1. Answer: The current outside the branches of a combination circuit is everywhere the same. The current inside of the branches is always less than that outside of the branches. When comparing the current of two parallel-­‐connected resistors, the resistor with the least resistance will have the greatest current. The current within a single branch will be the sameabove and below the resistor. a. The current at location A is equal to the current at location B. b. The current at location B is greater than the current at location E. c. The current at location G is less than the current at location F. d. The current at location E is greater than the current at location G. e. The current at location B is greater than the current at location F. f. The current at location A is equal to the current at location L. g. The current at location H is less than the current at location I. 2. Answer. The voltage drop across a resistor is dependent upon the current in the resistor and the resistance of the resistor. In situations in which the current is the same for both resistors (such as for series-­‐connected resistors), the resistor with the greatest resistance will have the greatest voltage drop. a. The electric potential difference (voltage drop) between points B and C is greater than the electric potential difference (voltage drop) between points J and K. b. The electric potential difference (voltage drop) between points B and K is greater than the electric potential difference (voltage drop) between points D and I. c. The electric potential difference (voltage drop) between points E and F is equal to the electric potential difference (voltage drop) between points G and H. d. The electric potential difference (voltage drop) between points E and F is equal to the electric potential difference (voltage drop) between points D and I. e. The electric potential difference (voltage drop) between points J and K is greater than the electric potential difference (voltage drop) between points D and I. f. The electric potential difference between points L and A is equal to the electric potential difference (voltage drop) between points B and K. For parallel-­‐connected resistors: 1/Req = 1/R1 + 1/R2 = 1 / (6 ) + 1 / (6 ) = 2 / (6 ) Req = 3 For series-­‐connected resistors: Req = R1 + R2 + R3 = 3 + 3 + 5 Req = 11 For parallel-­‐connected resistors: 1/Req = 1/R1 + 1/R2 = 1 / (12 ) + 1 / (6 ) = 3 / (12 ) Req = 4 For series-­‐connected resistors: Req = R1 + R2 + R3 = 9 + 4 + 5 Req = 18 For parallel-­‐connected resistors: 1/Req = 1/R1 + 1/R2 = 1 / (12 ) + 1 / (6 ) = 3 / (12 ) Req = 4 For series-­‐connected resistors: Req = R1 + R2 + R3 + R4 = 3 + 6 4 + 5 Req = 18 #4 The first step is to simplify the circuit by replacing the two parallel resistors with a single resistor with an equivalent resistance. The equivalent resistance of a 4 and 6 resistor placed in parallel can be determined using the usual formula for equivalent resistance of parallel branches: 1 / Req = 1 / R1 + 1 / R2 + 1 / R3 ... 1 / Req = 1 / (3 ) + 1 / (6 ) 1 / Req = 0.500 -­‐1 Req = 1 / (0.500 -­‐1) Req = 2.00 Based on this calculation, it can be said that the two branch resistors (R2 and R3) can be replaced by a single resistor with a resistance of 2 . This 2 resistor is in series with R1 and R4. Thus, the total resistance is Rtot = R1 + 2 + R4 = 2 + 2 + 4 Rtot = 8 Now the Ohm's law equation ( V = I • R) can be used to determine the total current in the circuit. In doing so, the total resistance and the total voltage (or battery voltage) will have to be used. Itot = Vtot / Rtot = (24 V) / (8 ) Itot = 3.0 Amp The 3.0 Amp current calculation represents the current at the battery location. Yet, resistors R1 and R4 are in series and the current in series-­‐connected resistors is everywhere the same. Thus, Itot = I1 = I4 = 3.0 Amp For parallel branches, the sum of the current in each individual branch is equal to the current outside the branches. Thus, I2 + I3 must equal 3.0 Amp. There are an infinite possibilities of I2 and I3 values which satisfy this equation. Determining the amount of current in either branch will demand that we use the Ohm's law equation. But to use it, the voltage drop across the branches must first be known. To determine the voltage drop across the parallel branches, the voltage drop across the two series-­‐connected resistors (R1 and R4) must