When R 1/R2=R4/R3, the pointer of the sensitive galvanometer in the figure will be 0; In fact, the principle is simple. The sensitive current (no matter which direction) is zero, that is to say, the potential difference between the two ends of the galvanometer is zero. Let the left end of the circuit branch be a and the right end be b; The connection point of R 1 and R4 is C, and the connection point of R2 and R3 is D; The current flowing through the branch R 1 is I 1, and the current flowing through the branch R2 is I2. When the potential difference between the two ends of the ammeter is zero, Uac=Uad, that is, I1xr1= i2xr2; R 1/R2=I2/I 1 Because the current flowing through the ammeter is zero, the current flowing through R4 is I 1 and the current flowing through R3 is I2; At the same time, I 1xR4=I2xR3,-> I2/I 1=R4/R3 Therefore, the condition that the galvanometer is zero is: R 1/R2=R4/R3. This conclusion can also be transformed into: R 1/R4=R2/R3 (just a simple mathematical transformation). Note: Answering at the next higher level is a very irresponsible way, which will only make you recite the formula without understanding it.