A 4 ohm resistor and 10 ohm resistor are connected in series. If this circuit is connected to a 12 V battery, how much current will flow through the circuit?
Connected in series means the resistances could be added.
Equivalent resistor = 4+10=14Ω
Voltage = 12 V
Assuming battery has zero internal resistance,
V=IR, or
I=V/R = 12/14 A = 6/7 A.
To find the current flowing through the circuit, we can use Ohm's Law, which states that the current (I) flowing through a circuit is equal to the voltage (V) divided by the resistance (R).
In this case, the total resistance in the circuit is the sum of the resistances of the two resistors connected in series. Therefore, the total resistance (R_total) is:
R_total = R1 + R2
= 4 ohm + 10 ohm
= 14 ohm
Now, we can use Ohm's Law to find the current (I):
I = V / R_total
= 12 V / 14 ohm
≈ 0.857 A
Therefore, approximately 0.857 Amps (or 857 mA) of current will flow through the circuit.
To find the current flowing through the circuit, you can use Ohm's Law, which states that the current (I) flowing through a circuit is equal to the voltage (V) divided by the resistance (R).
In this case, the resistors are connected in series, so the total resistance (RT) is equal to the sum of the individual resistances (R1 and R2):
RT = R1 + R2
Given that R1 = 4 ohms and R2 = 10 ohms, we can calculate the total resistance:
RT = 4 ohms + 10 ohms = 14 ohms
Now, we can use Ohm's Law to calculate the current (I):
I = V / RT
Given that the voltage (V) is 12 volts, we can substitute in values:
I = 12 V / 14 ohms
Dividing 12 volts by 14 ohms, we find that the current flowing through the circuit is approximately 0.857 Amps (A).