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...表的示数分别为I1、I2、I3,则有( )A.I1=I2+I3B.I1=I2=I3C.I1>I
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That will give you 2 equations for I1, I2, I3. The third equation is I1=I2 + I3, the current entering the top node has to equal the two currents leaving it. One of these three equations will be a first order differential equation (DE) in I3. Use the other two equations to eliminate I1 and I2 from this DE and then solve the DE.
如图所示电路中，电流表a1、a2、a3的示数分别为i1、i2、i3，它们的大小关系是（ ）a．i1=i2=i3b．i1＞三个灯泡为并联关系，l1与a2串联，表a2测的是l1上的电流；而l1与l2并联后电流流经a3，表a3测得是灯l1、l2的电
A priority encoder has four inputs I0 , I1 , I2 and I3 where I3 has the highest priority and I0 has the least priority. asked Mar 1 in Computer by DeepikaBaghel (29.5k points) bpsc; bpsc 2024 +1 vote. 1 answer (a) State all the energy levels of a symmetric top with principal moments of inertia I1 = I2 = I I3. ...
正确答案： b i1＞i2＞i3 由图知，三盏灯泡并联，a1测通过三盏灯的电流和，a2测通过l1和l2的电流，a3测通过l3的电流；所以a1的示数最大，其次是a2的示数，a3的示数最小．故选b．
根据并联电路中干路电流等于各支路电流之和可知：i1=i2+i3.由此可知：i1>i2,i1>i3,则i1=i2=i3,i1
如图所示，开关闭合后，三个电流表a 1 、a 2 、a 3 的示数分别为i 1 、i 2 、i 3 ，它们的大小关系是（ ） a．i 1 =i 2 =i 3 b．i 1 =i 2 +i 3 c．i 1 ＞i 2 ＞i 3 d．i 1 ＞i 3 ＞i 2
①. d ②. 能【详解】（1）由图可知，电流表a1测量的是总电流i1，电流表a2测量的是通过l1的电流i2，电流表a3测量的是通过l2、l2的电流i3，所以
Question: Using the ideal op - amp assumptions, determine I1,I2, and I3 in Fig. P4. 12. Show transcribed image text. There are 2 steps to solve this one. Solution.
解答：解：由电路图可知，两灯泡并联，电流表a 1 、a 2 分别测两支路电流，电流表a 3 测干路电流， 由并联电路特点可知，i 3 =i 1 +i 2 ，则i 1 =i 3-i 2 ，故ab错误，c正确；因为不知道两灯泡间的电阻关系， 因此不知道电流表a 1 、a 2 示数间的关系，故d错误； 故选c．
Using mesh analysis, find the currents I1, I2, and I3. Show transcribed image text. There are 2 steps to solve this one. Solution. 100 % ...