Chapter 6 DRECT CURRENT CRCUTS Recommended Problems: 6,9,,3,4,5,6,9,0,,4,5,6,8,9,30,3,33,37,68,7.
RESSTORS N SERES AND N PARALLEL - N SERES When two resistors are connected together as shown we said that they are connected in series. a R R t is clear that the total current is the same as that flows through R and R, i.e.. eq Since the potential difference between a and b equals the sum of the potential drop across each resistor we have ε b V eq V V eq R eq V V R eq R R
- N PARALLEL R When two resistors are connected together as shown we said that they are connected in Parallel. eq a R b ε The potential drops across R and R are equal and must equal to the potential drop across any equivalent resistor connected between a and b, that is V eq V V t is clear that the total current splits into two currents such eq V R eq eq V R V R R eq R R
Example 8.4 Four resistors are connected as shown in. (a) What is the equivalent resistance between points a & c.? (b) What is the current in each resistor if a potential difference of 4 V is maintained between a & c.? a a series 8.0 4.0 b 3 A 3 A 6.0 parallel 6.0 V A 6.0 V A 3.0 series 3 A b 3 A.0 6.0 V c c Solution a) a 3 A c b) 4 4 V
Example 8.5 What is equivalent resistance between a and b in the Figure shown? c.0 a 5.0.0 b a parallel.0 5.0 parallel.0 b.0.0.0 C,d.0 d Because of the symmetry in the circuit, the current in branches ac & ad must be equal, and hence the points c & d have the same potential (V cd =0), that is the circuit can be reduced as in the figure. a 0.5 series c,d 0.5 a b.0 b
Example 8.6 Three resistors are connected in parallel as shown. A potential difference of 8 V is applied across points a and b. a) Find the current in each resistor. b) Calculate the power delivered to each resistor and the total power delivered to the combination. Solution 8 V 3 6 a 3 a) Since the resistors are connected in parallel, the potential difference across each one is the same and equal to 8 V. Now b 9 8 3 R V 6A 8 6 R V 3A 8 9 3 3 3 R V A
b) For the power delivered to each resistor we apply 6 3 R P P 3 6 R P 9 3 3 R3 08 W 54 W 36 W P eq.64 98 W eq Req Note that P eq P P P3
Test Your Understanding () With the switch in the circuit shown closed the ammeter reads the current in the circuit. What happens to the reading on the ammeter when the switch is opened? (a) ncreased. (b) Decreased. (c) Not changed. (d) t is zero in both cases.
KRCHHOFF S RULES - The sum of the currents entering any junction must equal the sum of the currents leaving that junction. 3 (A junction is any point in a circuit where a current can split) 3 - The algebraic sum of the potential differences across all the elements around any loop must be zero. The first rule is an application of the conservation of charge principle, while the second rule is an application of the conservation of energy principle.
To apply the second rule we should know the following two remarks: Since the current through a resistor moves from the end of higher potential to that of lower potential. - The change in potential through any resistor is negative for a move in the direction of the current and positive for a move opposite to the direction of the current. - The change in potential through an ideal battery is positive for a move from the negative to the positive terminal of the battery and negative for a move in the opposite direction. R V=-R or V=R ε V=ε or V=-ε
Test Your Understanding () Kirchhoff s first rule is a consequence of (a) Conservation of energy. (b) Conservation of charge. (c) Ohm s law. (d) Newton s third law. While Kirchhoff s second rule is a consequence of (a) Conservation of energy. (b) Conservation of charge. (c) Ohm s law. (d) Newton s third law.
Strategy for solving problems using Kirchhoff s rules: - Draw a circuit diagram and label all quantities, known and unknown. - Assign a direction for the current in each part of the circuit. Do not bother if your guess of current direction is incorrect; the result will have a negative value. 3- Apply the first Kirchhoff s rule to any junction in the circuit. n general this rule is used one time fewer than the number of junctions in the circuit. 4- Choose any closed loop in the network, and designate a direction (clockwise or counterclockwise) to traverse the loop. 5- Starting from one point in the loop, go around the loop in the designated direction. Sum the potential differences across all the elements of the chosen loop to zero.
n doing so you should note the two remarks discussed before: the potential difference across an emf is + if it is traversed from the negative to the positive terminal and - if traversed in the opposite direction. the potential difference across any resistor is -R if this resistor is traversed in the direction of the assumed current and +R if traversed in the opposite direction. 6- Choose another loop and repeat the fifth step to get a different equation relating the unknown quantities. Continue until you have as many equations as unknowns. 7- Solve these equations simultaneously for the unknowns.
Example 8.8 n the circuit shown, find the current in the circuit and the power delivered to each resistor and the power delivered by the -V battery. R = 0 =6V R = 8 Solution =V The directions of the currents are assigned arbitrary as shown in the Figure. As it clear from the circuit there is one loop with no junctions. Now, we apply Kirchhoff s second rule to the loop and traverse the loop in the clockwise direction, obtaining ε R ε R 0 6 8 0 0 6 8 0 0.33A The minus sign indicates that the direction of is opposite the assumed direction.
