Current through R3 I=UABR? ②
Substitute ① and ② into the data, I= 1 A?
The current through R 1 I1= I2 = 0.5a. 。
(2)at 1× 10-2-2× 10-2？ S time d is in the off state, then the current flows through R3.
I′= ur 1+R3
Substitute data, I' = 0.8a.
Electric energy consumed by R3 within t =1se =12 (i2r3+I '2r3).
Substituting the data gives e = 4.92 j.
Answer: (1) The current through R 1 is 0.5ain0 ~1×10-2s; (2) The power consumed by resistor R3 in 2) 1s is 4.92J J.