# Alternate Solution to Example Problem on Series Circuits

Here is an alternate way of solving the example problem. Instead of first finding all the resistances, we can begin the problem by finding the emf voltage first.

By Ohm's Law, we know that the emf is equal to the product of the total current and the total resistance.

\(\varepsilon = I R\)

\(\varepsilon = (1.0)(30) = 30V\)

Now that we know the emf voltage, we also know the total voltage. Since \(\varepsilon =V\), where V is the total voltage, then \(V=30 V\).

Also, the total voltage, \(V\), is equal to the sum of the voltages across the resistors in this circuit (because this is a series circuit).

\(V = V_1 + V_2 + V_3 + V_4\)

We also know all the voltages across the resistors except for \(V_4\). So,

\(V_4 = V - (V_1 + V_2 + V_3)\)

\(V_4 = 30 - (5.0 + 8.0 + 7.0)\)

\(V_4 = 10V\)

The only unknowns now are the resistances. We know the voltages across each resistor. Also, because this is a series circuit, the current through each resistor is the same. Then, by Ohm's Law, we can find the resistances of each resistor:

\(R_1 = \frac {V_1}{I}, R_2 = \frac {V_2}{I}, R_3 = \frac{V_3}{I}\)

\(R_1 = \frac {5.0}{1.0} =5.0 \quad \Omega \quad R _2 = \frac {8.0}{1.0} = 8.0 \quad \Omega \quad R_3 = \frac{7.0}{1.0} = 7.0 \Omega\)