This document contains solutions to tutorial exercises on electronics circuits involving capacitors and other circuit elements. The first exercise involves determining currents and equivalent capacitance in a circuit with two capacitors. The second involves calculating capacitor voltage over time when a switch changes positions. The third determines capacitor voltage over time for virtually parallel capacitors driven by a voltage source. Subsequent exercises involve determining voltages, currents, and impedances in various RC, RL, and RLC circuits.

1. Esc201A: Introduction to Electronics
Tutorial #3 Solutions
Exercises
1. In the circuit shown below, determine the currents and . What is the equivalent
capacitance of the two capacitors?
Solution:
Applying KCL
I1 + I2 = os π 6
t mA

dt
tdv
C
dt
tdv
C
)()(
21 os π 6
t mA

dt
tdv
Ceq
)(
os π 6
t mA Here, eqC = C1 + C2 = 2 + 4 = 6µF
Integrating the above,

2. )(tv =
eq
6
C102
10

sin π 6
t volts, I1(t) =
3
1
os π 6
t mA, and I2(t) =
3
2
os π 6
t mA
2. For the circuit given below the switch moves from position A to B at = . Assuming
that the switch was in position A for a long time before moving to B, determine the
apa ito oltage as a fu tio of ti e. ‘epeat if esisto is kΩ a d apa ito is pF. Plot
the capacitor voltage versus time for both cases on the same plot.
Solution: At t< 0, switch is at A so the capacitance gets charged up to 4 V. So Vini = 4 V.
Afte o i g s it h to B, at t = ∞, oltage a oss apa ita e ill e Vf = 0 V (Fully
discharged). Now for t > 0,
Vc(t) = Vf + (Vini – Vf)exp(-t/τ = 0 + (4 - 0)exp(-t/ τ1)
= 4 exp(-t/τ1 olts, he e τ1= R1C1 = 8 ns
Similarly, Vc(t) = 4exp(-t/τ2 olts, he e τ2= R2C2 = 12 ns. Both voltages are plotted
in the figure above.

3. 3. The circuit shown below consists of virtually parallel capacitors being driven by a
voltage source of 5 V that is connected to the capacitors at = . Determine the
voltage �� and plot the same.
Solution:
At t=0-
,
At t = ∞
∴ �� −
= �
Under steady state DC conditions, capacitor behaves as an
open circuit. Here, both the capacitors have been connected
since t = -∞
�� ∞ = �
This is a steady state condition for
capacitors as well. So they behave as
open circuits.

4. Since, �� = �� ∞ + �� − �� ∞ exp − /�
�� = − exp − �⁄
Vi tuall pa allel apa ito s e ui ale t apa ita e is C +C so that time constant
� = � � + � = �s.
4. The circuit below represents a "forbidden combination" of elements. Explain the
unphysical nature of this circuit. If the capacitor is initially charged to voltage other than
� , will the combination be not forbidden? What if the capacitor is replaced by an
inductor?
Solution: For a capacitor, = �
�

5. This implies that voltage on a capacitor cannot change instantaneously since it will give rise
to infinite current which physically not possible. In the given circuit, the switching at t=0
induces a sudden change in voltage across C. Thus, the circuit is unphysical in nature.
If the capacitor is initially charged to some voltage other than vs, there will still be a sudden
change in voltage across capacitor which will imply infinite current. So, the circuit will be
unphysical.
If the capacitor is replaced by an inductor: Just after the switch is closed, the inductor
(uncharged previously) acts like an open circuit. The resistor will sink the entire current.
However, at steady state, inductor acts like a short circuit thus shorting the voltage source,
which is undesirable. Hence it is not advisable to connect this circuit.
Ex. 3.27: Consider the network in the Figure. At t = 0, the switch opens. Find v0(t) for t > 0.
Solution: Before t = 0, capacitor gets charged up to 4 V.
Vini = 4 V
As switch opens at t = 0, after infinite time capacitor will discharge up to 0 V.
Vf = 0 V
Now at t > 0,
Vc(t) = Vf + (Vini – Vf)exp(-t/τ he e τ = ‘C = se he e, ‘ = || + = Ω a d C = F
= 0 + (4 - 0 exp(-t/τ
Vc(t) = 4exp(-t/2) V
So, Vo(t) = 4/(2+4) Vc(t) V (applying voltage divider rule)

