two resistors , 47 ohm and 82 ohm are connected in series across a 45 V battery. what is the current in the circuit ?
Ok
The effective resistance of the two resistors in series is
Reff = 47 + 82 = 129 ohms.
The circuit current is
I = V/Reff = 45/129 = ___ Amps
To find the current in the circuit, you can use Ohm's Law, which states that the current (I) is equal to the voltage (V) divided by the resistance (R). In a series circuit, the current is the same through all the components.
1. Calculate the total resistance (RT) of the circuit by summing up the resistances of the two resistors connected in series:
RT = 47 ohm + 82 ohm
RT = 129 ohm
2. Once you have the total resistance, you can calculate the current (I) using Ohm's Law:
I = V / R
I = 45 V / 129 ohm
I ≈ 0.349 A
Therefore, the current in the circuit is approximately 0.349 Amperes.
To find the current in the circuit, you can use Ohm's Law which states that the current (I) flowing through a resistor is equal to the voltage (V) across the resistor divided by its resistance (R).
In this case, the resistors are connected in series, which means the same current flows through both resistors.
First, calculate the equivalent resistance (R_eq) of the series combination of the resistors using the formula:
R_eq = R1 + R2
where R1 is the resistance of the first resistor (47 ohm) and R2 is the resistance of the second resistor (82 ohm).
R_eq = 47 ohm + 82 ohm = 129 ohm
Next, use Ohm's Law to find the current (I):
I = V / R_eq
where V is the voltage across the resistors (45 V) and R_eq is the equivalent resistance.
I = 45 V / 129 ohm ≈ 0.35 A
Therefore, the current in the circuit is approximately 0.35 A.