Physics S.O.S
1.
2.
3.
4.
Electric Current and Direct Current Circuit| 3
Power and Electrical Energy
Terminal Voltage
Kirchhoff’s law
Potentiometer & Wheatstone Bridge
Theory
Electrical energy
Work done to move the charge
from one terminal to the other
W=qV
E =IVt
Electrical power
Rate of work done or rate of
energy transfer
P=
Power and Electrical Energy
Formula
Electrical Energy
E = IVt
Examples
Electrical Power
P = IV
P= I2R
P=
Power dissipated by a resistor
I = Current through the resistor
V = voltage across the resistor
R = Resistance of the resistor
1. Energy delivered to the
circuit in 1 minute
2. Power delivered to the
circuit
3. Power dissipated by the 4Ω
resistor
480 J,8W,1.77W
Terminal Voltage
V = ε - Ir
VT = V supplied
1. Current in the circuit
2. Determine the terminal
to the circuit
V = ε - Ir
Kirchhoff’s Law
Junction Rule
∑
1. Draw the direction of
current at the junction
2. Write the equation for the
junction rule
3. Draw the direction of e.m.f
on the battery.
∑
Loop/voltage Rule
∑
Step
∑
voltage of the battery
0,75A,4.5 V
4. Draw the loop, choose any
direction you like
5. Write down the equations
for both loops.
6. Solve the equation.
Loop = current & e.m.f
1. Current in 3Ω resistor
2. Total current in the circuit
3. Terminal voltage of the
battery
4. Internal resistance of cell
5. Power dissipated by 5Ω
6 A; 9 A; 63 V; 1;405 W
Physics S.O.S
Potential Divider
Electric Current and Direct Current Circuit| 3
 R1 
V
V1  
 R1  R2 
 R2 
V
V2  
 R1  R2 
 l 
V1   1 V
 l1  l2 
 l 
V2   2 V
 l1  l2 
Determine unknown e.m.f
Given that R1 = 6Ω and R2 = 8Ω
and e.m.f of the cell is 28 V.
Determine the voltage across
6Ω.
12V
Determine the internal
resistance of a cell
Determine the voltage across 8Ω
V8Ω = 5.33 V
Potentiometer
Determine the voltage across
30cm.
V30cm =2.16 V
Wheatstone Bridge
R1 R2

R3 Rx
Rx
R

LAL LJB
Determine:
1.
2. Potential difference across
the 6.7  resistor,
3. Power dissipated from the
1.2  resistor
I1  0.72 A; I 2  1.03 A; I  1.75 A
V  6.90 V
P  3.68 W
The wire has a length 100 cm and
the emf of the cell is 6 V ,Determine
the balance length
L = 50 cm
If the cell is replaced with another
cell, and the new balance is
achieved at 60 cm, determine the
e.m.f of the cell.
e.m.f. = 6.2 V
A 6Ω resistor is connected to the
circuit as shown above, Determine
1. New balance length
2. Internal resistance of the cell