first be determined. The Ohm's law equation ( V = I • R) can be used to determine the voltage drop across each resistor. These calculations are shown below. V1 = I1 • R1 = (3.0 Amp) • (2 ) = 6.0 V V4 = I4 • R4 = (3.0 Amp) • (4 ) = 12 V This circuit is powered by a 24-­‐volt source. Thus, the cumulative voltage drop of a charge traversing a loop about the circuit is 24 volts. There will be a 18.0 V drop (6.0 V + 12.0 V) resulting from passage through the two series-­‐connected resistors (R1 and R4). The voltage drop across the branches must be 6.0 volts to make up the difference between the 24 volt total and the 18.0 volt drop across R1 and R4. Knowing the voltage drop across the parallel-­‐connected resistors (R1 and R4) allows one to use the Ohm's law equation ( V = I • R) to determine the current in the two branches. I2 = V2 / R2 = (6.0 V) / (3 ) = 2.0 A I3 = V3 / R3 = (6.0 V) / (6 ) = 1.0 A #1A Answer a. -­‐ c. Location A is outside or before the branching locations; it represents a location where the total circuit current is measured. This current will ultimately divide into three pathways, with each pathway carrying the same current (since each pathway has the same resistance). Location D is a branch location; one-­‐third of the charge passes through this branch. Location B represents a location after a point at which one-­‐third of the charge has already branched off to the light bulb between points D and G. So at location B, there is two-­‐thirds of the current remaining. And location L is a location in the last branch; so one-­‐third of the charge passes through location L. d. -­‐ f. The current at every branch location and in the total circuit is simply equal to the voltage drop across the branch (or across the total circuit) divided by the resistance of the branch (or of the total circuit). As such, the current is directly proportional to the voltage. So a doubling of the voltage will double the current at every location. g. The current at a branch location is simply the voltage across the branch divided by the resistance of the branch. So the current at location G is inversely proportional to the resistance of the branch. Doubling the resistance will cause the current to be decreased by a factor of 2. h. The voltage drop across the first branch (or any branch) is simply equal to the voltage gained by the charge in passing through the battery. For a parallel circuit, the only means of altering a branch voltage drop is to alter the battery voltage. i. -­‐ k. Altering the resistance of a light bulb in a specific branch can alter the current in that branch and the current in the overall circuit. The current in a branch is inversely proportional to the resistance of the branch. So increasing the resistance of a branch will decrease the current of that branch and the overall current in the circuit (as measured at location A). However, the current in the other branches are dependent solely upon the voltage drops of those branches and the resistance of those branches. So while altering the resistance of a single branch alters the current at that branch location, the other branch currents remain unaffected. # 2A Answer: 9.375 x 1020 electrons The current (I) is the rate at which charge passes a point on the circuit in a unit of time. So I = Q/t. Rearranging this equation leads to Q = I•t. Recognizing that a current of 2.5 Amps is equivalent to 2.5 Coulombs per second and that 1 minute is equivalent to 60 seconds can lead to the amount of Coulombs moving pass the point. Q = I•t = (2.5 C/s)•(60 s) = 150 Coulombs The charge of a single electron is equal to 1.6 x 10-­‐19 C. So 150 Coulombs must be a lot of electrons. The actual number can be computed as shown: # electrons = 150 C • (1 electron / 1.6 x 10-­‐19 C) = 9.375 x 1020 electrons #3A Answer: 360 Ohms The power dissipated in a circuit is given by the equation P = I•
the resistance (R) to the voltage drop (
V. Substituting in V/R for the current can lead to an equation relating V) and the power (P). P = I•
V = (
V/R)•
