Electrical Circuits Calculations

Electrical Circuit Calculations

In electrical circuit calculations, we primarily use one key formula derived from a triangle relating voltage (V), current (I), and resistance (R).

Fundamental Formulas

Recap of the formulas:

  • To find Voltage (V): If you're looking for V, multiply the current (I) and resistance (R) since V is at the top of the triangle.
    V=I×RV = I \times R

  • To find Current (I): If you're looking for I, divide the voltage (V) by the resistance (R).
    I=VRI = \frac{V}{R}

  • To find Resistance (R): If you're looking for R, divide the voltage (V) by the current (I).
    R=VIR = \frac{V}{I}

Basic Circuit Calculations

Example 1

Determine the current (A) in a circuit:

Given:

  • Voltage (V) = 6 volts
  • Resistance (R) = 3 ohms

Using the formula: I=VRI = \frac{V}{R}

I=63=2 ampsI = \frac{6}{3} = 2 \text{ amps}

Example 2

Determine the current (A) and voltage (V) in a series circuit:

Circuit details:

  • Two resistors in series: R<em>1=3 ohmsR<em>1 = 3 \text{ ohms}, R</em>2=2 ohmsR</em>2 = 2 \text{ ohms}
  • Voltage across R1R_1 = 6 volts
  1. Determine A (Current):
    • Since the 6V is connected across the 3-ohm resistor, use these values to find the current.
      I=VR=63=2 ampsI = \frac{V}{R} = \frac{6}{3} = 2 \text{ amps}
  2. Determine V (Voltage across R2R_2):
    • In a series circuit, current remains constant.
    • Therefore, the current through R2R_2 is also 2 amps.
      V=I×R=2×2=4 voltsV = I \times R = 2 \times 2 = 4 \text{ volts}
Example 3

Calculate the current in a series circuit with multiple resistors:

Circuit details:

  • Two resistors in series: 3 ohms and 9 ohms.
  • Total voltage provided by the battery: 6 volts.

Since the resistors are in series, add them together:

Rtotal=3+9=12 ohmsR_{\text{total}} = 3 + 9 = 12 \text{ ohms}

Now, calculate the current:

I=VRtotal=612=0.5 ampsI = \frac{V}{R_{\text{total}}} = \frac{6}{12} = 0.5 \text{ amps}

Proper Circuit Examples

Series Circuit Example

Circuit details:

  • Two resistors in series: 2 ohms and 1 ohm.
  • Battery voltage: 3 volts.

  1. Calculate A1A_1 (Main current):

    • Since the battery voltage covers all resistors, use all resistors to calculate the total current.
    • The resistors are in series, so add them: R<em>total=2+1=3 ohmsR<em>{\text{total}} = 2 + 1 = 3 \text{ ohms}I=VR</em>total=33=1 ampI = \frac{V}{R</em>{\text{total}}} = \frac{3}{3} = 1 \text{ amp}

  2. Calculate V1V_1 (Voltage across the 2-ohm resistor):

    • Use the current that goes through the 2-ohm resistor (which is 1 amp) and its resistance.
      V=I×R=1×2=2 voltsV = I \times R = 1 \times 2 = 2 \text{ volts}

  3. Determine V2V_2 (Voltage across the 1-ohm resistor):

    • Similarly,
      V=I×R=1×1=1 voltV = I \times R = 1 \times 1 = 1 \text{ volt}

Note: The sum of the voltmeter readings (2 volts + 1 volt) equals the battery voltage (3 volts).

Example: Calculating Resistance and Voltages

Circuit Details:

  • A resistor R, a 1-ohm resistor, a voltage source V. A voltmeter (V2) reads 3 volts across the 1-ohm resistor, and an ammeter (A1) reads 3 amps in the main circuit.

