已知1000mA=1A,1h=3600s,n=400,n′=150,
那么锂电池可以做的电功为W=UIt=3.6V×1A×3600s=12960J,
则关闭液晶屏时,每拍一张照片所消耗的电能W 1=
W
n =
12960J
400 =32.4J,
打开液晶屏时,每拍一张照片消耗的电能W 2=
W
n′ =
12960J
150 =86.4J,
液晶屏消耗的电能W 3=W 2-W 1=86.4J-32.4J=54J.
故答案为:32.4J,54.0.
已知1000mA=1A,1h=3600s,n=400,n′=150,
那么锂电池可以做的电功为W=UIt=3.6V×1A×3600s=12960J,
则关闭液晶屏时,每拍一张照片所消耗的电能W 1=
W
n =
12960J
400 =32.4J,
打开液晶屏时,每拍一张照片消耗的电能W 2=
W
n′ =
12960J
150 =86.4J,
液晶屏消耗的电能W 3=W 2-W 1=86.4J-32.4J=54J.
故答案为:32.4J,54.0.