Answer:
25 V
Explanation:
It is convenient to use Kirchoff's current law (KCL), which tells you the sum of currents into a node is zero. The node of interest is the top left node.
The currents into it are ...
20 mA + (-5 -Vo)/(2kΩ) -(Vo/(5kΩ)) = 0
20 mA -2.5 mA = Vo(1/(2kΩ) +1/(5kΩ)) . . . . add the opposite of Vo terms
(17.5 mA)(10/7 kΩ) = Vo = 25 . . . volts . . . . divide by the coefficient of Vo
_____
You will notice that the equation resolves to what you would get if you drew the Norton equivalent of the voltage source with its 2k impedance. You have two current sources, one of +20 mA, and one of -2.5 mA supplying current to a load of 2k║5k = (10/7)kΩ. KCL tells you the total current into the node is equal to the current through that load (out of the node).
Answer:
1000 Mega Ohms.
Explanation:
One mega Ohm is equal to 1,000,000 ohms, which is the resistance between two points of a conductor with one ampere of current at one volt. The mega Ohm is a multiple of the Ohm, which is the SI derived unit for electrical resistance. In the metric system, "mega" is the prefix for 10 raised to the power of 6.
Answer:
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Answer:
V = 56.8 mV
Explanation:
When a current I flows across a circuit element, if we assume that the dimensions of the circuit are much less than the wavelength of the power source creating this current, there exists a fixed relationship between the power dissipated in the circuit element, the current I and the voltage V across it, as follows:
P = V*I
By definition, power is the rate of change of energy, and current, the rate of change of the charge Q, so we can replace P and I, as follows:
E/t = V*q/t ⇒ E = V*Q
Solving for V:
V = E/Q = 94.2 mJ /1.66 C = 56.8 mV
Answer:
The text file attached has the detailed solution of all the parts individually.
