Unit 6 DC Circuit Theory/Resistor Networks & Kirchhoff’s Laws
Assignment 1
From my diagram of the circuit; this is a series circuit with 2 parallel branches, I have
chosen values for my 5 resistors & for my cell. The values are:
- R1 - 100Ω
- R2 - 10Ω
- R3 - 22Ω
- R4 - 470Ω
- R5 - 330Ω
- V1 - 10V
The first part of this section was to work out what the total resistance (Rt) of the
circuit was; this was simple as you have to work out Rt in each branch. The first
branch consists of R1 & R2, & the second branch contains R3, R4 & R5.
To workout the Rt in a parallel circuit, the formula is
1
Rt = (
1
+
1 1
+ )
R 1 R 2 Rn
100
With this formula you sub in the values of R1 & R2 & you will get an answer of
11
Ω.
1
100
+ ) = 11 Ω
1 1
(
100 10
Next, you work out the Rt of branch 2, which is exactly the same method as before,
but you sub in the values for R3, R4 & R5.
1
3102
1 1 1 = Ω (19.76Ω)
( + + ) 157
22 470 330
Now that you have your results, you can simply re-draw the circuit as a series circuit
& add the 2 values together, which will give you the total Rt of the whole circuit.
100 3102
Rt = Ω+ Ω = 28.849Ω
11 157
Moving onwards with the next question; this is finding out the total current (It) in the
circuit. To work out the It you must know the Ohms law equation, which I have stated
Volts(v)
in my work sheet. So, we know that Current (I) =
Resistance ( Ω)
As we already worked out the Rt from the previous question, we can use that value, &
for voltage, we can use the value that I had selected for my circuit, which is 10V.
8635
Now we just sub in our values to find It, & that’ll give us Amps (A)
24911
10 8635
It = 28.849 = 24911 A (0.35A)
P1 Fahim Mohammed
Assignment 1
From my diagram of the circuit; this is a series circuit with 2 parallel branches, I have
chosen values for my 5 resistors & for my cell. The values are:
- R1 - 100Ω
- R2 - 10Ω
- R3 - 22Ω
- R4 - 470Ω
- R5 - 330Ω
- V1 - 10V
The first part of this section was to work out what the total resistance (Rt) of the
circuit was; this was simple as you have to work out Rt in each branch. The first
branch consists of R1 & R2, & the second branch contains R3, R4 & R5.
To workout the Rt in a parallel circuit, the formula is
1
Rt = (
1
+
1 1
+ )
R 1 R 2 Rn
100
With this formula you sub in the values of R1 & R2 & you will get an answer of
11
Ω.
1
100
+ ) = 11 Ω
1 1
(
100 10
Next, you work out the Rt of branch 2, which is exactly the same method as before,
but you sub in the values for R3, R4 & R5.
1
3102
1 1 1 = Ω (19.76Ω)
( + + ) 157
22 470 330
Now that you have your results, you can simply re-draw the circuit as a series circuit
& add the 2 values together, which will give you the total Rt of the whole circuit.
100 3102
Rt = Ω+ Ω = 28.849Ω
11 157
Moving onwards with the next question; this is finding out the total current (It) in the
circuit. To work out the It you must know the Ohms law equation, which I have stated
Volts(v)
in my work sheet. So, we know that Current (I) =
Resistance ( Ω)
As we already worked out the Rt from the previous question, we can use that value, &
for voltage, we can use the value that I had selected for my circuit, which is 10V.
8635
Now we just sub in our values to find It, & that’ll give us Amps (A)
24911
10 8635
It = 28.849 = 24911 A (0.35A)
P1 Fahim Mohammed
