Example 7.2: Equivalent resistance

Question: A $1\,\Omega$ and a $2\,{\rm\Omega}$ resistor are connected in parallel, and this pair of resistors is connected in series with a $4\,\Omega$ resistor. What is the equivalent resistance of the whole combination? What is the current flowing through the $4\,\Omega$ resistor if the whole combination is connected across the terminals of a $6\,{\rm V}$ battery (of negligible internal resistance)? Likewise, what are the currents flowing through the $1\,\Omega$ and $2\,\Omega$ resistors?
 
Answer: The equivalent resistance of the $1\,\Omega$ and $2\,\Omega$ resistors is

$$\frac{1}{R_{eq}'} = \frac{1}{1} + \frac{1}{2} = \frac{3}{2}\,\Omega^{-1},$$

giving $R_{\rm eq}' = 0.667\,{\rm\Omega}$. When a $0.667\,{\rm\Omega}$ resistor is combined in series with a $4\,\Omega$ resistor, the equivalent resistance is $R_{\rm eq} = 0.667 + 4 = 4.667\,\Omega$.

The current driven by the $6\,{\rm V}$ battery is

$$I = \frac{V}{R_{\rm eq}} = \frac{(6)}{(4.667)} = 1.29\,{\rm A}.$$

This is the current flowing through the $4\,\Omega$ resistor, since one end of this resistor is connected directly to the battery, with no intermediate junction points.

The voltage drop across the $4\,\Omega$ resistor is

$$V_4 = I\,R_4 = (1.29)\,(4) =5.14\,{\rm V}.$$

Thus, the voltage drop across the $1\,\Omega$ and $2\,\Omega$ combination is $V_{12} = 6-5.14 = 0.857\,{\rm V}$. The current flowing through the $1\,\Omega$ resistor is given by

$$I_1 = \frac{V_{12}}{R_1} = \frac{(0.857)}{(1)}= 0.857\,{\rm A}.$$

Likewise, the current flowing through the $2\,\Omega$ resistor is

$$I_2 = \frac{V_{12}}{R_2} = \frac{(0.857)}{(2)}= 0.429\,{\rm A}.$$

Note that the total current flowing through the $1\,\Omega$ and $2\,\Omega$ combination is $I_{12} = I_1 + I_2 = 1.29\,{\rm A}$, which is the same as the current flowing through the $4\,\Omega$ resistor. This makes sense because the $1\,\Omega$ and $2\,\Omega$ combination is connected in series with the $4\,\Omega$ resistor.