A 5-ohm resistor is connected in series with a 12-ohm resistor and a power supply. The power supply provides a constant potential difference of 24 volts. Calculate:
a) The total resistance of the circuit. b) The current flowing through the circuit. c) The power dissipated by each resistor.
Assume all resistors have negligible internal resistance.
a) The total resistance of a series circuit is the sum of the individual resistances. Therefore, the total resistance (R_total) can be calculated as:
R_total = R1 + R2
Given that R1 = 5 ohms and R2 = 12 ohms, we can substitute these values into the equation to find R_total:
R_total = 5 ohms + 12 ohms R_total = 17 ohms
Therefore, the total resistance of the circuit is 17 ohms.
b) The current (I) flowing through a circuit can be calculated using Ohm's Law, which states that the current is equal to the potential difference (V) across the circuit divided by the total resistance (R_total). Mathematically, this can be represented as:
I = V / R_total
Given that V = 24 volts and R_total = 17 ohms, we can substitute these values into the equation to find I:
I = 24 volts / 17 ohms I ≈ 1.41 amperes (rounded to two decimal places)
Therefore, the current flowing through the circuit is approximately 1.41 amperes.
c) The power dissipated by a resistor (P) can be calculated using the formula:
P = I^2 * R
where I is the current flowing through the resistor and R is the resistance of the resistor. Let's calculate the power dissipated by each resistor:
For the 5-ohm resistor: P1 = I^2 * R1 P1 = (1.41 amperes)^2 * 5 ohms P1 ≈ 9.95 watts (rounded to two decimal places)
For the 12-ohm resistor: P2 = I^2 * R2 P2 = (1.41 amperes)^2 * 12 ohms P2 ≈ 23.62 watts (rounded to two decimal places)
Therefore, the power dissipated by the 5-ohm resistor is approximately 9.95 watts, and the power dissipated by the 12-ohm resistor is approximately 23.62 watts.