To find the power delivered to each resistor, we use R P 0.87 W P R. W P 0.33 4W Note that P P 0.87. W That is half of the power supplied by the V-battery is delivered to the resistors and the other half is delivered to the 6V-battery.
Example 8.9 n the circuit shown, a) Find the current in each resistor. b) Calculate the potential difference V b V a. Solution a) f we apply Kirchhoff s first rule to the junction b we get 3 () a 4 0V 4V 6 b 3 Now applying Kirchhoff s second rule to the upper loop traversing it clockwise we get 4 6 0 4 0 () The bottom loop traversing it in the clockwise direction gives 0 6 3 0 (3) Solving the three equations we get A, 3A, 3 A
b) To find the potential difference between two points, start at the initial point and traverse the loop toward the final point, adding potential differences across all the elements you encounter. Starting at point a, we follow a path toward point b through the middle battery we obtain V b V a 0 6 0 6 V The minus sign here means that V a V b. a 4 0V 4V 6 b 3 Try to follow another paths from a to b to verify that they also give the same result.
The RC CRCUTS Charging Process The figure shows a capacitor, initially uncharged, connected in series with a resistor. f the switch S is thrown at point, the capacitor will begin to charge, creating a current in the circuit. Let be the current in the circuit at some instant during the charging process, and q be the charge on the capacitor at the same instant. From Kirchhoff s second rule, we obtain But dq dt R q C 0 dq q R 0 dt C S R dq C q dt RC C
dq C q dt RC ntegrating both sides of the last equation we get q t dq dt C q t 0 ln C q 0 RC C RC C q C t e RC q C t e RC where e is the base of the natural logarithm. To find the current as a function of time, we differentiate the last equation with respect to time to get with RC e R t RC e R t is called the time constant of the circuit. t is defined as the time required for the current to decrease to e - of its initial value.
Let us plot the charge and the current versus the time At t=0 the charge is zero and the current is maximum given by R that is, the capacitor acts as if it were a wire with negligible resistance (short circuit). After along time (t ), the charge is maximum and the current is zero. that is, the capacitor acts as if it were a an open switch (open circuit). Q m o q (a) (b) t t
Discharging Process Suppose that the capacitor is now fully charged such that its potential difference is equal to the emf. f the switch is thrown to point at a new time, the capacitor begin to discharge through the resistor. Let be the current in the circuit at some instant during this process, and q be the charge on the capacitor at the same instant. Applying Kirchhoff s rule to the loop, we get S R C q C R 0 Substituting for with dq dt
dq dt q RC ntegrating both sides and noting that at t=0 q=q m q Q dq q RC t m 0 dt ln q Qm t RC q Q m e t RC The current is the rate of decrease of the charge dq dt o e t RC
Test Your Understanding (3) For the circuit shown in the figure, just after closing the switch S, across which of the following is the voltage equal to the emf of the battery? (a) R. (b) C. S R C (c) Both R and C. (d) Neither R nor C. After closing the S for a long time, across which of the following is equal to the emf of the battery? (a) R. (b) C. (c) Both R and C. (d) Neither R nor C.
Example 8. A 8 0 5 -k resistor and a 5-F capacitor are connected, in series, with a -V battery as shown. The capacitor is initially uncharged, and the switch S is closed at t=0. V S 8 0 5 k 5 F a) Find the time constant of the circuit, the maximum charge on the capacitor, and the initial value of the current. b) What is the time required for the current to drop to half its initial value? c) After being closed for a long time, the switch is now opened at t=0, what is the time required for the charge and for the energy to decrease to one-fourth their maximum value.
Solution a) RC 5 6 8.00 5.00 4.0s Q m CV C C 6 5.00 60C at t=0, the capacitor acts as a short circuit o R 80 t RC 5 5A o b) Using o e and substituting for t e RC t o o ln t RCln.8s RC
c) n the discharging process, the charge varies with time according to q Substituting for q as Q m e t q RC Q 4 m t ln t RCln s 4 RC 4 5.5 For the energy we have 4 Q m Q m e t RC q Q U e C C Substituting for U as U t RC U 4 m U m e t RC t ln 4 t RC ln.8s RC 4
Example (Extra) n the circuit shown, the capacitor is initially empty and the switch S is closed at t=0. a) Find the current in each branch of the circuit at t=0. b) Calculate the maximum charge on the capacitor. Solution 3 V at t=0, the capacitor acts as a short circuit 6 S F the capacitor makes a short circuit across the 4- resistor. 3 4 3 0 and 3 6.0 5.3A
b) The maximum charge is attained after a long time (t ). At this time the capacitor is treated as if it were an open switch. 3 3 0 and 3 6.0 4.0 3.A 3 V To calculate the charge on the capacitor, we first want to find the potential difference across it. 6 S F Applying Kirchhoff s second rule to right loop we find that the potential difference across the capacitor is V C R 3.4.0 V C 3 4 Q CV 0 6.8.50 4