Comprehensive Study Guide on Ohm's Law, Electrical Circuits, Energy, and Heat Equivalents

Ohm's Law

  • Definition: At constant temperature, the current flowing in an electric circuit is directly proportional to the impressed electromotive force (emf) applied to the circuit and inversely proportional to the equivalent resistance of the said circuit.
  • Historical Context: Named after the German physicist, Georg S. Ohm (1787 – 1854).
  • Fundamental Formula:
    • I=VRI = \frac{V}{R}
    • Where VV represents the impressed voltage in volts (V\text{V}).
    • Where II represents the current drawn in amperes (A\text{A}).
    • Where RR represents the resistance in ohms (Ω\Omega).
  • Ohm's Law Triangle Quick Reference:
    • V=I×RV = I \times R
    • I=VRI = \frac{V}{R}
    • R=VIR = \frac{V}{I}

Ohm's Law Sample Calculations and Practice Problems

  • Sample Calculation 1: A circuit has a resistance of 8 Ω8\,\Omega. If a voltmeter connected across its terminals reads 10 V10\,\text{V}, how much current is flowing through the circuit?
    • I=VR=108=1.25 AI = \frac{V}{R} = \frac{10}{8} = 1.25\,\text{A}
  • Sample Calculation 2: A resistor of 15 Ω15\,\Omega has a current of 2.5 A2.5\,\text{A} flowing through it. What is the voltage across the resistor?
    • V=I×R=(2.5)(15)=37.5 VV = I \times R = (2.5)(15) = 37.5\,\text{V}
  • Practice Problem 1: A 24 V24\,\text{V} battery supplies a current of 4 A4\,\text{A} to an electric motor. Calculate the resistance of the motor.
  • Practice Problem 2: A circuit draws 0.5 A0.5\,\text{A} from a 12 V12\,\text{V} source. Find the total resistance of the circuit.
  • Practice Problem 3: A 9 V9\,\text{V} battery is connected to a 3 kΩ3\,\text{k}\Omega resistor. Find the current in milliamperes.
  • Practice Problem 4: A 4.7 kΩ4.7\,\text{k}\Omega resistor draws 5 mA5\,\text{mA} of current. What is the voltage across it?
  • Practice Problem 5: A 12 V12\,\text{V} car headlight draws 2.4 A2.4\,\text{A}. Calculate its resistance.

Electrical Power

  • Definition: Power is defined as the rate of energy transfer.
  • Unit of Measurement: Watt (W\text{W}) — a unit of electrical energy equal to one joule of energy consumed in one second. Named after the British engineer and inventor James Watt (1736 – 1819).
  • Power Formulas:
    • P=VI=I2R=V2RP = VI = I^2 R = \frac{V^2}{R}
    • Where PP represents electrical power in watts (W\text{W}).
    • Where VV represents voltage in volts (V\text{V}).
    • Where II represents current in amperes (A\text{A}).
    • Where RR represents resistance in ohms (Ω\Omega).

Electrical Power Sample Calculations and Practice Problems

  • Sample Calculation 1: A 200 V200\,\text{V} lamp has a hot resistance of 400 Ω400\,\Omega. The power rating in watts of the lamp is:
    • P=V2R=2002400=100 WP = \frac{V^2}{R} = \frac{200^2}{400} = 100\,\text{W}
  • Sample Calculation 2: An electric motor drives a mechanical load, taking 18.8 A18.8\,\text{A} from a 230 V230\,\text{V} source. Calculate the power input of the motor.
    • P=VI=(230)(18.8)=4324 WP = VI = (230)(18.8) = 4324\,\text{W}
  • Practice Problem 1: A 12 V12\,\text{V} car battery supplies 5 A5\,\text{A} to a light. Find the power consumed.
  • Practice Problem 2: A 220 V220\,\text{V} electric iron has resistance 48.4 Ω48.4\,\Omega. Calculate its power rating.
  • Practice Problem 3: A 10 Ω10\,\Omega resistor carries 3 A3\,\text{A} current. What is the power dissipated?
  • Practice Problem 4: A 60 W60\,\text{W} lamp draws 0.5 A0.5\,\text{A}. Find the supply voltage.
  • Practice Problem 5: A 1500 W1500\,\text{W} heater is connected to 230 V230\,\text{V} source. How much current does it draw?

