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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 supplies an emf of ε =0.30 V. What is the equivalentresistance, RP? What is the voltage drop across eachresistor? What is the current through each resistor? What is the powerexpended in each resistor?(i) Since the three resistors share two common points or nodes,the three resistors are 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 the same voltage drop astheir equivalent resistance, RP. The definition ofequivalent means that the three resistors could be replaced by RPwithout affecting any other aspect of the circuit. So the voltage dropacross RP, and thus across R1, R2, andR3, is ε = 0.30 V.

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

(iv) The power dissipated by each resistor is given by P = I2R= V2/R as shown 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 shown below, identify all nodes and branches.Are any nodes at the same potential? Which resistors, if any, are inseries? Which resistors, if any, are in parallel?

(a) | |

Nodes are where three or more wires come together. Thuspoints b, c, and e are nodes. Branches are thepaths between nodes. Therefore the branches are bafe, be,bc,ce,and cde. Any two nodes connected by just a straight wire (noresistors or batteries in between) are at the same potential. Hencenodes b and c are at the same potential. Resistors inthe branch carry the same current and are said to be in series. Herethe 6-Ω and 22-Ωresistors are in series even though there is a battery between them.Resistors which share the same two nodes are said to be in parallel. The20-Ω and 12-Ωresistors are in parallel because they have nodes c and ein common. The 10-Ω resistor is also inparallel with the 20-Ω and 12-Ω resistors because nodes b and care equivalent. | |

(b) | |

Nodes are where three or more wires come together. Thuspoints b, c, f, and g are nodes.Branches are the paths between nodes. Therefore the branches are bahg,bg,bc,bg,cf,cdef,and fg. Any two nodes connected by just a straight wire (noresistors or batteries in between) are at the same potential. Hencenodes b and c are at the same potential as are nodes fand g. Resistors in the same branch carry the same current andare said to be in series. Here the resistors R3 and R4are in series. Resistors which share the same two nodes are said to bein parallel. R1 and R5 are in parallel becausehave nodes b and g are equivalent to nodes c and f. |

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Find the equivalent resistance of the circuit shown below. Find the voltagedrop over, current through, and power dissipated by each resistor. Putyour results in a table.

The 40 Ω resistor is in parallel with just a wire of zero resistance - that is it isbeing short-circuited. The two arms may be replaced by a single wire sinceRP = <1/40 + 1/0>-1 = 0 . The 60 Ω resistor and the 120 Ω resistor are connectedin parallel; they can be replaced by a single resistor with valueRP = <1/60 + 1/120>-1 = 40 .Thus the equivalent circuit looks likeThus the equivalent resistance of the circuit is 40 Ω. The current from the battery isI = V/R = (8 V)/(40 Ω) = 0.20 A. The equivalent resistor is actually the 60Ω resistor and the 120 Ω resistor connected in parallel, sothey all have the same potential difference of 8 V. The 60 Ω being twice as small as the120 Ω resistor gets 2/3P of 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 a coil resistance of 250 Ωand requires a current of 1.5 mA for full-scale deflection. This deviceis to be used in an ammeter that has a full-scale current 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 I = 25.0 mA. We want the current through the coilto be IG = 1.5 mA.

Since the voltage drop across each arm 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 .You are watching: How much power is dissipated by the 12ω resistor in the figure? (figure 1)

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We would need the shunt to have a resistance of 16 Ω. Since the coil and shunt are is parallel, the equivalentresistance - the ammeter resistance - is