when the two resistance are connected in series their effective resistance 40 ohms and the same resistance are connected is parallel then the effective resistance is 10 ohms find the value of individual
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r = r1+ r2
40-r2= r1........(1)
1/rp= 1/r1 + 1/r2
1/10 = 1/40-r2 + 1/r2
1/10 = r2+40-r2/ 40r2-r4
40r2 - r4 = 400
r4-40r2 + 400=0
use and solve it
40-r2= r1........(1)
1/rp= 1/r1 + 1/r2
1/10 = 1/40-r2 + 1/r2
1/10 = r2+40-r2/ 40r2-r4
40r2 - r4 = 400
r4-40r2 + 400=0
use and solve it
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