Saved Bookmarks
| 1. |
Draw the circuit diagram of a Wheatstone bridge. Derive the balancing condition for the same. Name a device which works on the principle of Wheatstone bridge. |
|
Answer» Solution :This networkis used to determine the VALUE of an `R_1,R_2 ,R_3` and `R_4` connected in the form of a quadrilaterial . A cell of emf E is connected between A and `C_1` whilea galvanometer of resistance G is connected between B and C. If the resistances are adjusted such that the current through the galvanometer is zero, the NETWORK is said to be in a balanced condition. Using Kirchhoff.s law At node B, `rArr I_2= I_g+ I_4` or `I_4=I_2-I_g` At node D , `rArr I_1 +I_g =I_3` an `I_3=I_1 + I_g` Apply in a Kirchhoff.s voltage law for the loop ABDA, `I_2R_2 + I_gG-I_1R_1=0` ......(1) Apply in a Kirchhoff.s voltage law for the loopBCDB, `I_4R_4-I_3R_3-I_gG=0` Substiuting for `I_4` and `I_3` in then EQUATIONWE get , `(I_2-I_g)R_4-(I_1+I_g)R_3-I_gG=0` `I_2R_4-I_gR_4-I_1R_3 -I_gR_3-I_gG=0` ...(2) When the network is balanced the current through the galvanometer is zero i.e.,`I_g=0` Apply Ig=0 to the equations (1) and (2) we get `I_2R_2-I_1R_1=0` or `I_2R_2=I_1R_1` ....(3) `I_2R_4 -I_1R_3=0` or `I_2R_4=I_1R_3` ...(4) DIVIDING the equation (3) by (4) , equation `(cancelI_2R_2)/(cancelI_2R_4)=(cancelI_1R_1)/(cancelI_1R_3)` `R_2/R_4=R_1/R_3` A practical device using this principal is the meter bridge . 
|
|