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In Fig, `E =5` volt , `r = 1 Omega, R_(2) = 4 Omega, R_(1) = R_(3) = 1 Omega` and `C = 3 muF`. Then the numbercal value of the charge on each plate of the capacitor is A. 3`muC`B. 6`muC`C. 12`muC`D. 24`muC` |
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Answer» Correct Answer - 2 At steady state, No current will flow through capacitor, So I = `E/(R_(2)+r)=5/(4+1)=1A` Potential difference a cross capacitor = `ixxR_(2)` `=1xx4=4v` Charge on each capacitor =`(C/2)xx4` `=(3/2)xx4=6muC` |
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