V = V2 / R Rearrangement of the equation and substitution of known values of power (40 Watts) and voltage (120 V) leads to the following solution. R = V2 / P = (120 V)2 / (40 Watts) = 360 Ohms # 4A Determine the total monthly cost of using the following appliances/household wires for the given amount of time if each is plugged into a 120-­‐Volt household outlet. The cost of electricity is $0.13 / kW•hr. (Assume that a month lasts for 30 days.) Appliance Time Power Energy Cost (with info from labels) (hours/day) (Watts) Consumed ($) 0.10 1440 W 4.32 kW•h $0.56 0.10 1080 W 3.24 kW•h $0.42 8 100 W 24 kW•h $3.12 10 140 W 42 kW•h $5.46 0.25 996 W 7.47 kW•h $0.97 Hair Dryer (12 Amp) Coffee Percolator (9.0 Amp) Light Bulb (100 Watt) Attic Fan (140 Watt) Microwave Oven (8.3 Amps) Total $10.53 Answer: See table above. The power is either explicitly stated (as in the case of the light bulb) or calculated using P = I•
V. In this case, the voltage is 120 Volts. The energy consumed is the Power•time. It is useful to express this quantity in the same units for which one is charged for it -­‐ kiloWatt • hour. The calculation involves converting power in Watts to kiloWatts by dividing by 1000 and then multiplying by the time in hours/month and then multiplying by 30 days/month. The cost in dollars is simply the kiloWatt•hours of energy used multiplied by the cost of $0.13/kW•hr. # 5A. Determine the resistance of a 1500 Watt electric grill connected to a 120-­‐Volt outlet. Answer: 9.6 Ohms The power dissipated in a circuit is given by the equation P = I•
resistance (R) to the voltage drop (
V. Substituting in V/R for the current can lead to an equation relating the V) and the power (P). P = I•
V = (
V/R)•
V = V2 / R Rearrangement of the equation and substitution of known values of power (1500 Watts) and voltage (120 V) leads to the following solution. R = V2 / P = (120 V)2 / (1500 Watts) = 9.6 Ohms #6A. Four resistors -­‐ 2-­‐Ohms, 5-­‐Ohms, 12-­‐Ohms and 15-­‐Ohms -­‐ are placed in series with a 12-­‐Volt battery. Determine the current at and voltage drop across each resistor. Answer: See diagram below. The diagram below depicts the series circuit using schematic symbols. Note that there is no branching, consistent with the notion of a series circuit. For a series circuit, the overall resistance (RTot) is simply the sum of the individual resistances. That is RTot = R1 + R2 + R3 + R4 RTot = 2 ½ + 5 ½ + 12 ½ + 15 ½ = 34 ½ The series of three resistors supplies an overall, total or equivalent resistance of 34 Ohms. Since there is no branching, the current is the same through each resistor. This current is simply the overall current for the circuit and can be determined by finding the ratio of battery voltage to overall resistance (VTot/RTot). ITot = VTot/RTot = (12 Volt) / (34 Ohm) = 0.35294 Amps The current through the battery and through each of the resistors is ~0.353 Amps. The voltage drop across each resistor is equal to the I•R product for each resistor. These calculations are shown below. V1 = I1 • R1 = (0.35294 Amps) • (2 Ohms) = 0.71 V V2 = I2 • R2 = (0.35294 Amps) • (5 Ohms) = 1.76 V V3 = I3 • R3 = (0.35294 Amps) • (12 Ohms) = 4.24 V V4 = I4 • R4 = (0.35294 Amps) • (15 Ohms) = 5.29 V #7A. Four resistors -­‐ 2-­‐Ohms, 5-­‐Ohms, 12-­‐Ohms and 15-­‐Ohms -­‐ are placed in parallel with a 12-­‐Volt battery. Determine the current at and voltage drop across each resistor. Answer: See diagram below. The diagram below depicts the parallel circuit using schematic symbols. Note that there is a branching, consistent with the notion of a parallel circuit. For a parallel circuit, the reciprocal of overall resistance (1 / RTot) is simply the sum of the reciprocals of individual resistances. That is 1 / RTot = 1 / R1 + 1 / R2 + 1 / R3 + 1 / R4 1 / RTot = 1 / 2 ½ + 1 / 5 ½ + 1 / 12 ½ + 1 / 15 ½ = 0.850 / ½ RTot = 1.17647 ½ The series of three resistors supplies an overall, total or equivalent resistance of ~1.18 Ohms. This total resistance value can be used to determine the total current through the circuit. ITot = VTot/RTot = (12 Volt) / (1.17647 Ohm) = 10.2 Amps Since there is branching, the total current will be equal to the sum of the currents at each resistor. The current at each resistor is the voltage drop across each resistor divided by the resistance of each resistor. For series circuits, the voltage drop across each resistor is the same as the voltage gained by the charge in the battery (12 Volts in this case). The branch current calculations are shown below. I1 = V1 / R1 = (12 Volts) / (2 Ohms) = 6.00 Amp I 2 = V2 / R2 = (12 Volts) / (5 Ohms) = 2.40 Amp I 3 = V3 / R3 = (12 Volts) / (12 Ohms) = 1.00 Amp I 4 = V4 / R4 = (12 Volts) / (15 Ohms) = 0.80 Amp