  1. Calculate R:
    -The voltmeter reading 6 volts is connected to R
    R=VI=63=2 ohmsR = \frac{V}{I} = \frac{6}{3} = 2 \text{ ohms}

  2. Calculate V2V_2
    -The known values are Resistance = 1 ohm.
    V=I×R=3×1=3 voltsV = I \times R = 3 \times 1 = 3 \text{ volts}

  3. Calculating V:

Method 1:

V<em>battery=∑V</em>series+parallel branchesV<em>{\text{battery}} = \sum V</em>{\text{series}} + \text{parallel branches}

Vbattery=6+3=9 voltsV_{\text{battery}} = 6 + 3 = 9 \text{ volts}

Method 2:

V=I×RV = I \times R

Rtotal=2+1=3 ohmsR_{\text{total}} = 2 + 1 = 3 \text{ ohms}

V=3×3=9 voltsV = 3 \times 3 = 9 \text{ volts}

  1. Determine A2A_2
    -The circuit is in series, so current (A2) at any point will remain the same.

    A2=3 ampsA_2 = 3 \text{ amps}

Another Series Circuit Example

Circuit Details:

  • A 2-ohm resistor, a 1-ohm resistor, and a voltage source V. An ammeter (A1) measures the current in the circuit, and a voltmeter (V2) measures voltage across the 1-ohm resistor,

  1. Calculating A1:
    If we use the 4 volts resistor we need use the 2-ohm resistor (Since I=VRI = \frac{V}{R})

    I=42=2 ampsI = \frac{4}{2} = 2 \text{ amps}

  2. Calculate V2V_2

    • The resistance connected between the voltmeter V2V_2 is 1-ohm.

    V=I×R=2×1=2 voltsV = I \times R = 2 \times 1 = 2 \text{ volts}

  3. Calculate the Voltage
    Method 1:
    V<em>battery=∑V</em>series+parallel branchesV<em>{\text{battery}} = \sum V</em>{\text{series}} + \text{parallel branches}
    =4+2=6 volts= 4 + 2 = 6 \text{ volts}

Method 2:
V=I×RV = I \times R

Rtotal=2+1=3R_{\text{total}} = 2 + 1 = 3

V=2×3=6 voltsV = 2 \times 3 = 6 \text{ volts}

EMF

EMF (Electromotive Force) is the voltage of the battery.

Parallel Circuits

Example 1

Circuit Details:

A 2-ohm resistor, a 6-ohm resistor, and a voltage source where the total voltage is 2.

  1. Calculate A1(I=VR)A_1 (I = \frac{V}{R})

    Use the battery's voltage to calculate the resistors.

    1Rparallel=12+16=23\frac{1}{R_{\text{parallel}}} = \frac{1}{2} + \frac{1}{6} = \frac{2}{3}

    Rparallel=32=1.5 ohmsR_{\text{parallel}} = \frac{3}{2} = 1.5 \text{ ohms}

    I=21.5=1.33 ampsI = \frac{2}{1.5} = 1.33 \text{ amps}

  2. Calculate V1V_1
    Here, we are using the fomula that the batteries voltage is equal to the sum of the voltage is series that is added to the voltage in the parallel branches.

In this circuit there is no voltages in series, so that means the battery's voltage is equal to the voltage in the parallel branches

Therefore V1=2 VoltsV_1 = 2 \text{ Volts}

  1. Now calculate the rest of the Voltages in the circuit
    Here each parallel branch will have the same Voltage, so all the Voltages will = 2 Volts.

  2. Calculate A2A_2
    Since this ampmeter is connected on the green pathway, we will look at the green pathways voltages and resistors
    So therefore the voltage is 2 and the resistor is also 2

    I=VR=22=1 ampI = \frac{V}{R} = \frac{2}{2} = 1 \text{ amp}

  3. Calculate A3A_3
    Here we can use the same method as above (Ohm's law) However, we can also use the fact that the sum of the currents in the parallel parts are equal to the current in the main circuit.