Electrical Energy

  • Definition: Energy is defined as the capacity to do work.
  • Energy Formula:
    • W=PtW = Pt
    • Where WW represents electrical energy in joules (J\text{J}).
    • Where PP represents electrical power in watts (W\text{W}).
    • Where tt represents time in seconds (s\text{s}).
  • Standard Unit Conversions:
    • 1 calorie=4.186 Joules1\,\text{calorie} = 4.186\,\text{Joules}
    • 1 BTU=252 calories1\,\text{BTU} = 252\,\text{calories}
    • 1 kWh=3600 kJ=3413 BTU=860 kcal1\,\text{kWh} = 3600\,\text{kJ} = 3413\,\text{BTU} = 860\,\text{kcal}
    • 1 joule=1×107 ergs1\,\text{joule} = 1 \times 10^7\,\text{ergs}
    • 1 day=24 hours1\,\text{day} = 24\,\text{hours}
    • 1 month=30 days=720 hours1\,\text{month} = 30\,\text{days} = 720\,\text{hours}
    • 1 year=365 days=8760 hours1\,\text{year} = 365\,\text{days} = 8760\,\text{hours}

Electrical Energy Sample Calculations and Practice Problems

  • Sample Calculation 1: A 10 hp10\,\text{hp} motor runs at rated load for 5 hours5\,\text{hours}. How many kWh is consumed?
    • W=Pt=(10 hp×0.746 kW1 hp)(5 h)=37.3 kWhW = Pt = \left(10\,\text{hp} \times \frac{0.746\,\text{kW}}{1\,\text{hp}}\right)(5\,\text{h}) = 37.3\,\text{kWh}
  • Sample Calculation 2: A residential house has a lighting load of 1000 W1000\,\text{W} and a small appliance load of 2000 W2000\,\text{W}. If they are used at the same time, what will be the monthly bill at an energy cost of ₱0.400.40 per kilowatt-hour?
    • Step 1 — Total Power: P=1000 W+2000 W=3 kWP = 1000\,\text{W} + 2000\,\text{W} = 3\,\text{kW}
    • Step 2 — Monthly Energy Consumption: W=Pt=3 kW×720 h/month=2160 kWh/monthW = Pt = 3\,\text{kW} \times 720\,\text{h/month} = 2160\,\text{kWh/month}
    • Step 3 — Monthly Bill: Cost=2160 kWh×₱0.40/kWh=₱864.00 per month\text{Cost} = 2160\,\text{kWh} \times \text{₱}0.40/\text{kWh} = \text{₱}864.00\,\text{per month}
  • Sample Calculation 3: A 60 W60\,\text{W} electric fan runs continuously for 12 hours12\,\text{hours} a day. How much energy does it consume daily in kilowatt-hours?
    • W=PtW = Pt
    • W=0.06 kW×12 h=0.72 kWh per dayW = 0.06\,\text{kW} \times 12\,\text{h} = 0.72\,\text{kWh per day}
  • Sample Calculation 4: A 1500 W1500\,\text{W} air conditioner is used 8 hours8\,\text{hours} per day. At ₱8.508.50 per kWh, find the monthly electricity bill for 30 days30\,\text{days}.
    • Given: P=1500 W=1.5 kWP = 1500\,\text{W} = 1.5\,\text{kW}, tdaily=8 ht_{\text{daily}} = 8\,\text{h}, Rate=₱8.50/kWh\text{Rate} = \text{₱}8.50/\text{kWh}
    • Daily energy: Wdaily=1.5×8=12 kWhW_{\text{daily}} = 1.5 \times 8 = 12\,\text{kWh}
    • Monthly energy: Wmonthly=12×30=360 kWhW_{\text{monthly}} = 12 \times 30 = 360\,\text{kWh}
    • Monthly bill: 360×₱8.50=₱3060360 \times \text{₱}8.50 = \text{₱}3060
  • Practice Problem 1: Convert 5 kilowatt-hours5\,\text{kilowatt-hours} to joules.
  • Practice Problem 2: A 100 W100\,\text{W} light bulb consumes 2 kWh2\,\text{kWh} of energy. How many hours was it turned on?
  • Practice Problem 3: An appliance uses 3.6 MJ3.6\,\text{MJ} of energy in 1 hour1\,\text{hour}. Find its power rating in watts.
  • Practice Problem 4: A 5 hp5\,\text{hp} motor operates 6 hours6\,\text{hours} daily. Calculate weekly energy consumption in kWh.
  • Practice Problem 5: A household has: 4×20 W4 \times 20\,\text{W} LED lights (on 8 hrs/day8\,\text{hrs/day}), 1×150 W1 \times 150\,\text{W} TV (on 5 hrs/day5\,\text{hrs/day}). Electricity cost = ₱12.00/kWh12.00/\text{kWh}. Find monthly bill (30 days30\,\text{days}).
  • Practice Problem 6: A 500 W500\,\text{W} toaster is used for 2 minutes2\,\text{minutes}. How many joules of energy does it consume?

