As shown in the figure, the power supply voltage remains unchanged. When the switch S is turned off, the reading of V1 is 5V and V2 is 7V. After the switch is closed, the reading of V1 is 9V and V2 is

As shown in the figure, the power supply voltage remains unchanged. When the switch S is turned off, the reading of V1 is 5V and V2 is 7V. After the switch is closed, the reading of V1 is 9V and V2 is 5.4V.

I remember this saying, treat the voltmeter as a broken circuit and the ammeter as a wire. When the switch S is open, the three lights are connected in series. When S is closed, it is equivalent to short-circuiting L3. L1 and L2 are connected in series. After analyzing the opening and closing of S, V1 and V2 each show whose voltage it is. When S is turned on, the reading of V1 should be the voltage of L1 and L2. The V2 meter should be the reading of L1 and L3. When S is open: the series voltage divider U1 U2 U3=U total, (U1 U2) (U2 U3)=5V 7V; when S is closed, the display of V1 is still that of L1 L2, and the display of V2 is only that of L1. When S is closed: U1 U2 =9V, U2=5.4V. The voltage of the power supply remains unchanged, so when S is closed, the voltage of L1 and L2 is the total voltage of 9V. The final answer is: (1) When the switch is turned off, the voltage of L1 is =2V, the voltage of L2 is =3V, the voltage of L3 is =4V, and the total voltage =9V. (2) After closing, the voltage of L1 is =3.6V, the voltage of L2 is =5.4V, and the voltage of L3 is =0V