PHYS202 Spring Chapter 20 Examples

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Candace Wells
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1 HYS0 Spring 009 Chapter 0 Examples 1. particular electric circuit consists of a resistor (denote as ), and a light ul connected in series to an C source (wall outlet) that provides of rms 10. n a circuit without the resistor, the ul uses 100 W of power. a) Draw a circuit for the light ul and resistor in series with the C source. Determine the value of the resistance needed for resistor if the light ul is to use 75.0 W of power. First, find the resistance of the ul alone rmsrms Ω 100W Since the resistor and ul are in series, the current must e the same through oth and the voltage drop across oth is the sum of the voltage drops across each: + The net resistance in series is the sum of the resistances, The current can e found through Ohm s law. + + The power through the ul is + Solve for the value of resistor : + + ( + )

2 HYS0 Spring 008 ( + ) ( Ω) Ω 75W.3Ω ) Determine the power output of the C source and compare to the case without the resistor. With resistor: W ( 10) 86.6W 144Ω n the case the power reduction is not ¾, ut is the square root of ¾. f the output of the ul was reduced to x of the original value y adding a resistor in series, the attery power output is only decreased to the square root of x. c) second identical light ul replaces resistor. They are placed in series with the C generator (i.e. wall socket), then in parallel with the C generator. The C generator provides rms 10. Calculate the average power delivered y the C generator in each case to find the ratio of light emitted of oth cases. Series case: Since the uls are in series, the current must e the same through oth and the voltage drop across oth is the sum of the voltage drops across each: 1 1+ The net resistance in series is the sum of the resistances, + age of 8

3 HYS0 Spring 008 The power delivered y the C source is. arallel Case: Here, the uls are in parallel, the current must e the sum of the currents across each and the voltage drop across oth is the same as the voltage drops across each: + The net resistance in parallel is The power delivered y the C source is. t is easy to see that the power delivered in the parallel case is four times more than the serial case. Since the only items using power is the light uls. This is also the ratio of light emitted.. n Back to the Future, the time machine is said to need 1.1 gigawatts of power. Doc Brown runs a cale to the top of the town hall to make a connection to a wire hanging over the street thus providing a connection to the DeLorean as it speeds past. ssume the entire length (from the lightning rod to the car) is a 100 m iron cale of resistivity ρ 9.7 x 10 8 Ω m. a) The typical lightning olt has a current of aout 40.0 k. What is the voltage needed to transfer 1.1 gigawatts of power? W ) f the cale is 1.5 cm in diameter, how much power is lost with 40.0 k running through the cale? age 3 of 8

HYS0 Spring 008 Or aout 68 MW. L ρ L ρ ( 4.0 10 ) ( 9.61 10 m) ( 0.

83 10 W 100m c) ssume the resistance in part () was calculated at 0 C.

T0) 1 + α ( T T0) 0 3 7 1+ 5.0 10 / C 0 C 0 C 6.

4 HYS0 Spring 008 Or aout 68 MW. L ρ L ρ ( ) ( m) ( 0.015m) 4 8 Ω W 100m c) ssume the resistance in part () was calculated at 0 C. What would the power loss through the cale have een on a cold day when it is 0 C? (use α 5.0 x 10 3 / C) ρ ρ 1 + α T T L 0 0 From part () we have ρ, L ρ0 1 + α( T T0) 1 + α ( T T0) / C 0 C 0 C W Or aout 6 MW, a difference of aout 10% W 3. n the circuit drawing to the right, 4.0, Ω, 8.00Ω, Ω, 4 1.0Ω. a) Find the current through the attery. Make the equivalent circuits: age 4 of 8

HYS0 Spring 008 dding the three remaining resistors in series we find that the equivalent resistance across the attery is 40.8Ω. Using Ohm s Law: 4.0 equiv 40.8Ω 0.

60 40.8Ω 14.5W 4.3 W ) in the final 4. n the circuit drawing to the right, 1.0, 3 4.0, 1 1.00Ω,.00Ω. a) Find the three currents (i.e. the currents through the three atteries) in the circuit.

