Three identical resistors are connected in series to a battery.
The voltage drop across one of the resistors is ___ times the voltage of the source.
i the same through all
V = i Rtotal
V = i R1 + iR2 + iR3 = V1+V2+V3
R1 = R2 + R3
so V1=V2=V3
so
V = 3 V1 = 3V2=3V3
so
Vi = V/3
To find the voltage drop across one resistor in a series circuit, we need to consider the concept of voltage division. In a series circuit, the total voltage of the battery is divided equally among the resistors.
Since there are three identical resistors, the total voltage of the battery will be divided into three equal parts, with each resistor receiving an equal portion of the voltage.
Therefore, the voltage drop across one of the resistors will be 1/3 times the voltage of the source.
To find the voltage drop across one of the resistors, we need to know the voltage of the source and the total resistance in the circuit. Let's say the voltage of the source is V and the resistance of each resistor is R.
In a series circuit, the total resistance (R_total) is the sum of the resistances of the individual components. Since we have three identical resistors in series, the total resistance (R_total) is given by:
R_total = R + R + R = 3R
Now, according to Ohm's Law, the voltage drop (V_drop) across a resistor is directly proportional to its resistance. So, the voltage drop across each resistor (V_drop_resistor) can be calculated using:
V_drop_resistor = (V * R_resistor) / R_total
Since all the resistors are identical, we can substitute R_resistor with R. Therefore, the equation becomes:
V_drop_resistor = (V * R) / R_total
Substituting R_total = 3R:
V_drop_resistor = (V * R) / 3R
R cancels out:
V_drop_resistor = V / 3
So, the voltage drop across one of the resistors is one-third (1/3) of the voltage of the source.