Question 18.1: Four Resistors in Series GOAL Analyze several resistors conn...
Four Resistors in Series
GOAL Analyze several resistors connected in series.
PROBLEM Four resistors are arranged as shown in Figure 18.5a. Find (a) the equivalent resistance of the circuit and (b) the current in the circuit if the closed-circuit terminal voltage of the battery is 6.0 V. (c) Calculate the electric potential at point A if the potential at the positive terminal is 6.0 V. (d) Suppose the open circuit voltage, or emf {\mathcal{E}}, is 6.2 V. Calculate the battery’s internal resistance. (e) What fraction f of the battery’s power is delivered to the load resistors?
STRATEGY Because the resistors are connected in series, summing their resistances gives the equivalent resistance. Ohm’s law can then be used to find the current. To find the electric potential at point A, calculate the voltage drop ΔV across the 2.0-Ω resistor and subtract the result from 6.0 V. In part (d) use Equation 18.1
\Delta V={\mathcal{E}}-I r [18.1]
to find the internal resistance of the battery. The fraction of the power delivered to the load resistance is just the power delivered to load, IΔV, divided by the total power, I{\mathcal{E}}.
(a) Find the equivalent resistance of the circuit.
Apply Equation 18.4, summing the resistances:
R_{\mathrm{eq}}=R_{1}+R_{2}+R_{3}+R_{4}=2.0\:\Omega+4.0\:\Omega+5.0\:\Omega+7.0\:\Omega
= 18.0 Ω
R_{\mathrm{eq}}=R_{1}+R_{2}+R_{3}+… [18.4]
(b) Find the current in the circuit.
Apply Ohm’s law to the equivalent resistor in Figure 18.5b, solving for the current:
I=\frac{\Delta V}{R_{\mathrm{eq}}}=\frac{6.0\,\mathrm{V}}{18.0\,\Omega}=\mathrm{\small~{~0.33\,{A}}{~}}
(c) Calculate the electric potential at point A.
Apply Ohm’s law to the 2.0-Ω resistor to find the voltage drop across it:
\Delta V=I R=(0.33\,\mathrm{A})(2.0\,\Omega)=0.66\,\mathrm{V}
To find the potential at A, subtract the voltage drop from the potential at the positive terminal:
V_{A}=6.0\,\mathrm{V}-\,0.66\,\mathrm{V}=\ 5.3\,\mathrm{V}
(d) Calculate the battery’s internal resistance if the battery’s emf is 6.2 V.
Write Equation 18.1:
\Delta V={\mathcal{E}}-I r
Solve for the internal resistance r and substitute values:
r={\frac{\mathcal{E}~-~\Delta V}{I}}={\frac{6.2\,{V}~-~6.0\,{V}}{0.33\,{A}}}=~0.6~Ω
(e) What fraction f of the battery’s power is delivered to
the load resistors?
Divide the power delivered to the load by the total power output:
f={\frac{I\Delta V}{I \mathcal{E}}}={\frac{\Delta V}{\mathcal{E}}}={\frac{6.0\mathrm{~V}}{6.2\mathrm{~V}}}= 0.97
REMARKS A common misconception is that the current is “used up” and steadily declines as it progresses through a series of resistors. That would be a violation of the conservation of charge. What is actually used up is the electric potential energy of the charge carriers, some of which is delivered to each resistor.