Question 1

Write the 4 equations you would generally employ when solving calculations with series resistors only in the circuit.

Solution

Some of these properties are not so much equations, but statements that reflect the behaviour of series circuits.

The fundamentals of Kirchhoff's Voltage Law, Kirchhoff's Current Law, and Ohm's Law always apply, no matter what the connections of the circuit are. There are some useful special cases of these laws that are useful in a particular circuit.

  • The total resistance of a set of series resistors is the sum of each individual resistance. \( R_{\mathrm{T}} = R_1 + R_2 + \ldots \).
  • The total voltage drop of a set of series components is the sum of the voltage drop across each individual component. \( V_{\mathrm{s}} = V_1 + V_2 + \ldots \).
  • The current flow through a set of series components is the same as the current flow through each individual series component (and vice versa).
  • Ohm's Law applies to any combination of components in the circuit, from an individual component to the entire circuit taken as a whole.


Question 2

Write the 4 equations you would generally employ when solving calculations with parallel resistors only in the circuit.

Solution

Some of these properties are not so much equations, but statements that reflect the behaviour of series circuits.

The fundamentals of Kirchhoff's Voltage Law, Kirchhoff's Current Law, and Ohm's Law always apply, no matter what the connections of the circuit are. There are some useful special cases of these laws that are useful in a particular circuit.

  • The total resistance of a set of parallel resistors is the inverse of the sum of the inverse of each individual resistance. \( R_{\mathrm{T}} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2} + \cdots} \).
  • The current flow through a set of parallel components is the sum of the current flow through each individual component. \( I_{\mathrm{s}} = I_1 + I_2 + \ldots \).
  • The voltage drop across a set of parallel components is the same as the voltage drop across each parallel series component (and vice versa).
  • Ohm's Law applies to any combination of components in the circuit, from an individual component to the entire circuit taken as a whole.


Question 3

The circuit is shown below.


Calculate the following quantities:

  1. the total resistance of the load;
  2. the total current;
  3. the voltage at the battery terminals;
  4. the current through \( R_1 \), \( R_3 \), \( R_4 \), \( R_5 \) and \( R_7 \);
  5. the voltage drop across \( R_4 \) and \( R_6 \).


Solution

a) Total Resistance

The total resistance \( R_{\mathrm{T}} \) may be computed by inspection.

NOTES:

  • Do not worry about the way \( R_6 \) and \( R_7 \) are connected. They are still connected directly to node "B".
  • The network of \( R_1 \) through \( R_6 \) is better thought of as "\( R_1 \) in parallel with \( R_4 \) connected in series with the total resistance of \( R_2 \), \( R_3 \), \( R_5 \) and \( R_6 \)". The algebraic current (as in the current you would use in a KCL calculation) that flows through \( R_1 \) is not the same algebraic current that flows through \( R_2 \).

\( R_{\mathrm{T}} = R_7 \parallel ((R_5 + R_6) \parallel (R_2 + R_3) + R_1 \parallel R_4 )\)

Plugging in the component values yields the following.

\( R_{\mathrm{T}} = 35 \parallel ((15 + 22) \parallel (65 + 5) + 75 \parallel 18 = 35 \parallel (37 \parallel 70 + 75 \parallel 18 = 35 \parallel (24.2056 + 14.5161) = 35 \parallel 38.7217 = \) 18.3835 Ω (6 sf) or 18.38 Ω (4 sf) #.


b) Total Current

The total current \( I_\mathrm{s} \) is calculated by Ohm's Law. \( E \) is the battery EMF, and \( R_{\mathrm{int}} \) is the battery internal resistance.

\( I_\mathrm{s} = \frac{E}{R_{\mathrm{int}} + R_{\mathrm{T}}} = \frac{75}{0.5 + 18.3835} = \frac{75}{18.8835} = \) 3.97172 A (6 sf) or 3.97 A (3 sf) #.

The total current can also be calculated using Voltage division, then Ohm's Law.

The battery terminal voltage \( V_{\mathrm{s}} \) (the voltage between A and B) can be calculated using voltage division.

\( V_{\mathrm{s}} = E \cdot \frac{R_{\mathrm{T}}}{R_{\mathrm{int}} + R_{\mathrm{T}}} = 75 \cdot \frac{18.3835}{18.8835} = 75 \cdot 0.973522 = \) 73.0141 V (6 sf).

The total current can then be calculated using Ohm's Law.

\( I_\mathrm{s} = \frac{V_{\mathrm{s}}}{R_{\mathrm{T}}} = \frac{73.0141}{18.3835} = \) 3.97172 A (6 sf) or 3.972 A (4 sf) #.


c) Battery Terminal Voltage

The battery terminal voltage \( V_{\mathrm{s}} \) (the voltage between A and B) can be calculated using Ohm's Law.

\( V_{\mathrm{s}} = E - I_\mathrm{s} R_{\mathrm{int}} = 75 - 3.97172 \cdot 0.5 = 75 - 1.98586 = \) 73.0141 V (6 sf) or 73.01 V (4 sf) #.

Otherwise, the supply voltage was already calculated above in the Total Current section.

\( V_{\mathrm{s}} = \) 73.0141 V (6 sf) = 73.01 V (4 sf).


d) Resistor Currents: \( R_1 \), \( R_3 \), \( R_4 \), \( R_5 \) and \( R_7 \)

These currents can be computed by any combination of voltage division, current division, KCL, KVL, etc.

These solutions will focus on the most "direct" way.

The current through R_7 is given directly by Ohm's Law.

\( I_{R7} = \frac{V_{\mathrm{s}}}{R_7} = \frac{73.0141}{35} = \) 2.08612 A (3 sf) or 2.086 A (4 sf) #.

