A circuit is powered by a 5 V battery and has 0.2 A of current through it. If the circuit contains two resistors connected in series and one of the resistors has a resistance of 5 Ω, what is the resistance of the other resistor?
5 Ω * 0.2 A = 1 V
the other resistor is responsible for the remaining 4 V
... ? Ω * 0.2 A = 4 V
To determine the resistance of the other resistor in the circuit, we can use Ohm's Law, which states that resistance (R) is equal to voltage (V) divided by current (I).
In this case, we know that the battery voltage (V) is 5 V, and the current (I) through the circuit is 0.2 A.
First, let's find the total resistance of the circuit. Since the two resistors are connected in series, the total resistance (R_total) is the sum of the individual resistances.
R_total = R1 + R2
Given that R1 (the resistance of one resistor) is 5 Ω, we can substitute this value into the equation:
R_total = 5 Ω + R2
Now we can use Ohm's Law to find the resistance of the other resistor (R2).
R_total = V / I
Substituting the known values:
5 Ω + R2 = 5 V / 0.2 A
Simplifying the equation, we can solve for R2:
R2 = (5 V / 0.2 A) - 5 Ω
R2 = 25 Ω / 0.2 A - 5 Ω
R2 = 125 Ω / 1 A - 5 Ω
R2 = 125 Ω - 5 Ω
R2 = 120 Ω
Therefore, the resistance of the other resistor in the circuit is 120 Ω.