Question 18.2: Three Resistors in Parallel Goal Analyze a circuit having re...
Three Resistors in Parallel
Goal Analyze a circuit having resistors connected in parallel.
Problem Three resistors are connected in parallel as in Figure 18.7. A potential difference of 18 \mathrm{~V} is maintained between points a and b. (a) Find the current in each resistor. (b) Calculate the power delivered to each resistor and the total power. (c) Find the equivalent resistance of the circuit. (d) Find the total power delivered to the equivalent resistance.
Strategy We can use Ohm’s law and the fact that the voltage drops across parallel resistors are all the same to get the current in each resistor. The rest of the problem just requires substitution into the equation for power delivered to a resistor, \mathscr{P}=I^{2} R, and the reciprocal-sum law for parallel resistors.
(a) Find the current in each resistor. Apply Ohm’s law, solved for the current I delivered by the battery to find the current in each resistor:
\begin{aligned} & I_{1}=\frac{\Delta V}{R_{1}}=\frac{18 \mathrm{~V}}{3.0 \Omega}=6.0 \mathrm{~A} \\ & I_{2}=\frac{\Delta V}{R_{2}}=\frac{18 \mathrm{~V}}{6.0 \Omega}=3.0 \mathrm{~A} \\ & I_{3}=\frac{\Delta V}{R_{3}}=\frac{18 \mathrm{~V}}{9.0 \Omega}=2.0 \mathrm{~A} \end{aligned}
(b) Calculate the power delivered to each resistor and the total power.
Apply \mathscr{P}=I^{2} R to each resistor, substituting the results from part (a).
\begin{array}{ll} 3 \Omega: & \mathscr{P}_{1}=I_{1}{ }^{2} R_{1}=(6.0 \mathrm{~A})^{2}(3.0 \Omega)=110 \mathrm{~W} \\ 6 \Omega: & \mathscr{P}_{2}=I_{2}{ }^{2} R_{2}=(3.0 \mathrm{~A})^{2}(6.0 \Omega)=54 \mathrm{~W} \\ 9 \Omega: & \mathscr{P}_{3}=I_{3}{ }^{2} R_{3}=(2.0 \mathrm{~A})^{2}(9.0 \Omega)=36 \mathrm{~W} \end{array}
Sum to get the total power:
\mathscr{P}_{\text {tot }}=110 \mathrm{~W}+54 \mathrm{~W}+36 \mathrm{~W}=2.0 \times 10^{2} \mathrm{~W}
(c) Find the equivalent resistance of the circuit.
Apply the reciprocal-sum rule, Equation 18.6:
\begin{aligned} & \frac{1}{R_{\mathrm{eq}}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}} \\ & \frac{1}{R_{\mathrm{eq}}}=\frac{1}{3.0 \Omega}+\frac{1}{6.0 \Omega}+\frac{1}{9.0 \Omega}=\frac{11}{18 \Omega} \\ & R_{\mathrm{eq}}=\frac{18}{11} \Omega=1.6 \Omega \end{aligned}
(d) Compute the power dissipated by the equivalent resistance.
Use the alternate power equation:
\mathscr{P}=\frac{(\Delta V)^{2}}{R_{\text {eq }}}=\frac{(18 \mathrm{~V})^{2}}{(1.6 \Omega)}=2.0 \times 10^{2} \mathrm{~W}
Remarks There’s something important to notice in part (a): the smallest 3.0 \Omega resistor carries the largest current, while the other, larger resistors of 6.0 \Omega and 9.0 \Omega carry smaller currents. The largest current is always found in the path of least resistance. In part (b), the power could also be found with \mathscr{P}=(\Delta V)^{2} / R. Note that \mathscr{P}_{1}=108 \mathrm{~W}, but is rounded to 110 \mathrm{~W} because there are only two significant figures. Finally, notice that the total power dissipated in the equivalent resistor is the same as the sum of the power dissipated in the individual resistors, as it should be.