E-Tutorial Physics Book

Note:I have attached below the lesson for the two (2) activities. Thank you.

Activity 1.

Directions:Look at each circuit below and show the solution using Kirchhoff's Rules to find the required quantity. Show your loops in blue and assumed current in green.

Activity 1. Directions: Look at each circuit below and show the solution using Kirchhofl's Rules to nd the required quantity. Show your loops in blue and assumed current in green. Find the Current 1: Find the voltage across the 50 resistor. Activity 2. Directions: Write what are the key reminders whenever using the KVL and KCL in solving circuits. Give at least 5 tips. Tip 1 Tip 2 Tip 3 Tip 4 Tip SKirchhoff's Rules Kirchhoff's rules are named after German physicist Gustav Robert Kirchhoff who outlined a systematic way of solving multi-loop circuits. Note that we have already been applying these principles. The basis for how and why they work have already been laid out. Kirchhoff's rules keep everything organized as we analyze and solve more complicated networks. First, here are two terms that we will use often. A junction in a circuit is a point where three or more conductors meet. A loop is any closed conducting path. In Fig. 6a points a and b are junctions, but points c and d are not; in Fig. 6b the points a, b, c, and d are junctions, but points e and f are not. The lines (blue lines) inside and outside of the circuit in Figs. 6a and 6b show some possible loops in these circuits. (a) Junction (b) Loop 15 a -(2) Loop 2 Loop 3 RX R Not a Junction Not a junction junction Figure 6. Two networks that cannot be reduced to simple series-parallel combinations of resistors. Photo credit: University Physics with Modern Physics (13th Edition) Kirchhoff's rules are the following two statements: Kirchhoff's junction rule: The algebraic sum of the currents into any junction is zero. That is, (junction rule, valid at any junction) Kirchhoff's loop rule: The algebraic sum of the potential differences in any loop, including those associated with emfs and those of resistive elements, must equal zero. That is, [v = 0 (loop rule, valid for any closed loop) (a) Kirchhoff's junction rule (b) Water-pipe analogy Junction . The current leaving The flow rate of hthy a junction equals the water leaving the current entering it. pipe equals the flow rate entering it. Figure 7. Kirchhoff's junction rule states that as much current flows into a junction as flows out of it. Photo credit: University Physics with Modern Physics (13th Edition) The junction rule is based on conservation of electric charge. No charge can accumulate at a junction, so the total charge entering the junction per unit time must equal the total charge leaving per unit time (Fig. 7a). Charge per unit time is current, so if we consider the currents entering a junction to be positive and those leaving to be negative, the algebraic sum of currents into a junction must be zero.It's like a T branch in a water pipe (Fig. 7b); if you have a total of 1 liter per minute coming in the two pipes, you can't have 3 liters per minute going out the third pipe. We may as well confess that we used the junction rule (without saying so) in the previous part in the derivation of the equation for resistors in parallel. The loop rule is a statement that the electrostatic force is conservative. Suppose we go around a loop, measuring potential differences across successive circuit elements as we go. When we return to the starting point, we must nd that the algebraic sum of these differences is zero; otherwise, we could not say that the potential at this point has a definite value. Sign Conventions for the Loop Rule In applying the loop rule, we need some sign conventions. We rst assume a direction for the current in each branch of the circuit and mark it on a diagram of the circuit. Then, starting at any point in the circuit, we imagine traveling around a loop, adding emfs and IR terms as we come to them. When we travel through a source in the direction from to the emf is considered to be positive; when we travel from to the emf is considered to be negative (Fig. 8a). When we travel through a resistor in the same direction as the assumed current, the IR term is negative