Electrical & Heat Energy Equivalents

  • Heat Energy Formula:
    • Q=mcΔTQ = mc\Delta T
    • ΔT=t2−t1\Delta T = t_2 - t_1
    • Where QQ represents heat energy required.
    • Where mm represents mass of the material.
    • Where cc represents specific heat of the material.
    • Where ΔT\Delta T represents change in temperature.
    • Where t1t_1 represents initial temperature of the material.
    • Where t2t_2 represents final temperature of the material.
  • Unit Definitions:
    • Gram-calorie (calorie): The energy required to raise the temperature of 1 gram1\,\text{gram} of water by 1 ∘C1\,^\circ\text{C}.
    • British Thermal Unit (BTU): The energy required to raise the temperature of 1 pound1\,\text{pound} of water by 1 ∘F1\,^\circ\text{F}.
    • Joule: SI unit of energy and work. One (11) joule is the work done by a force of one (11) Newton acting through a distance of one (11) meter.
  • Specific Heat of Water (cc):
    • c=1 cal/g⋅∘Cc = 1\,\text{cal/g}\cdot^\circ\text{C}
    • c=1 kcal/kg⋅∘Cc = 1\,\text{kcal/kg}\cdot^\circ\text{C}
    • c=4186 J/kg⋅∘Cc = 4186\,\text{J/kg}\cdot^\circ\text{C}
    • c=1 BTU/lb⋅∘Fc = 1\,\text{BTU/lb}\cdot^\circ\text{F}

Electrical & Heat Energy Sample Calculations and Practice Problems

  • Sample Calculation 1: Calculate the heat needed to raise 500 g500\,\text{g} of water from 25 ∘C25\,^\circ\text{C} to 75 ∘C75\,^\circ\text{C}.
    • Given: m=500 gm = 500\,\text{g}, c=1 cal/g⋅∘Cc = 1\,\text{cal/g}\cdot^\circ\text{C}, t1=25 ∘Ct_1 = 25\,^\circ\text{C}, t2=75 ∘Ct_2 = 75\,^\circ\text{C}
    • Q=mcΔT=(500)(1)(75−25)=25000 cal=25 kcalQ = mc\Delta T = (500)(1)(75 - 25) = 25000\,\text{cal} = 25\,\text{kcal}
  • Sample Calculation 2: Find the heat energy required to heat 2 kg2\,\text{kg} of water from 30 ∘C30\,^\circ\text{C} to 100 ∘C100\,^\circ\text{C}.
    • Given: m=2 kgm = 2\,\text{kg}, c=4186 J/kg⋅∘Cc = 4186\,\text{J/kg}\cdot^\circ\text{C}, t1=30 ∘Ct_1 = 30\,^\circ\text{C}, t2=100 ∘Ct_2 = 100\,^\circ\text{C}
    • Q=mcΔT=(2)(4186)(100−30)=586.04 kJQ = mc\Delta T = (2)(4186)(100 - 30) = 586.04\,\text{kJ}
  • Sample Calculation 3: Determine the heat needed to warm 20 lb20\,\text{lb} of water from 60 ∘F60\,^\circ\text{F} to 120 ∘F120\,^\circ\text{F}.
    • Given: m=20 lbm = 20\,\text{lb}, c=1 BTU/lb⋅∘Fc = 1\,\text{BTU/lb}\cdot^\circ\text{F}, t1=30 ∘Ft_1 = 30\,^\circ\text{F}, t2=100 ∘Ft_2 = 100\,^\circ\text{F}
    • Q=mcΔT=(20)(1)(100−30)=1200 BTUQ = mc\Delta T = (20)(1)(100 - 30) = 1200\,\text{BTU}
  • Practice Problem 1: Energy to heat 3 kg3\,\text{kg} water from 30 ∘C30\,^\circ\text{C} to 100 ∘C100\,^\circ\text{C}.
  • Practice Problem 2: Warm 50 lb50\,\text{lb} water from 55 ∘F55\,^\circ\text{F} to 135 ∘F135\,^\circ\text{F}.
  • Practice Problem 3: 10,000 cal10,000\,\text{cal} raises water temp 25 ∘C25\,^\circ\text{C}. Find mass in grams.
  • Practice Problem 4: 20 kcal20\,\text{kcal} heats 4 kg4\,\text{kg} water. Find temperature rise.
  • Practice Problem 5: 600 cal600\,\text{cal} added to 120 g120\,\text{g} water at 30 ∘C30\,^\circ\text{C}. Find final temp.

