Click here to purchase the entire book in PDF format. Parallel circuits - from the point of view of the current Since we know the amount of voltage applied across each resistor (in this case, they're both 9 V) then we can again use Ohm's law to determine the amount of current flowing though each of the two resistors. $$I_{1} = \frac{V_{1} }{ R_{1}}$$ (3.64) $$= \frac{9 \textrm{ V} }{ 1.5 \textrm{ k}\Omega}$$ (3.65) $$= \frac{9 \textrm{ V} }{ 1500 \Omega}$$ (3.66) $$= 0.006 \textrm{ A}$$ (3.67) $$= 6 \textrm{ mA}$$ (3.68) $$I_{2} = \frac{V_{2} }{ R_{2}}$$ (3.69) $$= \frac{9 \textrm{ V} }{ 500 \Omega}$$ (3.70) $$= 0.018 \textrm{ A}$$ (3.71) $$= 18 \textrm{ mA}$$ (3.72) One way to calculate the total current coming out of the battery here is to calculate the two individual currents going through the resistors, and adding them together. This will work, and then from there, we can calculate backwards to figure out what the equivalent resistance of the pair of resistors would be. If we did that whole procedure, we would find that the reciprocal of the total resistance is equal to the sum of the reciprocals of the individual resistors. (huh?) It's like this... $$\frac{1}{R_{Total}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$$ (3.73) $$\frac{1}{R_{Total}} = \frac{1}{R_{1}} * \frac{R_{2}}{R_{2}} + \frac{1}{R_{2}} * \frac{R_{1}}{R_{1}}$$ (3.74) $$\frac{1}{R_{Total}} = \frac{R_{2}}{R_{1}R_{2}} + \frac{R_{1}}{R_{2}R_{1}}$$ (3.75) $$\frac{1}{R_{Total}} = \frac{R_{1} + R_{2} }{ R_{1}R_{2}}$$ (3.76) $$R_{Total} = \frac{R_{1}R_{2} }{ R_{1} + R_{2}}$$ (3.77)