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Consider the circuit shown in the figure below. (Assume R1 = 12.0 , R2 = 1.95 ,

ID: 1491787 • Letter: C

Question

Consider the circuit shown in the figure below. (Assume R1 = 12.0 , R2 = 1.95 , and V = 6.10 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. ________A

(f) What is the voltage drop across the R2 resistor? ______ V

(g) Subtracting the result of part (f) from the battery voltage, find the voltage across the 3.00- resistor. ______ V

(h) Calculate the current in the 3.00- resistor.
_________A

Explanation / Answer

R1 = 12.0

R2 = 1.95

V = 6.10 V

(a) The equivalent resistance of the R1 and 5.00- resistors connected in parallel R = ?

We know (1/R ) = (1/R1)+(1/5)

                        = (1/12)+(1/5)

                        = 0.2833

                    R = 1/0.2833

                       = 3.529

(b) the combined resistance of the R1, 5.00- and 4.00- resistors R ' = R + 4 = 3.529 + 4 = 7.529

(c) the equivalent resistance of the combined resistance found in part (b) and the parallel 3.00- resistor,R "= ?

We know (1/R") =(1/R') +(1/3)

                       =(1/7.529)+(1/3)

                      = 0.4661

                  R"= 1/0.4661

                     = 2.145

(d) the equivalent resistance found in part (c) with the R2 resistor Rq = R" + R2

   = 2.145 +1.95

   = 4.095

(e) the total current in the circuit i = V/Rq

                                                  = 6.1 /4.095

                                                  = 1.489 A

(f) the voltage drop across the R2 resistor is V' = iR2

                                                                     =1.489 (1.95)

                                                                     = 2.904 volt

(g) Subtracting the result of part (f) from the battery voltage,

Voltage across remaing circuit is V " = V - V ' = 6.1 volt - 2.904 volt

                                                      = 3.195 volt

the voltage across the 3.00- resistor is V " = 3.195 volt

(h) the current in the 3.00- resistor i ' = V"/3

                                                         = 3.195 /3

                                                         = 1.065 A