Math Problem Statement

An electrical circuit has four electrical loads in series. These loads measure 4.00 volts, 3.25 volts, 6.50 volts, and 7.75 volts. Calculate the source voltage.

An electrical circuit has a power source voltage of 14.5 volts. Within the series circuit, there are two electrical loads. The first load measures 9.6 volts. Calculate the voltage of the second load in the circuit.

496 technician is is gnosing an electrical circuit issue. The source voltage is 126 volts. There is only one electrical load built into the circuit, and it has a voltage drop of 11.4 volts. This voltage drop test indicates unwanted resistance in the circuit. Calculate the voltage drop caused by the unwanted resistance in the circuit.

A technician is using a multimeter to make electrical measurements. The resistance of a headlight circuit is 3.5 O and the amperage measures 4.05 amps. Calculate the voltage of this circuit.

A technician is using a multimeter to make electrical measurements. The resistance of a horn circuit is 2.15 & and the amperage measures 5.8 amps. Calculate the voltage of this circuit.

Solve these with this formula source voltage = voltage drop, + voltage drop, + voltage drop and take this example formula too for other problems which ever is needed 20 volts = 12 volts + x volts 20 volts - 12 volts = x volts 8 volts = x volts

And give rough work too in detail

Solution

Let's solve each problem step by step, showing detailed work and explanations:


1. Four Electrical Loads in Series

The formula is: Source Voltage=V1+V2+V3+V4\text{Source Voltage} = V_1 + V_2 + V_3 + V_4 Given: V1=4.00volts,V2=3.25volts,V3=6.50volts,V4=7.75voltsV_1 = 4.00 \, \text{volts}, \, V_2 = 3.25 \, \text{volts}, \, V_3 = 6.50 \, \text{volts}, \, V_4 = 7.75 \, \text{volts} Substitute values: Source Voltage=4.00+3.25+6.50+7.75\text{Source Voltage} = 4.00 + 3.25 + 6.50 + 7.75 Simplify: Source Voltage=21.50volts\text{Source Voltage} = 21.50 \, \text{volts}

Answer: Source Voltage = 21.50 volts


2. Voltage of the Second Load in Series

The formula is: Source Voltage=V1+V2\text{Source Voltage} = V_1 + V_2 Given: Source Voltage=14.5volts,V1=9.6volts\text{Source Voltage} = 14.5 \, \text{volts}, \, V_1 = 9.6 \, \text{volts} To find V2V_2: V2=Source VoltageV1V_2 = \text{Source Voltage} - V_1 Substitute values: V2=14.59.6V_2 = 14.5 - 9.6 Simplify: V2=4.9voltsV_2 = 4.9 \, \text{volts}

Answer: Voltage of the Second Load = 4.9 volts


3. Voltage Drop from Unwanted Resistance

The formula is: Source Voltage=Vload+Vunwanted resistance\text{Source Voltage} = V_{\text{load}} + V_{\text{unwanted resistance}} Given: Source Voltage=126volts,Vload=11.4volts\text{Source Voltage} = 126 \, \text{volts}, \, V_{\text{load}} = 11.4 \, \text{volts} To find Vunwanted resistanceV_{\text{unwanted resistance}}: Vunwanted resistance=Source VoltageVloadV_{\text{unwanted resistance}} = \text{Source Voltage} - V_{\text{load}} Substitute values: Vunwanted resistance=12611.4V_{\text{unwanted resistance}} = 126 - 11.4 Simplify: Vunwanted resistance=114.6voltsV_{\text{unwanted resistance}} = 114.6 \, \text{volts}

Answer: Voltage Drop from Unwanted Resistance = 114.6 volts


4. Voltage of the Headlight Circuit

Using Ohm's Law: V=I×RV = I \times R Given: R=3.5Ω,I=4.05ampsR = 3.5 \, \Omega, \, I = 4.05 \, \text{amps} Substitute values: V=4.05×3.5V = 4.05 \times 3.5 Simplify: V=14.175voltsV = 14.175 \, \text{volts}

Answer: Voltage of the Headlight Circuit = 14.18 volts (rounded)


5. Voltage of the Horn Circuit

Using Ohm's Law: V=I×RV = I \times R Given: R=2.15Ω,I=5.8ampsR = 2.15 \, \Omega, \, I = 5.8 \, \text{amps} Substitute values: V=5.8×2.15V = 5.8 \times 2.15 Simplify: V=12.47voltsV = 12.47 \, \text{volts}

Answer: Voltage of the Horn Circuit = 12.47 volts


Recap

  1. Source Voltage for Series Circuit = 21.50 volts
  2. Second Load Voltage = 4.9 volts
  3. Voltage Drop from Unwanted Resistance = 114.6 volts
  4. Headlight Circuit Voltage = 14.18 volts
  5. Horn Circuit Voltage = 12.47 volts

Would you like more details or further assistance with these calculations?


5 Related Questions:

  1. What happens if the total voltage drop in a series circuit exceeds the source voltage?
  2. How can Ohm's Law be applied in more complex circuits with parallel resistances?
  3. What tools can be used to measure unwanted resistance in a circuit?
  4. How does current change when resistance or voltage changes in a circuit?
  5. What are common causes of unwanted resistance in electrical systems?

Tip: Always double-check your measurements and ensure proper calibration of instruments like multimeters to avoid errors in circuit diagnostics!

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Math Problem Analysis

Mathematical Concepts

Electrical calculations
Series circuits
Ohm's Law
Voltage drop

Formulas

Source Voltage = Sum of voltage drops in series
Ohm's Law: V = I × R

Theorems

Ohm's Law

Suitable Grade Level

Grades 9-12