp.

Since current in series remains same, So

Current in 7 ohm = 4.42 Amp.

Current in 5 ohm = 4.42 Amp.

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