Напряжение на источнике или ЭДС источника (т.к внутренне сопротивление равно 0 )
В
А - сила тока до включения амперметра
Η=A/Q=(Nt)/(mq)=(Nt)/(<span>ρVq)
</span>η=(75•735•3600)/(750•0,0095•44•10⁶)=198450000/313500000=0,633
η=63,3%≈63%
F=kq1q2/r^2
F = 9*10^9 * 2*10^-9 * 6*10^-9/0,2*0,2 = 108*10^-9 / 0,04 = 2700*10^-9 H
<span>1.В,2.Б,3.В, </span>
<span>4 дано:</span>
І1=5А
<span>І2=10А</span>
<span>t</span>=0,1<span>c</span>
<em><span>ε</span></em><span /><span>=20</span><span>B</span>
<span> L</span><span>-?</span>
<em><span>ε</span></em><span /><span>=- </span><span>L</span><em><span><span> </span>∆ </span></em><em><span>I</span></em><em><span>∆</span></em><em><span>t</span></em><span>, </span><span><span> </span>L</span><span>=</span><em><span><span>ε</span></span></em><span>*</span><em><span>∆</span></em><span>t/</span><em><span>∆</span></em><span>I</span>
<span>L=20*0,1/10=0,2Гн</span>
<span>5дано:</span>
<span>L=3Гн</span>
<em><span>ε</span></em><span /><span>=</span><span>15</span><span>B</span>
<span>I=50A</span>
<em><span>∆</span></em><span>t-?</span>
<em><span>ε</span></em><span /><span>=- </span><span>L</span><em><span><span> </span>∆ </span></em><em><span>I</span></em><em><span>∆</span></em><em><span>t</span></em><span /><span>, </span><span><span> </span></span><em><span>∆</span></em><span>t=L*</span><em><span>∆</span></em><span>I/</span><em><span> ε</span></em><span /><span>, </span><em><span>∆</span></em><span>t=3*50/15=10c</span>
<span> </span>
