Find the current through each resistor and the voltage drop

Find the current through each resistor and the voltage drop across each resistor. b. Which of the bulbs will be the brightest? Justify your answer. c. Place an ammeter in the circuit diagram above that would measure the total current. Explain why it is placed there in terms of potential difference and current. d. If light bulb R_2 is broken, what is the current in the circuit? Justify your answer. If light bulb R_3 is broken, what is the current in the circuit? Justify your answer.

Solution

R2 and R3 are in parallel.

Resultant of these two resistors be R then

(1/R ) = (1/R2)+(1/R3)

         = (1/6)+(1/9)

      R = ( 6 x9) /(6+9)

         = 54 / 15

         = 3.6 ohm

R and R1 are in series.

So, equivalent resistance of the circuit R \' = R + R1

                                                             = 3.6 ohm + 3 ohm

                                                             = 6.6 ohm

Potential difference V = 36 volt

Current in the circuit i = V / R \'

                                = 36 / 6.6

                                = 5.4545 A

Current in R1 is i1 = i = 5.4545 A

Current in R2 is i2 = i [ R3/(R2+R3)]

                          = 5.4545 [9 /(6+9)]

                         = 3.272 A

Current in R3 is i3 = i -i2

                            = 5.4545 -3.272

                            = 2.1818 A

Potential drop across R1 is V1 = i1 R1

                                             = 5.4545 x 3

                                             = 16.3636 volt

Potential drop across R2 is V2 = i2R2

                                            = 3.272 x 6

                                            = 19.632 volt

Potentila drop across R3 is V3 = i3R3

                                             = 2.1818 x9

                                            = 19.632 volt

(b) R1 is brightest.Since current through R1 is maximum

(d).when R1 is broken then

R2 and R3 are in parallel.

Resultant of these two resistors be R then

(1/R ) = (1/R2)+(1/R3)

         = (1/6)+(1/9)

      R = ( 6 x9) /(6+9)

         = 54 / 15

         = 3.6 ohm

So, current in the circuit i \' = V / R

                                        = 36 / 3.6

                                        = 10 A

(e).When R3 is broken then ,

R1 andR2 are in series.

resultant resistance R = R1+R2

                                 = 3 ohm + 6 ohm

                                 = 9 ohm

Current in the circuit i \" = V / R

                                  = 36 / 9

                                 = 4 A