The current through the rest of \( R_1 \) through \( R_6 \) is given using Kirchhoff's Current Law. This current will be called \( I_8 \).

\( I_8 = I_{\mathrm{s}} - I_{R7} = 3.97172 - 2.08612 = \) 1.88560 A (6 sf).

The current through \( R_1 \), \( I_{R1} \) can be found through current division.

\( I_{R1} = I_8 \cdot \frac{R_1 \parallel R_4}{R_1} \)

We had previously calculated \( R_1 \parallel R_4 \) as 14.5161 Ω.

\( I_{R1} = 1.88560 \cdot \frac{14.5161}{18} = \) 1.52064 A (6 sf) or 1.521 A (4 sf) #.

The current through \( R_4 \) can be found using Kirchhoff's Current Law.

\( I_{R4} = I_8 - I_{R1} = 1.88560 - 1.52064 = \) 0.364960 A (6 sf) or 0.3650 A (4 sf) #.

The current through \( R_2 \) and \( R_3 \), \( I_{R3} \) can be found through current division.

\( I_{R3} = I_8 \cdot \frac{(R_2 + R_3) \parallel (R_5 + R_6)}{R_2 + R_3} \)

We had previously calculated \( R_2 + R_3 \) as 70 Ω; \( R_5 + R_6 \) as 37 Ω; and \( (R_2 + R_3) \parallel (R_5 + R_6) \) as 24.2056 Ω.

\( I_{R3} = 1.88560 \cdot \frac{24.2056}{70} = 1.88560 \cdot 0.345794 = \) 0.652030 A (6 sf) or 0.652 A (4 sf) #.

The current through \( R_5 \) can be found using Kirchhoff's Current Law.

\( I_{R5} = I_8 - I_{R3} = 1.88560 - 0.652030 = \) 1.23357 A (6 sf) or 1.234 A (4 sf) #.


e) Resistor Voltages: \( R_4 \), \( R_6 \)

These voltages can be computed by any combination of voltage division, current division, KCL, KVL, etc.

These solutions will focus on the most "direct" way.

The voltage drop across \( R_4 \) is the same as the voltage drop across \( R_1 \), as they are in parallel. It is also the same as the voltage drop as the voltage drop across the parallel combination of R_1 and R_4. All three methods of calculation are shown below.

\( V_{R4} = I_{R4} \cdot R_4 = 0.364960 \cdot 75 = \) 27.3720 V (6 sf) or 27.37 V (4 sf) #.

\( V_{R4} = I_{R1} \cdot R_1 = 1.52064 \cdot 18 = \) 27.3715 V (6 sf) or 27.37 V (4 sf) #.

\( V_{R4} = I_8 \cdot R_1 \parallel R_4 = 1.88560 \cdot 14.5161 = \) 27.3716 V (6 sf) or 27.37 V (4 sf) #.

The voltage drop across \( R_6 \) can be calculated using Ohm's Law. Remember that the same current flows through \( R_6 \) as flows through \( R_5 \).

\( V_{R6} = I_{R5} \cdot R_6 = 1.23357 \cdot 22 = \) 27.1385 V (6 sf) or 27.14 V (4 sf) #.


Question 4

The circuit is shown below.


Calculate the following quantities:

  1. the total current;
  2. the voltage drop across \( R_3 \);
  3. the voltage at the battery terminals.

Solution

We're given that \( R_1 \) has 4 A current flow through it. \( R_1 \) through \( R_3 \) form a network that has a total voltage that is also the same as the supply voltage \( V_{\mathrm{s}} \) (the required answer for c).

The total resistance of the \( R_1 \) "branch" will be called \( R_6 \). \( R_6 \) is calculated below.

\( R_6 = R_1 + R_2 \parallel R_3 = 32.75 + 80 \parallel 255 = 32.75 + 60.8955 = \) 93.6455 Ω (6 sf).

The voltage drop across \( R_6 \) is also the supply voltage.

\( V_{\mathrm{s}} = V_{R6} = I_{R1} R_6 = 4 \cdot 93.6455 = \) 374.582 V (6 sf).

The total current \( I_{\mathrm{L}} \) can now be calculated, knowing that \( R_4 \) and \( R_5 \) are in parallel across the supply voltage. The fact that they are "branched" first does not matter for the purposes of calculating the supply current.

\( I_{\mathrm{L}} = I_{R6} + I_{R4} + I_{R5} = 4 + \frac{V_{\mathrm{s}}}{88} + \frac{V_{\mathrm{s}}}{104} = 4 + 374.582 \cdot (\frac{1}{88} + \frac{1}{104}) = 4 + 374.582 \cdot 0.020979 = 4 + 7.85836 = \) 11.8584 A (6 sf).

The battery EMF \( E \) is calculated by adding the voltage drop across the internal resistance to the supply voltage.

\( E = V_{\mathrm{s}} + I_{\mathrm{L}} R_{\mathrm{Int}} = 374.582 + 11.8584 \cdot 1.2 = 374.582 + 14.2300 = \) 388.812 V (6 sf).


a) Total Current

The total current is calculated above as \( I_{\mathrm{L}} \): 11.86 A (4 sf) #.


b) Voltage Drop Across \( R_3 \)

The voltage drop across \( R_3 \) can be calculated via Ohm's Law.

\( V_{R3} = I_{R1} \cdot R_2 \parallel R_3 = 4 \cdot 80 \parallel 255 = 4 \cdot 60.8955 = \) 243.6 V (4 sf) #.


c) Voltage at the Battery Terminals

The battery terminal voltage is calculated above as \( V_{\mathrm{s}} \): 374.6 V (4 sf) #.

آخر تعديل: الأربعاء، 20 مايو 2020، 2:41 PM