Series Connected Resistors

  • Definition: A series circuit is one where the resistors are connected end to end.
  • Circuit Rules and Properties:
    • Total Resistance: Equal to the sum of the individual resistances.
      • Rt=R1+R2+R3+⋯+RnR_t = R_1 + R_2 + R_3 + \dots + R_n
    • Total Voltage: Equal to the sum of the voltage drops across each resistance.
      • Vt=V1+V2+V3+⋯+VnV_t = V_1 + V_2 + V_3 + \dots + V_n
    • Current Behavior: The current flowing through each resistance is the same or constant.
      • It=I1=I2=I3=⋯=InI_t = I_1 = I_2 = I_3 = \dots = I_n

Series Connected Resistors Sample Calculations and Practice Problems

  • Sample Calculation 1: A 10 Ω10\,\Omega and a 20 Ω20\,\Omega resistor are in series. Find total resistance.
    • Rt=R1+R2=10+20=30 ΩR_t = R_1 + R_2 = 10 + 20 = 30\,\Omega
  • Sample Calculation 2: 10 Ω+30 Ω10\,\Omega + 30\,\Omega in series across 12 V12\,\text{V}. Find total current.
    • Rt=R1+R2=10+30=40 ΩR_t = R_1 + R_2 = 10 + 30 = 40\,\Omega
    • I=VR=1240=0.3 AI = \frac{V}{R} = \frac{12}{40} = 0.3\,\text{A}
  • Sample Calculation 3: 20 Ω20\,\Omega and 30 Ω30\,\Omega in series; total current = 0.5 A0.5\,\text{A}. Find voltage across each resistor.
    • V1=It×R1=(0.5)(20)=10 VV_1 = I_t \times R_1 = (0.5)(20) = 10\,\text{V}
    • V2=It×R2=(0.5)(30)=15 VV_2 = I_t \times R_2 = (0.5)(30) = 15\,\text{V}
  • Sample Calculation 4: A 3 Ω3\,\Omega resistor and a 6 Ω6\,\Omega resistor are connected in series across a DC supply. If the voltage drop across the 3 Ω3\,\Omega resistor is 4 V4\,\text{V}, what is the voltage of the supply?
    • It=V1R1=43=1.333 AI_t = \frac{V_1}{R_1} = \frac{4}{3} = 1.333\,\text{A}
    • V2=It×R2=(1.333)(6)=8 VV_2 = I_t \times R_2 = (1.333)(6) = 8\,\text{V}
    • Vt=V1+V2=4+8=12 VV_t = V_1 + V_2 = 4 + 8 = 12\,\text{V}
  • Sample Calculation 5: A load of 10 Ω10\,\Omega was connected to a 12 V12\,\text{V} battery. The current drawn was 1.18 A1.18\,\text{A}. What is the internal resistance of the battery?
    • Rt=VtIt=121.18=10.17 ΩR_t = \frac{V_t}{I_t} = \frac{12}{1.18} = 10.17\,\Omega
    • Let rr = internal resistance
    • Rt=R+r  ⟹  10.17=10+r  ⟹  r=0.17 ΩR_t = R + r \implies 10.17 = 10 + r \implies r = 0.17\,\Omega
  • Practice Problem 1: A 10 Ω10\,\Omega resistor and a 30 Ω30\,\Omega resistor are connected in series across a 120 V120\,\text{V} DC source. Determine: (a) total resistance, (b) circuit current, (c) voltage drop across each resistor.
  • Practice Problem 2: Three resistors in series: 5 Ω5\,\Omega, unknown R2R_2, and 15 Ω15\,\Omega. The circuit draws 2.5 A2.5\,\text{A} from a 100 V100\,\text{V} supply. Find R2R_2.
  • Practice Problem 3: A 12 V12\,\text{V} battery with internal resistance 0.2 Ω0.2\,\Omega is connected to an 11.8 Ω11.8\,\Omega load. Find the terminal voltage (voltage across the load).
  • Practice Problem 4: Three resistors in series are connected to a 240 V240\,\text{V} source. The voltage drops across the first two are 60 V60\,\text{V} and 90 V90\,\text{V} respectively. If the third resistor is 25 Ω25\,\Omega, find its voltage drop and the total resistance of the circuit.
  • Practice Problem 5: A 5 Ω5\,\Omega and a 20 Ω20\,\Omega resistor are in series across 100 V100\,\text{V}. Calculate the power dissipated by each resistor and the total power.

