Electric circuit Watch
1. Since the galvanometer doesn't deflect, the voltage XN = the voltage of the galvanometer's cell. Suppose these two voltages are 8 V
2. The voltage along the potentiometer drops linearly from max at Y to 0 at X.
3. Suppose the battery at the top provides 15 V and that the drop across the variable resistor is 5 V. Then the drop across the pot. is 10 V.
4. The distance YN must give us a drop of 2 V so YN must be 2/10 = 1/5 of the length of the pot (by point 2 above).
5. We now increase the variable resistance so we get a drop of 6 V across it.
6. ... can you do the rest?
- Study Helper
a) The resistance of the wire in series with the variable resistor forms a potential divider. Ignoring the galvanometer reading for the moment, what happens the p.d. across the wire between XY if the variable resistor increases?
i.e. use the potential divider equation to do this.
b) Now, by sliding the resistance-wire contact, if the p.d. developed across the resistance-wire is the same as the p.d. across the terminals of the galvanometer battery, no current will flow through the galvanometer which will then read zero.
i.e. the contact point with the wire itself forms a potential divider between points XY. By sliding the contact, the point where the p.d. of the single cell battery and the p.d. across the XY potential divider is the same, will produce no p.d. across the galvanometer and hence it will read zero.