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Unformatted text preview: (a) 2 points For correct placement of voltmeter (b) 3 points (0) For correct application of Ohrn’s law _ 8 _ 20 V
R1 + R2 30 £2
For correct value of current
1 = E A
3 For correct value of voltage across R1 V1=IR= 2
3 (Alternate solution using voltage divider) For voltage divider equation V = R1 6
R1 + R2
For correct substitution
_ 10 Q
_ 10 £2 + 20 9
For correct answer
V = 6.67 V
i. 2 points For correct answer V=0 (1 point awarded for stating that no current ﬂows) —AXIOQ=6.67V X20V Distribution
of points 2 points 1 point 1 point 1 point (Alternate points 1 point I point 1 point 2 points Distribution
of points
E & M 2 (continued) (c) (continued)
ii. lpoint Q = CV
For correct substitutions and answer 1 point
Q = lSﬂFxZOV = 300/1C (d) 2 points For correct answer 2 points
V = 0
(1 point awarded for no current or realization of new initial conditions) (6) i. 2 points For correct application of Ohm’s law and answer 2 points I: a =2ov:_2_A
1121+}?2 309 3
20V 109 (1 point awarded for ) ii. 1 point For correct substitution in energy equation and correct answer 1 point 1 2 1 I 2 2
U, =—LI =—(2 H) — A =0.444J 2 2 3
(1 point also awarded if incorrect current in (e)i. (except zero) was substituted and answer was consistent with this current.)
(i) 2 points For correct equation 2 points 5 —1(R1 +R2)—L% = 0 (1 point only awarded if one sign incorrect) One point was subtracted from the ﬁnal score for two or more wrong or absent units for parts where the answer
was given, except where the answer was zero, in which case units were not counted. ...
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 Spring '09
 Park
 Physics

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