in the question it states the voltmeter is non ideal and has a resistance R, so u can basically assume it’s another resistor. therefore the circuit isn’t exactly series, and the total resistance would include the parallel combination of the voltmeter and 320 resistor + the 640 resistor. But since we don’t know the resistance of the voltmeter, we need to calculate the current in the 640 resistor as this will be the total circuit current.
hope that helped.
hope that helped.
Original post by user123569648
in the question it states the voltmeter is non ideal and has a resistance R, so u can basically assume it’s another resistor. therefore the circuit isn’t exactly series, and the total resistance would include the parallel combination of the voltmeter and 320 resistor + the 640 resistor. But since we don’t know the resistance of the voltmeter, we need to calculate the current in the 640 resistor as this will be the total circuit current.
hope that helped.
hope that helped.
So the current in the 640 resistor will be the same as the current through the voltmeter and 320 resistor?
Thanks!
Original post by MahmoodK
So the current in the 640 resistor will be the same as the current through the voltmeter and 320 resistor?
Thanks!
Thanks!
The current in the 640 resistor will have the most current, this current is split into the 320 and voltmeter branches. So if u think of the voltmeter as a resistor instead, it will have some current through it too, and the 320 will have the rest of the current, then they add together to make the total current, which will be the current in the 640 as the 640 part of the circuit is unbranched.
Original post by user123569648
The current in the 640 resistor will have the most current, this current is split into the 320 and voltmeter branches. So if u think of the voltmeter as a resistor instead, it will have some current through it too, and the 320 will have the rest of the current, then they add together to make the total current, which will be the current in the 640 as the 640 part of the circuit is unbranched.
How would you do BC
(edited 1 year ago)
Original post by MahmoodK
How would you do BC
I assume that you have computed the p.d. between A and B (VBA) and the p.d. between A and C (VCA). Since A is the “common point” for the two p.d.s, the difference between the 2 p.d. is the p.d. between B and C.
Please restrict to one problem per thread. If you have a new problem, start a new thread for the problem. Thanks.
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