DC Circuits

LessonAbout 1.5 h

DC circuits are electric circuits where current flows continuously in a single direction under constant voltage.

1. Ohm’s Law, Electric Power, and Energy

1.1 Ohm’s Law and DC Resistance

Ohm’s law states that electric current is directly proportional to voltage and inversely proportional to resistance.

Direct proportion: current increases as applied voltage increases
Inverse proportion: current decreases as circuit resistance increases
Linear conductor: constant resistance over varying voltage and current

Ohm’s Law

\[ I = \dfrac{V}{R} \qquad\qquad V = I R \qquad\qquad R = \dfrac{V}{I} \]

Example. A 12-V battery supplies a 10-ohm resistor. If the current drawn is 1.18 A, find the internal resistance of the battery.

Solution.

\[ r = \dfrac{E}{I} - R = \dfrac{12\ \text{V}}{1.18\ \text{A}} - 10\ \Omega = 0.169\ \Omega \]

Answer: \(0.169\ \Omega\)

1.2 Electric Power and Energy

Electric power is the time rate of energy transfer, while electric energy is total work converted.

Unit of power: watt (\(\text{W}\)) or kilowatt (\(\text{kW}\))
Unit of energy: joule (\(\text{J}\)) or kilowatt-hour (\(\text{kW}\cdot\text{hr}\))
Conversion: \(1\ \text{kW}\cdot\text{hr} = 3.6 \times 10^6\ \text{J}\)

Electric Power and Energy

\[ P = V I = I^2 R = \dfrac{V^2}{R} \]

\[ W = P t = V I t \]

Example. An arc lamp takes 10 A at 50 volts. A resistance R is placed in series so that the lamp burns correctly from a 110 V supply. Find the power wasted in this resistor.

Solution.

\[ V_R = V_T - V_L = 110\ \text{V} - 50\ \text{V} = 60\ \text{V} \]

\[ P_R = V_R I = (60\ \text{V})(10\ \text{A}) = 600\ \text{W} \]

Answer: \(600\ \text{watts}\)

1.3 Power Losses and Electrical Efficiency

Electrical efficiency is the ratio of useful power output delivered to a load to total input power supplied.

Copper loss: heat dissipated due to conductor resistance (\(I^2 R\))
Input power: sum of output load power and circuit power losses
Efficiency range: always less than 100% due to line loss

Electrical Efficiency and Power Loss

\[ \eta = \dfrac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \dfrac{P_{\text{out}}}{P_{\text{out}} + P_{\text{loss}}} \times 100\% \]

\[ P_{\text{loss}} = I^2 R \]

EE Board Exam October 1986

Example. A LRT car, 5 km distance from the Tayuman, takes 100 A over a 100 mm hard drawn copper trolley wire having a resistance of 0.270 ohm per km. The rail and ground return has a resistance of 0.06 ohm per km. If the station voltage is 750 V, what is the efficiency of transmission?

Solution.

\[ R_{\text{line}} = (0.270 + 0.06)(5) = 1.65\ \Omega \]

\[ V_{\text{car}} = V_s - I R_{\text{line}} = 750 - (100)(1.65) = 585\ \text{V} \]

\[ \eta = \dfrac{V_{\text{car}}}{V_s} \times 100\% = \dfrac{585}{750} \times 100\% = 78\% \]

Answer: \(78\%\)

2. Series and Parallel Resistor Circuits

2.1 Series Circuits and Voltage Division

Series circuits connect components end-to-end to form a single continuous path for current flow.

Current rule: identical current flows through every series element
Voltage rule: total voltage equals sum of individual voltage drops
Resistance rule: total resistance equals sum of series resistances

Series Resistance and Voltage Division

\[ R_T = R_1 + R_2 + \dots + R_n \]

\[ V_1 = V_T \dfrac{R_1}{R_1 + R_2} \qquad\qquad V_2 = V_T \dfrac{R_2}{R_1 + R_2} \]

EE Board Exam April 1981 / EE Board Exam October 1984

Example. Two 115-V incandescent lamps A and B are connected in series across a 230-V source. If lamp A is rated 75 watts and lamp B is rated 50 watts, determine the current drawn by the series connection.

Solution.

\[ R_A = \dfrac{V^2}{P_A} = \dfrac{115^2}{75} = 176.33\ \Omega \qquad\qquad R_B = \dfrac{115^2}{50} = 264.5\ \Omega \]

\[ R_T = R_A + R_B = 176.33 + 264.5 = 440.83\ \Omega \]

\[ I = \dfrac{V_T}{R_T} = \dfrac{230\ \text{V}}{440.83\ \Omega} = 0.52\ \text{A} \]

Answer: \(0.52\ \text{A}\)

2.2 Parallel Circuits and Current Division

Parallel circuits connect components across common nodes, providing multiple paths for electric current.

Voltage rule: identical voltage exists across all parallel branches
Current rule: total current equals sum of branch currents
Resistance rule: equivalent resistance is less than smallest branch resistance

Parallel Resistance and Current Division

\[ \dfrac{1}{R_T} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots + \dfrac{1}{R_n} \]

\[ I_1 = I_T \dfrac{R_2}{R_1 + R_2} \qquad\qquad I_2 = I_T \dfrac{R_1}{R_1 + R_2} \]

EE Board Exam March 1998

Example. Three resistors of 10, 15 and 20 ohms each are connected in parallel. What is the equivalent resistance?

