Consider the circuit shown in the figure below. (Assume R 1 = 11.00 ? , R 2 = 2.

Question

Consider the circuit shown in the figure below. (Assume R1 = 11.00 ?, R2 = 2.20 ?, and V = 7.00 V.)

Consider the circuit shown in the figure below. (Assume R1 = 11.00 ?, R2 = 2.20 ?, and V = 7.00 V.) (a) Calculate the equivalent resistance of the R1 and 5.00 - ? resistors connected in parallel. (b) Using the result of part (a), calculate the combined resistance of the R1, 5.00 - ?, and 4.00 - ? resistors. (c) Calculate the equivalent resistance of the combined resistance found in part (b) and the parallel 3.00 - ? resistor. (d) Combine the equivalent resistance found in part (c) with the R2 resistor. (e) Calculate the total current in the circuit. (f) What is the voltage drop across the R2 resistor? (g) Subtracting the result of part (f) from the battery voltage, find the voltage across the 3.00 - ? resistor. (h) Calculate the current in the 3.00 - ? resistor.

Explanation / Answer

Part A)

Equivalent resistance of resistors in parallel is found by

1/REQ = 1/R1 + 1/R2

1/REQ = 1/11 + 1/5

REQ = 3.44

Part B)

Equivalent resistance in series is found by

REQ = R1 + R2

REQ = (3.44) + (4)

REQ = 7.44

Part C)

Back to parallel

1/REQ = 1/7.44 + 1/3

REQ = 2.14

Part D)

Back to series

REQ = (2.14) + (2.2)

REQ = 4.34

Part E)

V = IR

(7) = (I)(4.34)

I = 1.61 A

Part F)

For R2, V = IR giving us V = (1.61)(2.2) = 3.54 V

The subtraction of that from the total circuit voltage gives us the voltage across the 3 resistor

7 - 3.54 = 3.46 V

Part E)

Using V = IR we get

3.46 = I(3)

I = 1.15 A

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