4 = 5 I1 Solve for I1 I1 = 4 / 5 = 0.8 A The two resistors are in series and therefore the same current passes through them. This article has been viewed 243,944 times. Just make sure to use quantities that refer to the same portion of the circuit. What is the resistance of the resistor in the circuit? Fill out the chart with all values provided in the problem. Ohm's Law works with values in the same row. Find the voltage across resistor R2 and the current passing through the same resistor. Use the properties of series circuits to fill blank spaces in columns: In the classroom, however, you do not need to find the power and energy unless the problem asks you to. V2 = R2 I2 = 10 (0.8) = 8 V, Example 3 Once the current through the resistor is known and you have found the resistance, you can now solve for voltage drop across the resistor (Voltage = Current x Resistance). Substitute R by 2 and V by 6 in Ohm's law V = R I. I = 6 / 2 = 3 A, Example 2 If you really can’t stand to see another ad again, then please consider supporting our work with a contribution to wikiHow. You know the values of V and R for the light bulb, so you can use Ohm's Law to solve for the current: I = 80V / 100Ω = 0.8 A (amps) Because the current is the same anywhere on a series circuit, the answer is 0.8 amps. © problemsphysics.com. The voltage across resistor R1 is equal to 4 V. Find the current passing through resistor R2 and the voltage across the same resistor. The graph below represents the voltage V across a resistor against the current I passing through the same resistor. Hence the current I2 through R2 is equal to 0.8 A. {"smallUrl":"https:\/\/www.wikihow.com\/images\/thumb\/1\/1e\/Solve-a-Series-Circuit-Step-1-Version-2.jpg\/v4-460px-Solve-a-Series-Circuit-Step-1-Version-2.jpg","bigUrl":"\/images\/thumb\/1\/1e\/Solve-a-Series-Circuit-Step-1-Version-2.jpg\/aid484350-v4-728px-Solve-a-Series-Circuit-Step-1-Version-2.jpg","smallWidth":460,"smallHeight":345,"bigWidth":"728","bigHeight":"546","licensing":"