Example 4.3.4. Parallel resistors.
Consider the series circuit in Figure 4.3.3.
If the battery supplies 100 V, and the resistors are \(R_1 = 100\ \Omega\text{,}\) \(R_2 = 200\ \Omega\text{,}\) and\(R_3= 500\ \Omega\text{,}\) determine
- The total current \(I\) in the circuit, and
- The total current if branch 3 is turned off by opening the third switch.
Answer.
For all three branches in operation: I = 1.7 A.
For just the first two branches I = 1.5 A.
Solution 1.
The current in each branch may be found using Ohms Law
\begin{equation*}
I = E/R\text{.}
\end{equation*}
Use Ohm’s Law \(I = E/R\) to calculate the current through each resistor.
\begin{align*}
I_1 \amp= E/R_1 = \frac{100 \text{ V}}{100\ \Omega} = 1.0 \text{ A} \\
I_2 \amp= E/R_2 = \frac{100 \text{ V}}{200\ \Omega}= 0.5 \text{ A}\\
I_3 \amp= E/R_3 = \frac{100 \text{ V}}{500\ \Omega}= 0.2 \text{ A}
\end{align*}
The total current for all three branches is
\begin{equation*}
I = I_1 + I_2 +I_3 = 1.7 \text{ A}
\end{equation*}
When the switch in branch 3 is opened current \(I_3\) drops to zero, so the total current drops to 1.5 A
Solution 2.
An alternate solution method is to first find the equivalent resistance using (4.3.1), then use that resistance to find the current.
For all three resistors
\begin{equation*}
R_t = \frac{1}{\frac{1}{100}+\frac{1}{200} + \frac{1}{500}} = 58.8\ \Omega\text{,}
\end{equation*}
and the corresponding current is
\begin{equation*}
I =\frac{E}{R_t}= \frac{100 \text{ V}}{58.8\ \Omega} = 1.7 \text{ A}
\end{equation*}
Repeating the calculations for the first two resistors only we get
\begin{gather*}
R_t = \frac{1}{\frac{1}{100}+\frac{1}{200}} = 66.7\ \Omega\\
I = \frac{E}{R_t} = \frac{100 \text{ V}}{66.7\ \Omega} = 1.5 \text{ A}
\end{gather*}