Parallel Connected Resistors

  • Definition: A parallel circuit is one where the resistors are connected across each other.
  • Circuit Rules and Properties:
    • Total Resistance: Equal to the reciprocal of the sum of the reciprocals of the individual resistances.
      • 1Rt=1R1+1R2+1R3+⋯+1Rn\frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots + \frac{1}{R_n}
    • Total Voltage: The voltage drop across each resistor is equal to the total voltage.
      • Vt=V1=V2=V3=⋯=VnV_t = V_1 = V_2 = V_3 = \dots = V_n
    • Total Current: Equal to the sum of the currents flowing in each resistance.
      • It=I1+I2+I3+⋯+InI_t = I_1 + I_2 + I_3 + \dots + I_n
      • It=V1R1+V2R2+V3R3+⋯+VnRnI_t = \frac{V_1}{R_1} + \frac{V_2}{R_2} + \frac{V_3}{R_3} + \dots + \frac{V_n}{R_n}
    • Total Conductance: Equal to the sum of the individual conductances.
      • Gt=1Rt=1R1+1R2+1R3+⋯+1RnG_t = \frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots + \frac{1}{R_n}

Parallel Connected Resistors Sample Calculations and Practice Problems

  • Sample Calculation 1: A 5 Ω5\,\Omega resistance is connected in parallel with a 10 Ω10\,\Omega resistance. What is the equivalent resistance?
    • 1Rt=1R1+1R2=15+110=310  ⟹  Rt=103 Ω\frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{5} + \frac{1}{10} = \frac{3}{10} \implies R_t = \frac{10}{3}\,\Omega
  • Sample Calculation 2: A 20 Ω20\,\Omega resistor is connected in parallel with an unknown resistor RxR_x. Total resistance = 12 Ω12\,\Omega. Find R_x$.\n * \frac{1}{12} = \frac{1}{20} + \frac{1}{R_x} \implies \frac{1}{R_x} = \frac{1}{12} - \frac{1}{20} = \frac{2}{60} \implies R_x = 30\,\Omega\n* **Sample Calculation 3**: A 10\,\Omegaandaand a15\,\Omegaresistorareconnectedinparallelacrossaresistor are connected in parallel across a30\,\text{V} source. Find total resistance, total current, and current in each resistor.\n * \frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{10} + \frac{1}{15} = \frac{5}{30} \implies R_t = 6\,\Omega\n * I_1 = \frac{30}{10} = 3\,\text{A}\n * I_2 = \frac{30}{15} = 2\,\text{A}\n * I_t = 3 + 2 = 5\,\text{A}\n* **Practice Problem 1**: 4\,\text{k}\Omega,,6\,\text{k}\Omega,and, and12\,\text{k}\Omegaresistorsareinparallelacrossresistors are in parallel across24\,\text{V}. Find total current in mA.\n* **Practice Problem 2**: Three identical 60\,\Omegaresistorsareconnectedinparallelacrossresistors are connected in parallel across120\,\text{V}. Find total resistance and total current.\n* **Practice Problem 3**: Two resistors — 5\,\Omegaandand20\,\Omega—areinparallelacross— are in parallel across100\,\text{V}. Find total current, total resistance, and total power.\n* **Practice Problem 4**: A 60\,\Omegaandaand a40\,\Omegaresistorareinparallelacrossresistor are in parallel across120\,\text{V}. Find total resistance and total current.\n* **Practice Problem 5**: 10\,\Omega,,20\,\Omega,,30\,\Omega,and, and60\,\Omegaareinparallelacrossare in parallel across60\,\text{V}$$. Find total current.