Solution.

\[ \dfrac{1}{R_T} = \dfrac{1}{10} + \dfrac{1}{15} + \dfrac{1}{20} = 0.2167\ \Omega^{-1} \]

\[ R_T = \dfrac{1}{0.2167} = 4.62\ \Omega \]

Answer: \(4.62\ \text{ohms}\)

2.3 Series-Parallel Combination Circuits

Series-parallel circuits combine series-connected and parallel-connected resistor groups within a single network.

Reduction step 1: simplify purely series or parallel branches first
Reduction step 2: replace simplified groups with equivalent single resistances
Reduction step 3: repeat network reduction until total resistance is found

Series-Parallel Equivalent Resistance

\[ R_T = R_1 + \dfrac{R_2 R_3}{R_2 + R_3} \]

EE Board Exam March 1998

Example. Two resistances of 10 and 15 ohms each respectively are connected in parallel. The two are then connected in series with a 5-ohm resistance. What is the equivalent resistance?

Solution.

\[ R_p = \dfrac{10 \times 15}{10 + 15} = 6\ \Omega \]

\[ R_T = R_s + R_p = 5 + 6 = 11\ \Omega \]

Answer: \(11\ \text{ohms}\)

3. Delta-Wye Transformations and Bridge Circuits

3.1 Delta-Wye and Wye-Delta Conversions

Delta-wye transformation simplifies three-terminal resistor networks that cannot be reduced by simple series-parallel rules.

Delta network (\(\Delta\)): three resistors connected in a closed triangular loop
Wye network (\(\text{Y}\)): three resistors connected from three terminals to a common star node
Equivalence rule: terminal resistances remain identical between corresponding nodes

Delta-to-Wye Conversion

\[ R_A = \dfrac{R_b R_c}{R_a + R_b + R_c} \qquad\qquad R_B = \dfrac{R_a R_c}{R_a + R_b + R_c} \qquad\qquad R_C = \dfrac{R_a R_b}{R_a + R_b + R_c} \]

Wye-to-Delta Conversion

\[ R_a = \dfrac{R_A R_B + R_B R_C + R_C R_A}{R_A} \]

EE Board Exam April 1997

Example. A circuit consisting of three resistors rated: 10 ohms, 15 ohms and 20 ohms are connected in delta. What would be the resistances of the equivalent wye connected load?

Solution.

\[ R_A = \dfrac{(10)(20)}{10 + 15 + 20} = 4.44\ \Omega \]

\[ R_B = \dfrac{(10)(15)}{10 + 15 + 20} = 3.33\ \Omega \qquad\qquad R_C = \dfrac{(15)(20)}{10 + 15 + 20} = 6.66\ \Omega \]

Answer: \(3.33,\ 4.44\ \text{and}\ 6.66\ \text{ohms}\)

3.2 Wheatstone Bridge and Balanced Bridge Circuits

A Wheatstone bridge is a four-arm resistor bridge network used for measuring unknown electrical resistances accurately.

Ratio arms: two fixed resistor arms forming a reference ratio (\(R_1 / R_2\))
Rheostat arm: adjustable standard resistor adjusted until balance is reached
Balance condition: zero current through galvanometer when opposite arm products equal

Balanced Wheatstone Bridge

\[ R_x = R \left(\dfrac{R_1}{R_2}\right) \]

Where: \(R_x\) = unknown resistance, \(R\) = rheostat arm resistance, \(R_1/R_2\) = ratio arms

Example. An unknown resistor \(R_x\) is to be measured by means of a Wheatstone bridge. When ratio arms are arranged \(R_1 = 10\ \Omega\) and \(R_2 = 1200\ \Omega\), balance is obtained when rheostat arm is equal to \(1256\ \Omega\). What is the value of the unknown resistor?

Solution.

\[ R_x = R \left(\dfrac{R_1}{R_2}\right) = 1256 \left(\dfrac{10}{1200}\right) = 10.47\ \Omega \]

Answer: \(10.47\ \Omega\)

Formula Summary

Formula Name	Formula
Ohm’s Law	\(I = \dfrac{V}{R} \quad V = I R \quad R = \dfrac{V}{I}\)
Electric Power and Energy	\(P = V I = I^2 R = \dfrac{V^2}{R} \quad W = P t = V I t\)
Electrical Efficiency and Power Loss	\(\eta = \dfrac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \dfrac{P_{\text{out}}}{P_{\text{out}} + P_{\text{loss}}} \times 100\% \quad P_{\text{loss}} = I^2 R\)
Series Resistance and Voltage Division	\(R_T = R_1 + R_2 + \dots + R_n \quad V_1 = V_T \dfrac{R_1}{R_1 + R_2} \quad V_2 = V_T \dfrac{R_2}{R_1 + R_2}\)
Parallel Resistance and Current Division	\(\dfrac{1}{R_T} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots + \dfrac{1}{R_n} \quad I_1 = I_T \dfrac{R_2}{R_1 + R_2} \quad I_2 = I_T \dfrac{R_1}{R_1 + R_2}\)
Series-Parallel Equivalent Resistance	\(R_T = R_1 + \dfrac{R_2 R_3}{R_2 + R_3}\)
Delta-to-Wye Conversion	\(R_A = \dfrac{R_b R_c}{R_a + R_b + R_c} \quad R_B = \dfrac{R_a R_c}{R_a + R_b + R_c} \quad R_C = \dfrac{R_a R_b}{R_a + R_b + R_c}\)
Wye-to-Delta Conversion	\(R_a = \dfrac{R_A R_B + R_B R_C + R_C R_A}{R_A}\)
Balanced Wheatstone Bridge	\(R_x = R \left(\dfrac{R_1}{R_2}\right)\)