First, we need to find the total resistance in the circuit. Since the resistors are in series, we simply add their resistances together:

R_total = R1 + R2 = 4Ω + 8Ω = 12Ω

Next, we use Ohm's Law, which states that the current (I) is equal to the voltage (V) divided by the resistance (R). In this case, the voltage is 12V and the total resistance is 12Ω:

I = V / R = 12V / 12Ω = 1A

Since the resistors are in series, the current through each resistor is the same. Therefore, the current through the 4Ω resistor is also 1A.

Question

First, we need to find the total resistance in the circuit. Since the resistors are in series, we simply add their resistances together:

R_total = R1 + R2 = 4Ω + 8Ω = 12Ω

Next, we use Ohm's Law, which states that the current (I) is equal to the voltage (V) divided by the resistance (R). In this case, the voltage is 12V and the total resistance is 12Ω:

I = V / R = 12V / 12Ω = 1A

Since the resistors are in series, the current through each resistor is the same. Therefore, the current through the 4Ω resistor is also 1A.

Knowee AI · Accepted Answer

First, we need to find the total resistance in the circuit. Since the resistors are in series, we simply add their resistances together:

R_total = R1 + R2 = 4Ω + 8Ω = 12Ω

Next, we use Ohm's Law, which states that the current (I) is equal to the voltage (V) divided by the resistance (R). In this case, the voltage is 12V and the total resistance is 12Ω:

I = V / R = 12V / 12Ω = 1A

Since the resistors are in series, the current through each resistor is the same. Therefore, the current through the 4Ω resistor is also 1A.

If you have a 12V supply and resistors of 4Ω and 8Ω in series, what is the current through the 4Ω resistor?

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