circuit analysiscurrentresistors
I'm having trouble with this circuit. I'm trying to find the current through and voltage drop across the resistors when the switch is open and closed.
When the switch is open, I've found that the voltage drop across the upper resistor and lower resistor is 10V each. I'm not sure how to deal with the second current source when the switch is closed.
Any help is appreciated!
If the sources are ideal, you don't have to put any internal resistance on them. Also, I think you are mistaken in assigning a resistance value at the voltage source. It is true that you have a voltage and a current, but the voltage is constant while the current depends on the circuit.
In order to solve the problem, you correctly found the current on the voltage source, and since the text says that it produces 3.333W, it must be the inverse of the arrow.
Therefore, if you apply the KCL at the upper node, you get that \$I_2 = I_S - I_1\$, but since I1 is negative you have to sum the absolute values. I actually don't now how you got the 23/6 A value.
Since you know the voltage on Is, you can calculate the voltage on the two resistances, and since you know all the currents you can obtain the resistance values.
Follow the arrows below. The first transition is due to making a Thevenin equivalent out of \$V_1\$, \$R_1\$, and \$R_2\$. The second transition is just summing the remaining two resistors:
simulate this circuit – Schematic created using CircuitLab
There will therefore be about \$600\:\textrm{mV}\$ subtracted away from the Thevenin voltage, leaving \$5\:\textrm{V}\$ across the total resistance of \$10\:\textrm{k}\Omega\$. About \$500\:\mu\textrm{A}\$.
Best Answer
Q) What do you know about current sources?
A) That the current at their terminals is always the same as the number written beside them.
Q) So what is the current through the top 1k resistor, whether the switch is open or closed?
A) As it's connected only to the 10mA source, it's always 10mA.
Q) So where does the extra 5mA flow when you close the switch?
A) The only place it can flow ...