6. Vo(t) =(8/3) exp(-t/2) V
Ex. 3.28 Find vc(t) for t>0 in the network in Figure.
Solution: Before t = 0, capacitor will be ope i uit a d ha ged up to kΩ 2mA = 8 V =
Vini. S it h losed at t = , afte i fi ite ti e apa ito ill e dis ha ged full th ough 6kΩ
resistance only. So Vf = 0 V.
Now at t >0,
Vc(t) = Vf + (Vini – Vf)exp(-t/τ V he e τ = ‘C = .6 se he e, ‘ = 6kΩ and C = 100µF)
= 0 + (8 – 0) exp(-t/τ V
Vc(t) = 8exp(-t/0.6) V
Ex. 4.24: Determine the voltage across the current source in network in the Figure.
Solution: Overall impedance of the network is,
Z = /Y + /Y + /Y + /Y
Z = 7 + 4j
F o Oh s la V = IZ):

7. � = . ∠ . V
Ex. 4.25: Fine the current I in the network in Figure.
Solution: Overall impedance (Z) of the network can be calculated as
/Z = /Z + /Z + /Z + /Z
/Z = . + . j
F o Oh s la � = ��), � = . ∠ . A
Problems
1. In the circuit shown below, the switch is closed at = . Assuming that the
capacitor is initially uncharged and � = � is the dc voltage applied to the circuit,
determine the voltage across ��. Sketch ��as a function of time.
Solution: At t < 0, Vini = 0 V.
At t= 0, switch is closed and capacitor will start charging and it will charge up to Vs, so Vf =
Vs = V0 V.
Now for t > 0, Vc(t) = Vf + (Vini – Vf)exp(-t/τ V he e τ = ‘C and R = Ri|| Rs
v1 = Vc(t) = V0(1- exp(-t/τ ) V. Hence, VL(t) = gv1( Ro|| RL)

8. VL(t) = g V0(1- exp(-t/τ ( Ro|| RL) V
2. In the circuit shown below, the voltage � is zero for times < and for > �s.
The voltage source is connected to the RL network at = s (not shown in Figure). The
voltage � = � for ≤ < �s and then becomes � during ≤ < �s. Determine
and plot �� for > . Assume that the inductor is initially uncharged.
Solution: The circuit can be analyzed in three stages for the inductor current iL and
voltage vL.
a) for ≤ < �s: RL circuit with zero initial condition
b) for ≤ < �s : RL circuit with initial condition( initial current iL
−
. )
c) for �s ≤ : a RL circuit with initial condition ( source removed & initial current iL
−
a) for ≤ < �s
� = , � ∞ =
�
For the circuit � =
�
= 1 µsec.

9. For t>0, � = � ∞ − ( � ∞ − � )�−
�
�
=
�
− �− /�
�� = � =
��
�
�− /�
��
−
= �−
= . ��
At t=2µsec, �
+
= �
−
= −
− �−
= . �
b) for � ≤ < �s :
Since the current through the inductor cannot change instantaneously, so the voltage
across inductor will change to compensate the same, after applying V2=5 volts,
� =
�
�
−
�
�
− �
+
�− /�
�� =
��
��
�− /�
−
�
� �
+
�− /�
��
+
= − . ��
Similarly,
��
−
= � �−
− . × �−
= − . ��
c) For t> 4 µsec :
At t=4µsec, �
+
= �
−
=
�
−
�
− . × −
�−
�
� = . �
Also since the source is removed, � ∞ = , which gives,
�� = −
�
� �
+
�− /�
��
+
= −
�
�
× �
+
�−
�
� = − . volts

10. 3. Determine the equivalent impedance Z for each of the circuits shown below.
Solution:
Equivalent impedance Z =
jωC
|| R + jωL =
+ ��
+ � �−� ��
Substituting values we obtain Z = . ∠ . Ω
Equivalent impedance Z = � + �� +
+ ��
+ ��
= . + . = . ∠ . Ω

11. Equivalent impedance Z = � +
��
− � ��
+
R
+j��
= � +
��
− � ��
+
R −j��
+ ��
= . × – .1419 = . × ∠ − . Ω
Equivalent impedance Z = 6 Z1 || 8Z1 =
Z × Z
Z
= . �
=
.
+j
= . ∠ − Ω
Q4. For the circuit shown below, dete i e the oltage vc . Assu e ω = , ad/s.
Solution:
At node A,
∠ −
A B

12. ∠ − =
��
+ �� ; �� = ∠ −
and voltage across 2�� capacitor
� = �� ×
�
= × ∠ − ×
× 4× × −6
= − ∠ − = ∠ −
� = �� � −
Q5. The circuit shown below is operating in the sinusoidal steady state. The voltage across
the capacitor is given by �̅ = ∠ . o
at ω = ad/s. Dete i e i(t) and v(t).
v(t)
i(t)
v1(t)
3Ω
3Ω-j2.5
j6
Solution: From KVL � = + � and KCL at node A, =
�
− .
+
�
+
=
∠ .
− .
+
∠ .
+
; = . ∠
From KVL we obtain � = × . ∠ + ∠ . = . ∠ .
A