    I=VR=26=0.33 ampsI = \frac{V}{R} = \frac{2}{6} = 0.33 \text{ amps}

Example 2

  1. Calculate A2A_2

    • Determine A2, consider it is part of only one branch. Use V=3 volts and R=3 ohmsV = 3 \text{ volts and } R = 3 \text{ ohms}
      I=VR=33=1 ampI = \frac{V}{R} = \frac{3}{3} = 1 \text{ amp}

  2. Determine V2V_2

    • In parallel, all voltages are the same, so V2=3 voltsV_2 = 3 \text{ volts}

  3. Calculate A3A_3

    • Calculate A3 using V=3 volts and R=6 ohmsV = 3 \text{ volts and } R = 6 \text{ ohms}
      I=VR=36=0.5 ampsI = \frac{V}{R} = \frac{3}{6} = 0.5 \text{ amps}

  4. Calculate A1A_1

    • A1 is the sum of currents in parallel, so A<em>1=A</em>2+A<em>3A<em>1 = A</em>2 + A<em>3A</em>1=1+0.5=1.5 ampsA</em>1 = 1 + 0.5 = 1.5 \text{ amps}

  5. Calculate A<em>4A<em>4A</em>4=A1=1.5 ampsA</em>4 = A_1 = 1.5 \text{ amps}

  6. EMF
    -Here is a method to find the battery Voltage

    The easiest way to find the Voltage is to remember the fact that voltage of the battery is equal to the parallel branches.

    So this means the voltage of the battery = 3 volts!

  7. Calculate EMF with total resistance
    -Now we can use the total Resistance to find the battery Voltage.
    We already know the total current = 1.5 amps.

    -1R<em>parallel=13+16\frac{1}{R<em>{\text{parallel}}} = \frac{1}{3} + \frac{1}{6}=12= \frac{1}{2}R</em>parallel=2 ohmsR</em>{\text{parallel}} = 2 \text{ ohms}
    V=IRV = IR
    =(1.5)(2)= (1.5)(2)
    =3 Volts= 3 \text{ Volts}

Series and Parallel Combined

  1. Now we can find the the parallel to the series.
    Here we can see this particular circuit has a parallel followed up something in series. The A2 has a total of 3 volts and with a resistor = 3ohms.
    I=VR=33=1 ampI = \frac{V}{R} = \frac{3}{3} = 1 \text{ amp}

  2. Calculate V<em>1V<em>1 Since we know this a circuit has parallel, so the voltmeters voltage, in series, has a voltage across V</em>1=3 voltsV</em>1 = 3 \text{ volts}
    Calculate A3A_3
    To find the value
    I=VR=36=0.5 ampsI = \frac{V}{R} = \frac{3}{6} = 0.5 \text{ amps}

  3. Calculate A<em>1A<em>1 To find the value. A</em>1=A<em>2+A</em>3A</em>1 = A<em>2 + A</em>3
    =1+0.5=1.5 amps= 1 + 0.5 = 1.5 \text{ amps}

  4. Calculate V<em>3V<em>3 In order to solve this we must use the following formula. V</em>battery=V<em>series+V</em>parallel branchesV</em>{\text{battery}} = V<em>{\text{series}} + V</em>{\text{parallel branches}}
    The voltage of the battery = 9
    The voltages in series is equal to V<em>3V<em>3 The parallel branches, remember that all voltmeters has the same value that is across the parallel branches. = 3. So we must one of them.   ⟹  9=V</em>3+3\implies 9 = V</em>3 + 3
    Thus to find the value of V<em>3V<em>3V</em>3=9−3=6 voltsV</em>3 = 9 - 3 = 6 \text{ volts}

  5. Calculate the resistance.
    from our formulas.
    R=VIR = \frac{V}{I}
    In the resistor is connected with a voltmeter. Thus we can use a value of voltage 6 and the current that flows throw the resistor is 1.5.
    We will use the same method as last time
    R=61.5=4 ohmsR = \frac{6}{1.5} = 4 \text{ ohms}