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## Physics 1100: DC Circuits Solutions

In the diagram below, R1 = 5 Ω,R2 = 10 Ω, and R3 = 15 Ω. The battery provides an emf the ε =0.30 V. What is the equivalentresistance, RP? What is the voltage drop across eachresistor? What is the existing through every resistor? What is the powerexpended in each resistor?  (i) since the three resistors re-publishing two typical points or nodes,the three resistors are in parallel. Because that parallel resistors, theequivalent resistance is

1/RP = 1/R1 + 1/R2 + 1/R3= 1/(5 Ω) + 1/(10 Ω)+ 1/(15 Ω) = 11/30 Ω-1.

So the equivalent resistance is RP = 30/11 = 2.727 Ω .

(ii) Resistors in parallel each have actually the exact same voltage fall astheir identical resistance, RP. The meaning ofequivalent method that the 3 resistors can be changed by RPwithout affecting any other aspect of the circuit. So the voltage dropacross RP, and also thus across R1, R2, andR3, is ε = 0.30 V.

(iii) utilizing Ohm"s Law, the current through every resistor isgiven by i = V/R. The results are given below.

(iv) The strength dissipated by each resistor is given by p = I2R= V2/R as presented below.

 Resistor (Ω) V (Volts) I=V/R (Amps) P=V2/R (Watts) 5 0.30 0.060 18.0 × 10-3 10 0.30 0.030 9.0 × 10-3 15 0.30 0.020 6.0 × 10-3 30 0.30 0.110 33.0 × 10-3

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for the circuits presented below, identify all nodes and branches.Are any kind of nodes at the very same potential? which resistors, if any, room inseries? i m sorry resistors, if any, are in parallel?

 (a) Nodes are where three or more wires come together. Thuspoints b, c, and also e space nodes. Branches space thepaths between nodes. Thus the branches are bafe, be,bc,ce,and cde. Any type of two nodes associated by just a right wire (noresistors or batteries in between) are at the same potential. Hencenodes b and c space at the same potential. Resistors inthe branch bring the exact same current and are stated to be in series. Herethe 6-Ω and also 22-Ωresistors room in series even though there is a battery between them.Resistors i beg your pardon share the exact same two nodes are said to it is in in parallel. The20-Ω and 12-Ωresistors space in parallel because they have actually nodes c and also ein common. The 10-Ω resistor is also inparallel v the 20-Ω and also 12-Ω resistors because nodes b and care equivalent. (b) Nodes room where three or more wires come together. Thuspoints b, c, f, and g room nodes.Branches space the paths in between nodes. Therefore the branches space bahg,bg,bc,bg,cf,cdef,and fg. Any type of two nodes associated by simply a straight wire (noresistors or battery in between) space at the exact same potential. Hencenodes b and also c space at the exact same potential as room nodes fand g. Resistors in the exact same branch bring the same current andare said to it is in in series. Right here the resistors R3 and R4are in series. Resistors which share the very same two nodes are said to bein parallel. R1 and R5 space in parallel becausehave nodes b and g are indistinguishable to nodes c and f.

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Find the equivalent resistance that the circuit presented below. Discover the voltagedrop over, present through, and power dissipated by every resistor. Putyour outcomes in a table. The 40 Ω resistor is in parallel with simply a wire of zero resistance - the is the isbeing short-circuited. The 2 arms may be replaced by a solitary wire sinceRP = <1/40 + 1/0>-1 = 0 . The 60 Ω resistor and the 120 Ω resistor space connectedin parallel; they can be changed by a single resistor v valueRP = <1/60 + 1/120>-1 = 40 .Thus the tantamount circuit watch like Thus the identical resistance of the circuit is 40 Ω. The present from the battery isI = V/R = (8 V)/(40 Ω) = 0.20 A. The identical resistor is in reality the 60Ω resistor and also the 120 Ω resistor linked in parallel, sothey all have actually the exact same potential distinction of 8 V. The 60 Ω being twice as little as the120 Ω resistor it s okay 2/3P of the current, 0.133 A, if the 120 Ωresistor the rest, 0.067 A. ResistorPotential distinction (V)Current (A)Power (W)60 Ω80.1331.067120 Ω80.0670.53340 Ω000 ">

A galvanometer has a coil resistance that 250 Ωand requires a present of 1.5 mA for full-scale deflection. This deviceis come be provided in one ammeter that has a full-scale present of 25.0 mA.What is the value of the shunt resistance? What is the equivalentresistance of the ammeter?

The shunt resistance is the resistor connected in parallel withthe coil as shown in the diagram below. The current entering andleaving the branch is ns = 25.0 mA. We want the present through the coilto it is in IG = 1.5 mA. Since the voltage drop throughout each eight is the same

IGRC = (I - IG)RS .

Solving for RS, we find

RS = IGRC / (I - IG)= (1.5 mA)(250 Ω)/(25 mA - 1.5 mA) = 16 Ω .

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We would need the shunt to have actually a resistance that 16 Ω. Because the coil and also shunt room is parallel, the equivalentresistance - the ammeter resistance - is