✔ 最佳答案
After switch S is closed, resistor R2 comes into effect. The equivalent resistant, say R', of R1 and R2 in parallel must be smaller than R1. Hence, the total resistance of the circuit (R' + R3) is now smaller than that before S is closed (R1 + R3). The current delivered by the battery thus increases. Since voltage across R3 = current x R3. An increase of current leads to an increase of voltage across R3. Because one end of R3 (i.e. your point A) is earthed (always maintained at zero volt), the potential of point Q needs to decrease.
emf of batter = voltage across the (R1, R2) combination + voltage across R3,
The emf of the battery is fixed, an increase of voltage across R3 thus leads to a decrease of voltage across the (R1, R2) combination. Again, because one end of the resistor combination is always maintained at zero volt, potential at the other end (point P) needs to decrease in order to achieve a drop in voltage.
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Your explanation is not convincing. You did not give sufficient reason for the following statement.
"After S is closed the p.d. of A and Q increases with the e.m.f. supply of the battery."
Why "p.d. of A and Q increases"? How does it relate with "the emf supply of the battery"?
