# Chapter 19 DC Circuits

```Chapter 19
DC Circuits
EMF and Terminal Voltage
• Devices which transfer one type of energy to electrical
energy are called the sources of electromotive force (EMF).
• The potential difference between terminals of such source
when no current is flowing to the external circuit is called the
emf of the source ( Ɛ).
Remember, A battery is not a constant current source but can
be assumed to be a constant voltage source.
When electron is flowing within the battery, it faces some
resistance due to the molecules present on it path. Hence, it is
fair enough to say that every battery has its own resistance,
which we call the Internal resistance (r).
Hence if we use a multimeter to
measure the terminal voltage,
i.e., voltage across the battery
terminals, we would never
measure the emf of the cell.
=  −
Unless otherwise stated, the
internal resistance is considered
to be negligible.
Problem 1:
From figure below, calculate
current, terminal voltage, power
dissipated in ‘R’ and in ‘r’.
Resistors in Series Connection
- Same current flows through each
resistors.
- The sum of the voltage across
each of the resistor is equal to
the total voltage of the battery.
= 1 + 2 + 3 = 1 + 2 + 3
For the equivalent circuit, Ohm’s Law
yields:
=
Hence, comparing the two equations,
we get
=  +  +
Equivalent Circuit
Resistors in Parallel Combination
- Resistors are connected so that the
current splits
= 1 + 2 + 3
- Voltage across each resistor is the same.
Hence,
= 1 1 = 2 2 = 3 3
=

1
+

2
+

3
=
1
1
+
1
2
+
Also, in the equivalent circuit, I =
Thus,
1
1
1
1
=
+
+
1 2 3
1
3
…. (1)

… (2)
Problem 2:
Determine (a) equivalent resistance and (b) the voltage across
each resistor, and (c) current through each resistor.
Solution:
help me !!
a) 357 + 990 Ohm
b) 8.8, 3.2 V
c) 8.91, 4.7,4.2 mA
Remember:
Series - Current same, voltage add up
Parallel - Voltage same, current add up
Problem 3:
Find net resistance of the circuit. Find current through each resistor
if R= 3.25k- Ohm and V=12V. Also what is the potential difference
between point A and B.
Solution:
a) 5/8R + R
b) I1=I2= 0.284mA
I3= 0.568 mA
I4= 0.852mA
I5= 1.42mA
I6=2.27mA
c) 1.85V
Kirchhoff’s Law:
- Used to solve circuits with multiple
loops and voltage source.
Law of current:
At any junction, the sum of all currents
entering must be equal to sum of all
currents leaving.
For the circuit above:
1 + 2 = 3 + 4 + 5
Law of Voltage:
The sum of the changes in potential
around any closed loop of a circuit must
be zero.
Problem 4:
Calculate the magnitude and direction of the currents in each resistor.
Approach:
1) Choose current direction with some thoughts
2) Write current equation at the junctions
3) Write voltage equation for the loop
56 Ω: 0.16 A
110 Ω: 0.090 A
74 Ω: 0.25 A
25 Ω: 0.090 A
Capacitors in Series and Parallel
=
Parallel:
• Total charge gets split so that
= 1 + 2 + 3
= 1  + 2  + 3
So,  =  +  +
Series:
• Same charge flows
• Voltage splits
= 1 + 2 + 3

= + +
1 2 3
So,

=

+

+

Problem 5:
Calculate (a) Voltage across each capacitor (b) charge on each capacitor.
Approach:
(a) Q = CV
- Voltage across C3 = 21 V
- Voltage across C12 (C1 series to C2)= 21 V. Since Q is
the same in series:
C1V1=C2V2 with V= V1 + V2
So,
3V1 = 4(21-V1)
V1 = 12 V , and V2 = 9V.
(b)
Charge across C3, Q3= C3V=2x21
Charge across C1, Q1 = C1V1 = 3x12
Charge across C2, Q2 = C2V2 = 4X9
RC Circuits:
- Car windshield, Pacemaker,
camera flashes
Charging:
As switch is turned on, electron
moves from -ve terminal of the
battery to one of the capacitor
plate, which incudes +ve charge
on the other plate. A –ve free
charge on the second plate
moves to the +ve terminal of the
battery. The charge continues to
deposit on the capacitor until
Ɛ = Vc.
=  1
−
−
Since Q= CV, above equation can
be written as:
−

= 0 1 −
=  (Time Constant)
Time constant is the measure of
how quickly capacitor gets
charged.
- e is exponential function, it’s
value is 2.71. For t = RC, we
can see Vc = 0.63.

Also, Ɛ = Vc + VR . So, VR =   −  .
Above figure shows this variation.
RC Circuit Discharging: Read Yourself !!!
Problem 6:
Total Resistance is 15 K Ω, emf 24 V. If the time
constant is measured to be 18 μs, calculate (a)
total capacitance and (b) time it takes for voltage
across the resistor to reach 16V after the switch is
closed.
Solution:
= .
18 x 10-6 = 15 x 1000 x C
So, C = 1.2 x 10-9 F.
---------------------------------------------
−
During Charging: VR =    .
− 18×10−6

Hence, 16 = 24
Divide both sides by 24, and take log:
ln(0.66) = -t/18x10-6
-0.415 x 18x 10-6 = -t
Hence, t = 7.48 x 10-6 s.
```