一架数码相机,用的锂电池电压为3.6V,容量为1000mA•h(相当于用1000mA的电流释放可放电1小时,用500mA

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  • 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.