because the current goes in the direction of decreasing potential. When we travel through a resistor in the direction opposite to the assumed current, the IR term is positive because this represents a rise of potential (Fig. 8b). Kirchhoff's two rules are all we need to solve a wide variety of network problems. Usually, some of the emfs, currents, and resistances are known, and others are unknown. We must always obtain from Kirchhoff's rules a number of independent equations equal to the number of unknowns so that we can solve the equations simultaneously. Often the hardest part of the solution is not understanding the basic principles but keeping track of algebraic signs! Use these sign conventions when you apply Kirchhoff's loop rule. In each part of the figure \"Travel\" is the direction that we imagine going around the loop, which is not necessarily the direction of the current. (a) Sign conventions for cmfs (b) Sign convemions for resistors +8: Travel direction - E: Travel direction +18: Travel 0pm!!! '38: Travel in from-tor: frm'n-r-to-z mcmdirarion: curmudhucriun: lerI _- ._' Inn-cl Imwl 'p - l'rzncl E E R E Figure 8. {a} Sign conventions for emfs and {19) sign conventions for resistors Photo credit: University Physics with Modern Physics {139' Edition} Strategy for Applying Kirchhoff's Rules to Multi-loop Networks a. Name and label the direction of the current in each branch. Based on these directions, label the high and low-voltage ends of each resistor. b. Pick a loop direction for every loop minus 1. c. Apply the junction rule to every junction minus 1. d. Apply the loop rule to every loop you picked in step 2. Solve the equations simultaneously. You should have as many equations as there are unknowns. Check your answers by using the loop rule on the extra loop you did not use in step 4. Sample Problem Using Kirchhoff's rules. Calculate the currents 11,12, and 13 in the three branches of the circuit in Fig. 9. Approach and Solution 1. Label the currents and their directions Fig. 9 uses the labels I 1, 1'2, and I3 for the current in the three separate branches. Since (positive) current tends to move away from the positive terminal of a batter, we choose 12 and 13 to have the directions shown in Fig. 9. The direction of II is not obvious in advance, so we arbitrarily chose the direction indicated. If the current actually ows in the opposite direction, our answer will have a negative sign. Identify the unknowns. We have three unknowns and therefore we need three equations, which we get by applying Kirchhoff's junction and loop rules. Junction rule: We apply Kirchhoff's junction rule to the currents at point a, where 13 enters and 12 and 11 leave: 13 =11 +12 (:1) This same equation holds at point d, so we get no new information by writing an equation for point d. Figure 9. Currents can be calculated using Kir'chhoff's rules. Photo credit: Giancoli Physics {5m} Loop rule: We apply Kirchhoff's loop rule to two different closed loops. First, we apply it to the upper loop ahdcba. We start (and end) at point a. From a to h we have a potential decrease VM = (11)(30f1). From h to d there is no change, but from d to c the potential increases by 45 V: that is, ME = (11)(40.Q + 1.0) = (41I).)I3. Thus we have VM + Vcd + Vac = 0, or 3011 + 45 4113 = 0, (b) Where we have omitted the units. For our second loop, we take the outer loop ahdefga. (We could have chosen the lower loop abcdefga instead.) Again we start at point a and have Vha = (.'1) (30(1), and Van = 0. But when we take our positive test charge from d to e, it actually is going uphill, against the current or at least against the assumed direction of the current, which is what counts in this calculation. Thus Ved = 12 (20(1) has a positive sign. Similarly, er = 12(10). From f to g there is a decrease in potential of 80V since we go from the high potential terminal of the battery to the low. Thus ng = 80V. Finally, Vag = 0, and the sum of the potential changes around this loop is then 30.r1 + (20 + 1).!2 30 = 0. (c) 5. Solve the equations. We have three equations labeled (a), (b), and (c) and three unknowns. From Eq. (c) we have 80 + 3011 12 = T = 3.8 + 1.411. (d) From Eq. (b) we have 45 + 3011 13 = T = 1.1 0.7311. (3) We