An active two terminal network with open circuit voltage of 12V and short circuit current of 2A has been measured_____ ?

An active two terminal network with open circuit voltage of 12V and short circuit current of 2A has been measured_____ ?


(12V/2A)=6 12/(6+2)=1.5A



As shown in the figure, when 100V voltage is applied to a and B ends of the circuit, the open circuit voltage at C and D ends is 80V, then the resistance value of resistance R is_ Ω
5、 Question 5 in this week's self test


80, 40 very simple (1) when 100V voltage is applied at both ends of a and B, the open circuit voltage at both ends of C and D is 80V, which means that the voltage at both ends of R is 80V, so that the voltage at both ends of the 20 Ω (as a whole) resistor is 20V, so using the deduction of Ohm's Law (U1 / R1 = U2 / r2), we can know that the resistance of R is 80 Ω (2)



A capacitor and a resistor are connected in series on a 220 V AC power supply. Given that the voltage drop on the resistor is 120 V, what is the voltage on the capacitor?
Fundamentals of Electrical Engineering


220 V power supply voltage will not change, resistance voltage 120 V, capacitance at both ends of course is 220 V - 120 V = 100 v



As shown in the figure, R1 = 10, R2 = 120, when both ends of a and B are connected with 100V power supply, and the reading of ideal voltmeter connected at both ends of C and D is 80V, the resistance R=
Here are 22 questions


As shown in the figure, R1 = 10, R2 = 120, when both ends of a and B are connected with 100V power supply, the reading of ideal voltmeter connected at both ends of C and D is 80V, then resistance R = R2 is equivalent to wire, so the voltage measured by Voltmeter is the voltage at both ends of R, so R1 voltage is 100-80 = 20V, I = U1 / R1 = 20 / 10 = 2A, R2 = U1 / I = 80 / 2 = 40