1.

If the charge on the capacitor is 1 mC in the given circuit,then(R_(1)R_(2))/(R_(3))= . . . . . Omega .

Answer»

6
`0 . 4`
`0 . 6`
10

Solution :Charge on cap = 1 mC
`V_("capacitro")= (Q)/(C) =(1 xx 10^(-3))/(5 xx 10^(-6)) = (1000)/(5)`
= 200 V
So,current THROUGHT `R_(5)`
`I=(V)/(R)`
`=(200)/(10)`
20 A
Hence, we have following currentdistribution .
In loop BCDB,
`- 15 R_(3) + 10 xx 5 + 5R_(2) = 0 "". . . (i)`
In loop ABCDA,
` - 200 - 15 R_(3) + 50 + 250 = 0 `
`15 R_(3) = 100`
`R_(3) = (100)/(15) OMEGA`
From Eq. (i), weget
`5R_(2) = 50`
`R_(2) = 10 Omega`
and from loop ADCA,
` - 250 - 50n+ 310 - 25 R_(1) = 0 `
` 25 R_(1) = 10`
`rArr R_(1) = (10)/(25) Omega`
So, ` (R_(1) xx R_(2))/(R_(3)) = ((10)/(25)xx10)/(((100)/(15)))`
`= (15)/(25) = (3)/(5) = 0 . 6 `


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