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(1) Let the tolerance of the equal difference series be d, then.
a3=a1+2d=10, s11=11a1+11 10 2d=11, the solution is a1=16, d=-3
The general formula for the series is: an=a1+(n-1)d=16-3(n-1)=19-3n
2) From (1) know an=19-3n, a1=16
Therefore sn=n(a1+an) 2=(19-3n+16)n 2=-3 2n 2+35 2n=-3 2(n-35 6) 2+1225 24
This parabolic opening is downward, and when n = 35 6, the sn is maximum. But because n is a positive integer, when n=5, sn=50;When n=6, sn=51>50
The maximum value of sn is 51, and n is 6
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3. Solution: When the voltage source acts alone, the current source is open, as shown in the figure below
u'ab=10×6/(6+6)=5(v)。
When the current source acts alone, the voltage source is short-circuited, the left of the figure below, and the equivalent is redrawn to the right of the figure below
u"ab=2×6∥6=2×3=6(v)。
Superposition theorem: UAB=U'ab+u"ab=5+6=11(v)。
4. Solution: There are only two closed loops in the circuit: 2a i - 10v - 4.
So: i=2a, umb=6 i=6 2=12(v).
uam=10+4×3i=10+4×3×2=34(v)。
Therefore: uoc = uab = uam + umb = 12 + 34 = 46 (v).
Short-circuit the voltage source and open the current source. Apply voltage U to ports A and B, and set the incoming current to I. Obviously: i=i. Below:
According to KVL: U=4 (3I+I)+6 I=12I+4I+6I=22I.
So: req=u i=22( )
That is, the equivalent circuit parameters of Thevenin are: UOC=46V, REQ=22.
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(1) Let the tolerance of the equal difference series be d, then.
a3=a1+2d=10, s11=11a1+11 10 2d=11, the solution is a1=16, d=-3
The general formula for the series is: an=a1+(n-1)d=16-3(n-1)=19-3n
2) From (1) an=19-3n, a1=16, so sn=n(a1+an) 2 =(19-3n+16)n 2=-3 2n 2+35 2n=-3 2(n-35 6) 2+1225 24
This parabolic opening is downward, and when n = 35 6, the sn is maximum. But because n is a positive integer, when n=5, sn=50;When n=6, sn=51>50
The maximum value of sn is 51, and n is 6
-
A, 6 (two) + 10 (two) = 1 catty.
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