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Questions: 1 2 3 4 5 6 7 8 9 10 11

Physics 1100: DC Circuits Solutions

In the diagram below, R1 = 5 Ω,R2 = 10 Ω, and R3 = 15 Ω. The battery offers an emf that ε =0.30 V. What is the equivalentresistance, RP? What is the voltage drop throughout eachresistor? What is the present through each resistor? What is the powerexpended in every resistor?
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(i) because the three resistors re-superstructure two common points or nodes,the three resistors room in parallel. For 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 very same voltage autumn astheir identical resistance, RP. The definition ofequivalent method that the 3 resistors can be changed by RPwithout affecting any other aspect of the circuit. So the voltage dropacross RP, and thus throughout R1, R2, andR3, is ε = 0.30 V.

(iii) using Ohm"s Law, the present through each resistor isgiven by ns = V/R. The outcomes are given below.

(iv) The power dissipated by every resistor is given by ns = I2R= V2/R as displayed 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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because that the circuits presented below, determine all nodes and branches.Are any nodes at the exact same potential? which resistors, if any, room inseries? i m sorry resistors, if any, room in parallel?

(a)
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Nodes room where 3 or more wires come together. Thuspoints b, c, and also e room nodes. Branches space thepaths between nodes. Thus the branches space bafe, be,bc,ce,and cde. Any two nodes connected by just a straight wire (noresistors or battery in between) room at the very same potential. Hencenodes b and also c are at the exact same potential. Resistors inthe branch lug the same current and also are claimed to be in series. Herethe 6-Ω and 22-Ωresistors are in collection even though there is a battery between them.Resistors i m sorry share the very same two nodes are stated to be in parallel. The20-Ω and also 12-Ωresistors room in parallel since they have actually nodes c and also ein common. The 10-Ω resistor is likewise inparallel v the 20-Ω and also 12-Ω resistors because nodes b and care equivalent.

(b)
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Nodes room where 3 or an ext wires come together. Thuspoints b, c, f, and g are nodes.Branches room the paths between nodes. Thus the branches are bahg,bg,bc,bg,cf,cdef,and fg. Any type of two nodes associated by simply a directly wire (noresistors or batteries in between) space at the exact same potential. Hencenodes b and also c room at the exact same potential as are nodes fand g. Resistors in the exact same branch carry the same existing andare claimed to it is in in series. Right here the resistors R3 and also R4are in series. Resistors i beg your pardon share the exact same two nodes are stated to bein parallel. R1 and R5 are in parallel becausehave nodes b and g are tantamount to nodes c and f.

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Find the equivalent resistance that the circuit shown below. Find the voltagedrop over, current through, and power dissipated by every resistor. Putyour outcomes in a table.

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The 40 Ω resistor is in parallel with simply a cable of zero resistance - that is the isbeing short-circuited. The 2 arms may be changed by a single wire sinceRP = <1/40 + 1/0>-1 = 0 . The 60 Ω resistor and the 120 Ω resistor room connectedin parallel; they have the right to be changed by a solitary resistor through valueRP = <1/60 + 1/120>-1 = 40 .Thus the identical circuit look at like
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Thus the equivalent resistance that the circuit is 40 Ω. The current from the battery isI = V/R = (8 V)/(40 Ω) = 0.20 A. The indistinguishable resistor is in reality the 60Ω resistor and the 120 Ω resistor associated in parallel, sothey all have actually the same potential difference of 8 V. The 60 Ω being double as little as the120 Ω resistor it s okay 2/3P the the current, 0.133 A, while the 120 Ωresistor the rest, 0.067 A. ResistorPotential difference (V)Current (A)Power (W)60 Ω80.1331.067120 Ω80.0670.53340 Ω000 ">

A galvanometer has actually a coil resistance the 250 Ωand calls for a present of 1.5 mA because that full-scale deflection. This deviceis to be supplied 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 the the ammeter?

The shunt resistance is the resistor associated in parallel withthe coil as presented in the diagram below. The existing entering andleaving the branch is ns = 25.0 mA. We desire the existing through the coilto be IG = 1.5 mA.

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Since the voltage drop across each eight is the very same

IGRC = (I - IG)RS .

Solving for RS, we uncover

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 the 16 Ω. Due to the fact that the coil and shunt are is parallel, the equivalentresistance - the ammeter resistance - is