Question

Find the Equivalent Resistance

Four resistors are connected as shown in figure (a), below. (Let R=3.00 Ω.)

The original network of resistors is reduced to a single equivalent resistance.

(a) Find the equivalent resistance between points \(a\) and \(\mathrm{c}\).

solution

Conceptualize Imagine charges flowing into and through this combination from the left. All charges must pass from a to \(b\) through the first two resistors, but the charges split at \(b\) into Categorize Because of the simple nature of the combination of resistors in the figure, we categorize this example as one for which we can use the rules for \([\)-Select- \(>\) combinations of resistors.

Analyze The combination of resistors can be reduced in steps as shown in the figure.

Find the equivalent resistance (in \(\mathrm{Q})\) between a and \(b\) of the \(3.00 \mathrm{R}\) and \(4.00 \mathrm{R}\) resistors, which are in series (left-hand red-brown circles);

\(R_{e q, a b}=\square\)

\(R_{\text {eq, bc }}-\frac{6.00 \Omega}{3}=\square\)

\(R_{\text {eq, ac }}=\square\)

This resistance is that of the single equivalent resistor in figure (c)

(b) What is the current in each resistor if a potential difference of \(36.00 \mathrm{~V}\) is maintained between \(a\) and \(c\) ?

SOLUTION

The currents in the \(3.00 \Omega\) and \(4.00 \Omega\) resistors are the same because they are in -Select--V . In addition, they carry the same current that would exist in the equivalent resistor subject to the \(36.00 \mathrm{~V}\) potential difference. Use the equation \(R=\Delta V / I\) and the result from part (a) to find the current (in A) in the \(3.00 \Omega\) and \(4.00 \Omega\) resistors:

\(I=\frac{\Delta V_{a c}}{R_{e q, a c}}=\frac{36.00 \mathrm{~V}}{R_{\mathrm{eq}, a c}}=\square \mathrm{A}\)

Set the voltages across the resistors in parallel in figure (a) equal to find a relationship between the currents:

$$ \Delta V_{1}=\Delta V_{2} \rightarrow(6.00 \Omega) I_{1}=(3.00 \Omega) I_{2} \rightarrow \frac{I_{2}}{I_{1}}= $$

Use \(I_{1}+I_{2}=I\) to find \(I_{1}\) (in A):

\(I_{1}+I_{2}=I \rightarrow I_{1}+2 I_{1}=I \rightarrow I_{1}=\square \mathrm{A}\)

Find \(I_{2}(\) in \(A)\) :

$$ I_{2}=2 I_{1}= $$

A

Finalize As a final check of our results, note that \(\Delta V_{b c}=(6.00 \Omega) I_{1}=(3.00 \Omega) I_{2}=8.00 \mathrm{~V}\) and \(\Delta V_{a b}=(7.00 \Omega) I=28.00 \mathrm{~V}\); therefore, the voltage \(\Delta V_{a c}(\mathrm{in} \mathrm{V})\) is \(\Delta V_{a c}=\Delta V_{a b}+\Delta V_{b c}=[\mathrm{l}\)

\(\mathrm{V}_{\text {, as it }}\) must.

EXERCISE

Suppose that for the figure in the Example, all the resistors are equal to R. The total current in the circuit is 2.60 A, and a 42.0 V battery is connected between a and c. What is R( in Ω) ?

Answer 1