[attachmentid=125][attachmentid=125][attachmentid=125][attachmentid=125][attachmentid=125][attachmentid=125]I would be so pleased if somebody can help me with this problem presented in the attachment file
HI, Is there some figure for the resistance of the ammeter? If it's a classroom perfect ammeter, then it has no resistance, and so Vo is zero.
just use node voltage analysis node with 2 resistors = V1 node at current source = Vo KCL at Vo (Vo - V1) / 10000 + (Vo / 10000) - 0.004 = 0 2Vo - V1 = 40 Current Ix can be expressed in terms of Vo and V1 Ix = (Vo - V1) / 10000 Now u can express V1 in terms of Ix V1 = -4000Ix (note the reference of the dependant voltage source) V1 = -4000 * (Vo - V1) / 10000 2 unknowns, 2 equations.. I think this is right.. V1 = -10V, Vo = 15V
Hi, The ammeter is in parallel with the output resistor. The combination of the 10K resistor and the internal resistance of the ammeter has to be much less than 10K. Ammeters don't measure accurately if they have more than negligable resistance. Let's say 10 ohms, for an example. In parallel with the 10K resistor, the value of Rt will be something over 9 ohms. With 4 mills current, the figure for Vo will be on the order of 36 millivolts.