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1) A battery consists of a 12V battery connected to three resistors (25,30, and

ID: 1835983 • Letter: 1

Question

1) A battery consists of a 12V battery connected to three resistors (25,30, and 35 Ohms). Calculate A. The equivalent of the series circuit B. The current across each resistor.
2) If the three resistors in problem #1 are connected in parrallel, calculate A. The equivalent resistance B. The current across each resistor.

1) A battery consists of a 12V battery connected to three resistors (25,30, and 35 Ohms). Calculate A. The equivalent of the series circuit B. The current across each resistor.
2) If the three resistors in problem #1 are connected in parrallel, calculate A. The equivalent resistance B. The current across each resistor.

A. The equivalent of the series circuit B. The current across each resistor.
2) If the three resistors in problem #1 are connected in parrallel, calculate A. The equivalent resistance B. The current across each resistor.

Explanation / Answer

(1).Emf of the battery E = 12 V

Resistances R = 25 ohm

                  R ' = 30 ohm

                 R " = 35 ohm

Three are in series.So, equivalent resistance Req = R + R ' + R "

Req = 25 ohm + 30 ohm + 35 ohm

       =90 ohm

(B). Current across each resistor i = E / Req     Since in series circuit the current is same in all resistors

                                                  = 12 / 90

                                                  = 0.13333 A

(2) In parallel combination ,

(1/Req) =(1/R ) +(1/R ') +(1/R ")

           =(1/25) +(1/30)+(1/35)

           = 0.1019

    Req = 1/0.1019

           = 9.813 ohm

Current in circuit i = E / Req

                           = 12 / 9.813 = 1.2228 A

(B). ratio of currents in R,R ',R " are = (1/R) :(1/R ') :(1/R ")

                                                    =(1/25):(1/30):(1/35)

                                                    = (1050/25):(1050/30):(1050/35)

                                                    =42 : 35 : 30

So, current in R is = i [ 42 /(42+35+30)]

                           = 1.2228 [ 42 /107]

                           = 0.48 A

current in R ' is = i [ 35 /(42+35+30)]

                           = 1.2228 [ 35 /107]

                           = 0.4 A

current in R " is = i [ 30 /(42+35+30)]

                           = 1.2228 [ 30 /107]

                           = 0.3428 A