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In the circuit in fig. If no current flows through the galvanometer when the key k is closed, the bridge is balanced. The balancing condition for bridge is A. `C_(1)/C_(2) = R_(1)/R_(2)`B. `C_(1)/C_(2)=R_(2)/R_(1)`C. `(C_(1)^(2)/C_(2)^(2)=R_(1)^(2)/R_(2)^(2))`D. `C_(1)^(2)/C_(2)^(2)=R_(2)/R_(1)` |
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Answer» Correct Answer - B b) In the steady state, no current is passing through capacitor. Let the charge on each capacitor be q. Since, the current through galvanometer is zero. `therefore I_(1)=I_(2)` The potential difference between ends of galvanometer will be zero. `therefore V_(A)-V_(B)=V_(A)-V_(D)` `therefore I_(1)R_(1)=q/C_(1)` similarly, `V_(B)-V_(C) = V_(D)-V_(C)` `I_(2)R_(2)=q/C_(2)` ..............(ii) On dividing Eq. (i) by Eq. (ii), we get `therefore (I_(1)R_(1))/(I_(2)R_(2))=(q//C_(1))/(q//C_(2))= C_(2)/C_(1)` `C_(1)/C_(2) = R_(2)/R_(1)` |
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