With three resistors in series (Diagram 3) , how does the sum of v1 , v2 and v3 compare to vT
To determine how the sum of v1, v2, and v3 compares to vT in a series circuit with three resistors, we need to understand the concept of voltage in a series circuit.
In a series circuit, the total voltage (vT) across the resistors is equal to the sum of the voltage drops across each individual resistor. Each resistor experiences a voltage drop proportional to its resistance value. This means that the sum of the voltages across the resistors (v1, v2, and v3) should be equal to the total voltage (vT).
To find the relationship between the voltages, you can calculate the voltage drops across each resistor using Ohm's Law. Ohm's Law states that the voltage drop (V) across a resistor is equal to the current (I) flowing through the resistor multiplied by its resistance (R). Mathematically, it can be expressed as V = I × R.
In a series circuit, the current flowing through all the resistors is the same. Therefore, the voltage drops across each resistor can be calculated by multiplying the current (I) by the respective resistance value (R1, R2, and R3).
Once you have calculated the voltage drops across each resistor, you can compare their sum (v1 + v2 + v3) to the total voltage (vT). If the sum of the voltage drops across the resistors is equal to the total voltage, it means that the voltages are balanced or equal. If the sum is greater than the total voltage, it indicates a violation of Kirchhoff's Voltage Law.
So, in conclusion, in a series circuit with three resistors, the sum of v1, v2, and v3 should be equal to the total voltage (vT) if the circuit is properly connected and follows the rules of Kirchhoff's Voltage Law.