SOLVED: E110: Kirchhoff's Rules For numbers 22 to 24, consider the circuit below In this circuit VDc1 = 11 V; VDcz = 5 V Ri 30 Q, Rz = 20 Q and R3 = 10 Q Complete the table of the missing information Computed Ii 1z I3 IDc1 Ipcz (22) A (23) A (24) A 2.09 A 0.27 A

您所在的位置:网站首页 vdcz SOLVED: E110: Kirchhoff's Rules For numbers 22 to 24, consider the circuit below In this circuit VDc1 = 11 V; VDcz = 5 V Ri 30 Q, Rz = 20 Q and R3 = 10 Q Complete the table of the missing information Computed Ii 1z I3 IDc1 Ipcz (22) A (23) A (24) A 2.09 A 0.27 A

SOLVED: E110: Kirchhoff's Rules For numbers 22 to 24, consider the circuit below In this circuit VDc1 = 11 V; VDcz = 5 V Ri 30 Q, Rz = 20 Q and R3 = 10 Q Complete the table of the missing information Computed Ii 1z I3 IDc1 Ipcz (22) A (23) A (24) A 2.09 A 0.27 A

2023-04-01 13:23| 来源: 网络整理| 查看: 265

Transcript

Welcome to this numera tetorial. So if the data set pertains to a series circuit, then the the current will be constant and then the total resistance of all resisters will just be added to 60 s total resistance. We then have a total current between both dc current sources of 2.36 amps, so this means the current across i 1, i 2 and i 3 will remain constant at 2.36 amps. So i could redraw this so the batteries are connected in series. So this means the amperes will obey a cumulative are added to function while the voltage can fluctuate across the resisters. But that's not what the questions asking it's asking for the current values at r, 1, r, 2 and r 3. So that's really! I r 1! I r 2 and i r 3, and it's just 2.36 amps across all 3 resistors. The data set pertains to a series circuit. Now. If the data set pertains to a parallel circuit, the voltage is going to be constant. So then we have to combine or add or combine the resisters based on the inverse some rule to all. You do take 1 over 30 ohms plus 1 over 20 s plus 1 over 10, ms and then this entire expression 1 over the expression. So you just take it to the negative 1 power, so you just add 1, he 38 plus onde, 28 plus 110, and then whatever answer you get, take it to the negative 1 power. So it's 1 over the fraction addition sum, and so this will then compute the resisters when combined in parallel it. If the resisters are in serious, we just add them…



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