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spring 04Exam2_Solutions

# spring 04Exam2_Solutions - Problem 1(30 points You are...

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Unformatted text preview: Problem 1 (30 points) You are given 3 mm x 3 hurt sheets of aluminum with wires attached. You are also given 3 mm x 3 mm sheets of strontium titanate (SrTi03, K = 23 3) of thickness 10 pm. You construct parallel plate capacitors by sandwiching the SrTiO; between the aluminum plates. a.) What is the capacitance of a single capacitor that you can construct, C? (5 pts) b.) You construct ﬁve of these capacitors. How would you arrange these 5 capacitors to get the minimum net capacitance? (4 pts) c.) What is the minimum capacitance you can get with all ﬁve capacitors, Cmin?(5 pts) e.) The capacitor circuit shown above is connected to a battery of 12 V at time t = 0. A resistor of R = 10 k9 is in the circuit as shown. What is the charging time of the circuit, 1:? (4 pts) f.) What is the total energy stored in the capacitors at time t >> ‘E, s? (5 pts) - 7. _q a.) C a Ken}: c. 233(82's'xlo “0%) E Worm) : L95 no F ._ J _6 ol 1‘ Kayla W1 :‘-?§n; It“ Sencg —.l l l ' — C — 0 2:? F —- -t- __ l l _ _ ._ - h C") CM“: (<1 c+g+E+EU 5 “a. e%u.uolen‘l- (opomxnnic oil- 4Mree. rafﬂe-7"va tn Parallellli‘ C. 15th SP‘HE3. m l‘l’h +030 addu‘auwol- Capos: “0‘45 -l = —L .L L —_ i :03? ml: Chair (3 -l- + > C Place your answers here. \ (“x \ " t. React : many—war): may": ( I L a.) C= i_(5)_ 1?) E = gcv —._- i(o.?%r‘}(n®2: stays”: b.) arrangement 4(4)— c.) Cmin = (5) Problem 2 (40 points) A circuit is composed of two emf sources with voltages V. = 12.0 V and V2 = 9.0 V, and two resistors R1 = 100 kQ and R2 = 200 k!) as shown below. a.) In terms ofl], 12, I; as drawn and in terms ofR., R2, V. and V2 write Kirchof‘f’s Junction Rule for point B , Kirchoﬁ‘s Loop Rule for loop ABEFA and Kirchoff’s Loop Rule for loop BCDEB. (13 pts) b.) Determine the currents ll, 12 and [3. (12 pts) c.) How much power does battery V3 supply, P2? (5 pts) (1.) Resistor R2 consists of a wire that is hanging out doors. Ice accumulates on the wire and starts to stretch it. If the length is increased by a factor of 4 and the diameter of the wire is decreased by a factor of two, what is the new resistance of this wire, Rncw? (Hint: the resistivity of the wire material doesn’t change. How do you determine the resistance from the resistivity?) (10 pts) R,= 100 kQ A B C | 'l Place your answers here. a.) KJR (B): _(_)_3 5—._.——- :le/lru) Lolfj Ao\& An»: Loki [lolgL KLR(ABEFA): (6) KLR(BCDEB): _LJ_4 b.) 11 = —_LJ_4 I» = r41 .2 Lo LII-mu) Acid "--"" ‘__—_- Adlaﬂ“ Lol\$— Anew) D.) ueterrnlne me currents 11,13 anu l3. [11.pr c.) How much power does battery V; supply, P2? (5 pts) d.) Resistor R2 consists of a wire that is hanging out doors. Ice accumulates on the wire and starts to stretch it. If the length is increased by a factor of 4 and the diameter of the wire is decreased by a factor of two, what is the new resistance of this wire, Rnew? (Hint: the resistivity of the wire material doesn’t change. How do you determine the resistance from the resistivity?) (10 pts) A R1=100k£2 V1 12.0 V 07) TL. +11, :15. o Lms Agar/i: —1‘.’Ri — V2. + V. Loop BQDEﬁ 1 13RL+ Va. :0 n 0 Place your answers here. a.) KJR (B): —_(_)_3 KLR(ABEFA)J I61 KLR(BCDEB)I (4;)— I] = —_(_)_..4 ...
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spring 04Exam2_Solutions - Problem 1(30 points You are...

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