substitute Eqs. (d) and (e) into Eq. (a): 11 :13 12 = 1.1 0.7311 3.8 1.411. We solve for 11, collecting terms: 3.111 = 2.7 [1 = 0.87 A. The negative sign indicates that the direction of 11 is actually opposite to that initially assumed and shown in Fig. 9. Note that the answer automatically comes out in amperes because all values were in volts and ohms. From Eq. (d) we have 12 = 3.8 + 1.411 = 3.8 + 1.4(0.87) = 2.6 A, and from Eq. (e) 13 = 1.1 + 0.7311 = 1.1 0.73(0.87) = 1.7 A. This completes the solution. Note: The unknowns in different situations are not necessarily currents. It might be that the currents are given and we have to solve for unknown resistance or voltage. RC Circuits Capacitors and resistors are often found together in a circuit. Such RC circuits are used to control a car's windshield wipers and the timing of trafc lights; they are used in camera ashes, in heart pacemakers, and many other electronic devices. Charging a capacitor In RC circuits, we are not so interested in the nal \"steady state\" voltage and charge on the capacitor, but rather in how these variables change in time. A simple example is shown in Fig. 10a. We now analyze this RC circuit. Figure I 0. For the RC circuit Shown in {a}, the voltage across the capacitor increases with time, as shown in {b}, after the switch S is closed. Photo credit: ems Physics {5m} When the switch S is closed, current immediately begins to ow through the circuit. Electrons will ow out from the negative terminal of the battery, through the resistor R, and accumulate on the upper plate of the capacitor. And electrons will ow into the positive terminal of the battery, leaving a positive charge on the other plate of the capacitor. As charge accumulates on the capacitor, the potential difference across it increases (V = g), and the current is reduced until eventually the voltage across the capacitor equals the emf of the battery, .8. There is then no potential difference across the resistor, and no further current ow. The potential difference across the capacitor, which is proportional to the charge on the capacitor (VC- = 3), thus increases in time, as shown in Fig. 10b. The actual shape of this curve is a type of exponential. It is given by the formula VC = e (1 e't'RC), where we use the subscript C to remind us that Va is the voltage across the capacitor and is given here as a function of time t. [The constant e, known as the base for natural logarithms, has a value e = 2.718 . Do not confuse this e with e for the Charge on the eictron.] We can write a similar formula for the charge Q(= CV5) on the capacitor: Q = Q0(1 _ et/RC)' where 00 represents the maximum charge. Time Constant The product of the resistance R times the capacitance C, which appears in the exponent, is called the time constant I of the circuit: r = RC. The time constant is a measure of how quickly the capacitor becomes C charged. [The units of RC are 0. - F = (E) (E) = E = 3.] Specically, it can be shown that the product RC gives the time required for the capacitor's voltage (and charge) to reach 63% of the maximum. This can be t checked using any calculator with ex key: e1 = 0.37, so for t=RC, then 1 eE = (1 e'l) = (1 0.37) = 0.63. In a circuit, for example, where R=2002 and C = 3.0uF, the time constant is (2.0x105)(3.0x10'6F) = 0.60 s. If the resistance is much smaller, the time constant is much smaller and the capacitor becomes charged almost instantly. This makes sense, since a lower resistance will retard the ow of charge less. All circuits contain some resistance (if only in the connecting wires), so a capacitor can never be charged instantaneously when connected to a battery. Discharging a Capacitor The circuit just discussed involved the charging of a capacitor by a battery through a resistance. Now let's take a look at another situation: a capacitor is already charged (say, to a voltage V0 and charge 00), and it is then allowed to discharge through a resistance R as shown in Fig. 11a. (In this case there is no battery.) When the switch S is closed, charge begins to ow through resistor R from one side of the capacitor decreases, as shown in Fig. 1 1b. This \"exponential decay\" curve is given by VC = Ve'l/RC, where V5 is the initial voltage