If you know any two of these values, use Ohm’s Law to solve for the third. For example, if you know the resistance and voltage of a circuit, rearrange V = IR to I = V / R, and plug in the known values to solve for I, the current. Always use values form the same part of the circuit. If you are trying to solve for the resistance of a single resistor, you will need to know the voltage and current for that resistor. Do not use the voltage for the whole circuit.

Example 1: A series circuit has two resistors. One resistor R1 has 3Ω (ohms) of resistance, and the second resistor R2 has 6Ω of resistance. Find the total resistance. The total resistance of the circuit is equivalent to the sum of the two individual resistances:Rtotal=R1+R2=3+6=9{\displaystyle R_{total}=R_{1}+R_{2}=3+6=9}Ω. On a circuit diagram, a resistor looks like a zig-zag in the wire.

Example 2: A series circuit is powered by a 9 volt battery and has two resistors R1 and R2. The voltage drop across R1 is 5V. What is the voltage drop across R2?Vtotal=V1+V2{\displaystyle V_{total}=V_{1}+V_{2}}9=5+V2{\displaystyle 9=5+V_{2}}V2=9−5=4{\displaystyle V_{2}=9-5=4} volts. Old-fashioned textbooks may use E to represent voltage instead of V. You may also see ΔV, meaning “change in voltage. " The symbol Δ is the Greek letter delta, and means “change. "

Example 3: A series circuit plugged into a 220V source is connected to several light bulbs. You measure the voltage drop across a light bulb with resistance 100 Ω and get a result of 80V. How much current flows through this circuit?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. Be careful: you cannot use the circuit’s total voltage drop 220V. Ohm’s Law only works if you use values for the same portion of the circuit, and this problem does not tell you the total resistance of the circuit.

[V_I_R_A:VaIaRaB:VbIbRbC:VcIcRcTotal:VtotalItotalRtotal]{\displaystyle {\begin{bmatrix}&{\underline {\mathbf {V} }}&{\underline {\mathbf {I} }}&{\underline {\mathbf {R} }}\\mathbf {A} :&V_{a}&I_{a}&R_{a}\\mathbf {B} :&V_{b}&I_{b}&R_{b}\\mathbf {C} :&V_{c}&I_{c}&R_{c}\\mathbf {Total} :&V_{total}&I_{total}&R_{total}\end{bmatrix}}} Fill out the chart with all values provided in the problem. Ohm’s Law works with values in the same row. For example, Vb=IbRb{\displaystyle V_{b}=I_{b}R_{b}}. Use this to complete any rows that have 2 out of 3 cells filled. Use the properties of series circuits to fill blank spaces in columns: Rtotal=Ra+Rb+Rc{\displaystyle R_{total}=R_{a}+R_{b}+R{c}} Vtotal=Va+Vb+Vc{\displaystyle V_{total}=V_{a}+V_{b}+V{c}} Itotal=Ia=Ib=Ic{\displaystyle I_{total}=I_{a}=I_{b}=I_{c}}

In the classroom, however, you do not need to find the power and energy unless the problem asks you to. If the problem only tells you to fill out a circuit diagram, use the method above to find resistance, voltage, and current.

All the formulas in this section work for the circuit as a whole, or for individual components. Just make sure to use quantities that refer to the same portion of the circuit.

All the formulas in this section work for the circuit as a whole, or for individual components. Just make sure to use quantities that refer to the same portion of the circuit.

We know V = IR from Ohm’s Law, so we can replace V with IR in other equations:P=VI{\displaystyle P=VI} → P=(IR)(I)=I2R{\displaystyle P=(IR)(I)=I^{2}R} Rearrange Ohm’s Law to I = V / R and use the same trick:P=VI{\displaystyle P=VI} → P=(V)(VR)=V2R{\displaystyle P=(V)({\frac {V}{R}})={\frac {V^{2}}{R}}}

We know V = IR from Ohm’s Law, so we can replace V with IR in other equations:P=VI{\displaystyle P=VI} → P=(IR)(I)=I2R{\displaystyle P=(IR)(I)=I^{2}R} Rearrange Ohm’s Law to I = V / R and use the same trick:P=VI{\displaystyle P=VI} → P=(V)(VR)=V2R{\displaystyle P=(V)({\frac {V}{R}})={\frac {V^{2}}{R}}}

We know V = IR from Ohm’s Law, so we can replace V with IR in other equations:P=VI{\displaystyle P=VI} → P=(IR)(I)=I2R{\displaystyle P=(IR)(I)=I^{2}R} Rearrange Ohm’s Law to I = V / R and use the same trick:P=VI{\displaystyle P=VI} → P=(V)(VR)=V2R{\displaystyle P=(V)({\frac {V}{R}})={\frac {V^{2}}{R}}}

The equations above give you a power result in watts. Multiply by seconds to get an energy result in Joules.