[填空题]
The potential differlawk(qxem/uzhg0 xwsd)f* v t xj;0) i:s+r4 2enc*fttxw+e nw4; e ($V$) across a resistor is measured with a digital voltmeter to be $0.10 \; V$ . What is the percentage uncertainty in $V^2$ ? % (please only input numerical value without units and without the $\pm$ symbol.)
参考答案: 20.00
本题详细解析: The voltmeter is a digital device and so the uncertainty is the smallest value z / w.gb:s kw4;hp8gbiwi4w ;h/bg pg8w:zbs.khich it can record, $0.01 \mathrm{~V}$. This is a $10 \%$ uncertainty on the $0.10 \mathrm{~V}$ measurement. The uncertainty in $V^2$ will be twice this, giving $20 \%$.
Post time 2024-8-21 21:38:19