You are given Circuit 1 with unknown values R1, R2, R3, and C1. You cannot access the nodes
within the dotted box. Your job is to solve for their values using measurements from the
accessible nodes: a, b, and c.
You are given the following measurement:
A) When a 10 V source is connected across nodes a and b and node c is left unconnected, the
current drawn from node a through node b is 10/3 mA.
B) When a 10 V source is connected across nodes a and c and node b is left unconnected, the
current drawn from node a through node c is 5/2 mA.
C) When a 10 V source is connected across nodes b and c and node a is left unconnected, the
current drawn from node b through node c is 2 mA.
D) A 50 Hz square wave which goes from 0 V to 4 V is connected to nodes a and c. This
configuration is shown below in Circuit 2.
The plots for the voltage over time of the input square wave (Va) is given in blue, and the plot
for the output waveform (Vb) is given in orange in Figure 1 (an enlarged version is given in
Figure 2). Plots are given below. Both plots are with respect to ground being node c. Assume the
square wave is generated by an ideal source.
Determine the values of R1, R2, R3, and C1.
Hint 1: determine R1, R2, and R3 using measurements A, B, and C first.
Hint 2: Make sure the R in your R-C time constant is the equivalent R.
edit :
Link to the graphs above :
http://i62.tinypic.com/ek26it.jpg
http://i58.tinypic.com/103v08w.jpg
Solution
(1) From th efirst case A, if we redraw the circuit and aplly kcl

we will get an equation i.e, R1+R3=3000.................................(1)
From the second case (b) as above if we redraw and aplly kcl
we will get an equation i.e, R1+(R2//C)=4000..............................(2)
From the third case (c)
we will get equation i.e, R3+(R2//C)=5000.................................(3)
FROM THE EQUATION (2) R2//C=4000-R1
substitute this value in (3)
then we will get R3+(4000-R1)=5000.............................(4)
SOLVING (1) (4) WE GET values of R1 andR3 as
R1=1000
R3=2000
FROM EQUATION (3) SUBSSTITUTE VALUE OF R3
then the value of (R2//C)=3000
C=3000R2/(3R2-3000).................................(5)
NOW APPLY KCL IN CASE (B) CIRCUIT
i.e, -10+(0.002*R3)+(0.002*R2//C)=0
substitute value of R3 obtained and c from (5)
then R2=1000
FROM EQUATION (5) C=1.5F
(for the circuits figure to observe go through this pic i.e,http://i59.tinypic.com/xpy0z9.jpg
(2) FOR THE CASE D TO SOLVE THE GRAPHS GIVEN ARE NOT CLEAR AND THE
NUMERICAL VALUES ARE NOT VISIBLE