across the capacitor. The voltage falls 63% of the way to zero (to 0.37Vg) in a time T = RC. Because the charge Q on the capacitor is Q=CV, we can write Q = Que-W for a discharging capacitor, where Qo is the initial charge. Figure 11. For the RC circuit shown in (a), the voltage Vc on the capacitor decreases with time, as shown in (b), after the switch S is closed. The charge on the capacitor follows the same curve since Q x V. Time Photo credit: Giancoli Physics (6th) Table 1: Charging and Discharging Equations for RC Circuits Discharging Charging Charge loe-t/t Qo(1 - e-t/T) Current loe-t/t loe-t/T Voltage Voe-t/T Vo (1 - e-t/T The charging curve for a RC charging circuit is exponential and not linear. This means that in reality the capacitor never reaches 100% fully charged. So, for all practical purposes, after five times constants (5T) it reaches 99.3% charge, so at this point the capacitor is considered to be fully charged. As the voltage across the capacitor Vc changes with time, and is therefore a different value at each time constant up to 5T, we can calculate the value of capacitor voltage, Vc at any given point, for example. We can show in the following table the percentage voltage and current values for the capacitor in a RC charging circuit for a given time constant. Table 2. RC Charging Table Time Percentage of Maximum RC Value Constant Voltage Current 0.5 time constant 0.5T - 0.5RC 39.3% 60.7% 0.7 time constant 0.7T - 0.7RC 50.3% 49.7% 1.0 time constant 1T - 1RC 63.2% 36.8% 2.0 time constants 2T - 2RC 86.5% 13.5% 3.0 time constants 3T - 3RC 95.0% 5,0% 4.0 time constants 4T - 4RC 98.2% 1.8% 5.0 time constants ST - 5RC 99.3% 0.7%Also, we can show in the following table the percentage voltage and current values for the capacitor in a RC discharging circuit for a given time constant. Table 3. RC Discharging Table Current 0.5 time constant 0.51 a 0.5RC 39.396 0.7 time constant 071 = DJRC 49.796 \" .495 2.0 time constants 36.5% 4.0 time constants 5.0 time constants Note that as the discharging curve for a RC discharging circuit is exponential, for all practical purposes, after ve time constants the capacitor is considered to be fully discharged. So, an RC Circuit's time constant is a measure of how quickly it either charges or discharges. Sample Problem 1 Calculate the RC time {- o R . 'm constant, I of the following circuit. Vs - Sir I + ' _' Figure 12. RC circuit with 1000mcroF, Vt ' : C . R=47ko, t=0, Vs=5V. - . Photo credit: Electronic Tutorials The time constant, I is found using the formula T = R x C in seconds. Therefore, the time constant it is given as: T = R x C = 47k x l OOOuF = 47 Secs a) What will be the value of the voltage across the capacitors plates at exactly 0.7 time constants? At 0.7 time constants ( 0.7T) Vc = 0.5Vs. Therefore, Vc = 0.5 X 5V = 2.5V 1:] What value will be the voltage across the capacitor at 1 time constant? At 1 time constant ( 1T) Vc = 0.63Vs. Therefore, Vc = 0.63 X 5V = 3.15V c) How long will it take to \"fully charge" the capacitor from the supply? We have learnt that the capacitor will be fully charged after 5 time constants, (5T). 1 time constant ( 1T ) = 47 seconds, (from above). Therefore, 5T = 5 x 47 = 235 secs d) The voltage across the Capacitor after 100 seconds? The voltage formula is given as Vc = V(1 - ett/RC)) so this becomes: Vc = 5(1 - e(-100/47)) Where: V = 5 volts, t = 100 seconds, and RC = 47 seconds from above. Therefore, Vc = 5(1 - e(-100/47)) = 5(1 - e-2.1277) = 5(1 - 0.1 191) = 4.4 volts Sample Problem 2 In the circuit given in Fig. 12, find the C = 6 UF a. Current the instant the switch is closed R = 2kQ (maximum current), b. Time constant, c. Maximum charge stored in the capacitor, and d. Charge stored after 10ms has elapsed. Given: & = 9V, R = 2kQ, and C = 6/F. 9V Solution: a. The very first current is the maximum current, given by Ohm's law: E 1= R 2000 0 = 4.5X10-3A = 4.5 mA b. The time constant r is the product RC: T = RC = (20000) (6x10-F) = 0.012 s = 12 ms c. We can get the maximum charge from the definition of capacitance. Qmax = CE = (6x10-F) (9V) = 5.4X10-C = 54 MIC d. To get the charge after t=10ms, we use Equation: Q = Qo(1 - e-t/RC) = (5.4x10-5c) (1 -e 0.012 s -10X10-3 s = 3.05x10-5C = 30.5 JC