A series resistive circuit has two resistors. R1 is 570 ohms and R2 is 560 ohms.
The total circuit current is 17.9 milliamps.
Find the voltage drop across R1 in volts.

Answer:

10.203 Volts

Explanation:

For this problem, we need to understand that a series resistive circuit is simply a circuit with some type of voltage source and some resistors, in this case, R1 and R2.

First, we need to find the voltage in the circuit.  To do this, we need to find the total resistance of the circuit.  When two resistors are in series, you sum the resistance.  So we can say the following:

R_Total = R1 + R2

R_Total = 570 Ω + 560 Ω

R_Total = 1130 Ω

Now that we have R_Total for the circuit, we can find the voltage of the circuit by using Ohm's law, V = IR.

V_Total = I_Total * R_Total

V_Total = 17.9 mA * 1130 Ω

V_Total = 20.227 V

Now that we have V_Total, we can find the voltage drop across each resistor by using Ohm's law once more.  Note, that since our circuit is series, both resistors will have the same current (I.e., I_Total = I_1 = I_2).

V_Total = V_1 + V_2

V_Total = V_1 + I_2*R2

V_Total - I_2*R2 = V_1

20.227 V - (17.9 mA * 560 Ω) = V_1

20.227 V - (10.024 V) = V_1

10.203 V = V_1

Hence, the voltage drop across R1 is 10.203 Volts.

Cheers.

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