5 HYS0 Spring 008 dding the three remaining resistors in series we find that the equivalent resistance across the attery is 40.8Ω. Using Ohm s Law: 4.0 equiv 40.8Ω 0.60 equiv ) Find the power expended y the attery and resistor 4. Since the current through the equivalent diagram is the same through all resistors diagram, we can find the power. equiv equiv 4 4 ( Ω 14.5W 4.3 W ) in the final 4. n the circuit drawing to the right, 1.0, 3 4.0, Ω,.00Ω. a) Find the three currents (i.e. the currents through the three atteries) in the circuit. There are two junction points and two loops. t should e clear that the junction points and B provide the same information. Let us denote the currents as l(eft), r(ight) and m(iddle). B Taking Kirchhoff s junction rule for point, let us have the current in the left and right ranches flow inward and in the middle ranch flow outward : Taking the two loops, we get: and + l r m 0 1 l 1 m l l 1 m 0 3 r 1 m Let us add these equations together. 3 r 1 m r l r 1 m From the junction rule we know l + r m age 5 of 8

6 HYS0 Spring 008 m m 1 m + + m 1 3 ( ) Ω+ Ω l 0.33 m The minus sign indicates that the flow is in the opposite direction of the original assumption. We sustitute the current for the middle ack into the loop equations for the left and right sides l 1 m 3 r 1 m 0 l 1 1 m r 1 3 m l 1 m 3 m r Ω Ω r.0ω.0ω l r gain, the minus sign for the current l, means it is going in the opposite direction of the original assumption. So we have the currents as left ranch: 0.67 down Middle ranch: 0.33 up ight ranch: 0.33 up ) Find the voltage difference etween points and B ( a ) The difference in voltage etween points and B is found y finding the voltage drop through any path from to B. Note that we use the correct direction for the current now! Left path: B l1 1 l Ω Ω 3.3 Middle path: ight path: B B + m Ω B B + r1 1+ m Ω Ω 3.3 B age 6 of 8

HYS0 Spring 008 n all three cases we have 3.3, which indicates that it is a drop of 3.3 from to B. 5. n the circuit drawing to the right, 100, C 6.5 μf, 1 3 7.3 x 10 5 Ω.

Let 1,, and 3 e the currents through S resistors 1,, and 3 flowing as shown in the diagram. 1 will flow through the attery and 3 will flow through the capacitor C.

7 HYS0 Spring 008 n all three cases we have 3.3, which indicates that it is a drop of 3.3 from to B. 5. n the circuit drawing to the right, 100, C 6.5 μf, x 10 5 Ω. With C completely uncharged, the switch S is closed (t 0) 3 C a) Determine the currents through each resistor for t 0 and t. Use q/c to solve with Kirchhoff s rules with a capacitor. Let 1,, and 3 e the currents through S resistors 1,, and 3 flowing as shown in the diagram. 1 will flow through the attery and 3 will flow through the capacitor C. Kirchhoff s junction rule gives, The loop rules give the following. q C 1 Since all resistors are the same: q C q ( 1 ) 0 + ( 3) 0 C t t 0, we have no charge on the capacitor, so it provides no change in voltage: ( ) 0 3 Which gives 0.55 m, so, m and m t t, the capacitor is fully charged and there is no current flowing through it, so y definition 3 0, so the junction rule gives 1 and only the lower loop is of concern m ) Determine the voltage across resistor for t 0 and t. age 7 of 8

8 HYS0 Spring t Ω 400 ( 4 )( 5 ) t Ω 600 c) fter a long enough period, the capacitor is fully charged. f the switch is now opened, how long will it take for the capacitor to reach 5% of its full charge? tc / q tc / q q0e e q0 t q ln ( + 3) C q0 q q 0 t ( + 3) Cln ( )( F) ln q0 q0 t 8 s age 8 of 8

4 Electric circuits. Serial and parallel resistors V 3 V 2 V Serial connection of resistors:

4 Electric circuits. Serial and parallel resistors V 3 V 2 V Serial connection of resistors: 4 lectric circuits PHY67 Spring 006 Serial and parallel resistors Serial connection of resistors: As the current I through each of serially connected resistors is the same, one can use Ohm s law and write...