Example  4
 

 Multimedia



The formula to determine the total resistance, \(R\), in ohms \((\Omega) \), of a parallel electrical circuit with four resistors is \(R = 1 \div \left( {\frac{1}{{R_1 }} + \frac{1}{{R_2 }} + \frac{1}{{R_3 }} + \frac{1}{{R_4 }}} \right)\).


  1. Rewrite the formula in simplest form.

    \(R = 1 \div \left( {\frac{1}{{R_1}} + \frac{1}{{R_2}} + \frac{1}{{R_3}} + \frac{1}{{R_4}}} \right)\)

    Step 1: Identify the NPVs.

    \(\begin{align}
     R_1&\ne 0 \\
     R_2&\ne 0 \\
     R_3&\ne 0 \\
     R_4&\ne 0 \\
     \end{align}\)

    Step 2: Expand and simplify.

    \[\begin{align}
     R &= 1 \div \left[ {\left( {\frac{1}{{R_1}}} \right)\left( {\frac{{R_2 R_3 R_4 }}{{R_2 R_3 R_4 }}} \right) + \left( {\frac{1}{{R_2}}} \right)\left( {\frac{{R_1 R_3 R_4 }}{{R_1 R_3 R_4 }}} \right) + \left( {\frac{1}{{R_3}}} \right)\left( {\frac{{R_1 R_2 R_4 }}{{R_1 R_2 R_4 }}} \right) + \left( {\frac{1}{{R_4}}} \right)\left( {\frac{{R_1 R_2 R_3 }}{{R_1 R_2 R_3 }}} \right)} \right] \\
     R &= 1 \div \left( {\frac{{R_2 R_3 R_4 + R_1 R_3 R_4 + R_1 R_2 R_4 + R_1 R_2 R_3 }}{{R_1 R_2 R_3 R_4 }}} \right) \\
     R &= 1 \cdot \left( {\frac{{R_1 R_2 R_3 R_4 }}{{R_2 R_3 R_4 + R_1 R_3 R_4 + R_1 R_2 R_4  + R_1 R_2 R_3 }}} \right) \\
     R &= \frac{{R_1 R_2 R_3 R_4 }}{{R_2 R_3 R_4 + R_1 R_3 R_4 + R_1 R_2 R_4 + R_1 R_2 R_3 }},\thinspace R_1 \ne 0,\thinspace R_2 \ne 0,\thinspace R_3 \ne 0,\thinspace R_4 \ne 0 \\
     \end{align}\]


  2. Decide which formula to use to determine the total resistance of the circuit if \(R_1 = 3\), \(R_2 = 5\), \(R_3 = 2\), and \(R_4 = 3\). Explain why that formula is better, and solve for the total resistance of the circuit, rounding to the nearest hundredth of an ohm.

    Answers will vary. The first formula is still the simplest to use because it requires fewer calculations.

    Start by substituting the resistance of each resistor into the formula.

    \[\begin{align}
     R &= 1 \div \left( {\frac{1}{{R_1}} + \frac{1}{{R_2}} + \frac{1}{{R_3}} + \frac{1}{{R_4}}} \right) \\
      &= 1 \div \left( {\frac{1}{3} + \frac{1}{5} + \frac{1}{2} + \frac{1}{3}} \right) \\
     \end{align}\]

    The LCD of \(\frac{1}{3} + \frac{1}{5} + \frac{1}{2} + \frac{1}{3}\) is \(3 \cdot 5 \cdot 2 = 30\).

    \[\begin{align}
     R &= 1 \div \left[ {\left( {\frac{1}{3}} \right)\left( {{\color{red}\frac{{10}}{{10}}}} \right) + \left( {\frac{1}{5}} \right)\left( {{\color{red}\frac{6}{6}}} \right) + \left( {\frac{1}{2}} \right)\left( {{\color{red}\frac{{15}}{{15}}}} \right) + \left( {\frac{1}{3}} \right)\left( {{\color{red}\frac{{10}}{{10}}}} \right)} \right] \\
      &= 1 \div \left( {\frac{{10}}{{30}} + \frac{6}{{30}} + \frac{{15}}{{30}} + \frac{{10}}{{30}}} \right) \\
      &= 1 \div \left( {\frac{{10 + 6 + 15 + 10}}{{30}}} \right) \\
      &= 1 \div \frac{{41}}{{30}} \\
     \end{align}\]

    Once the denominator is simplified, perform the division.

    \[\begin{align}
     R &= 1 \div \frac{{41}}{{30}} \\
      &= 1 \times \frac{{30}}{{41}} \\
      &= \frac{{30}}{{41}} \\
      &= 0.731... \\
       &\doteq 0.73 \thinspace \Omega  \\
     \end{align}\]