Question

section 5 - 7 voltage dividers
*37. the total resistance of a circuit is 560 ω. what percentage of the total voltage appears across a 27 ω resistor that makes up part of the total series resistance?

1. determine the voltage between points a and b in each voltage - divider of figure 5 - 80.

Explanation:

Step1: Recall voltage - divider formula

In a series - circuit, the voltage across a resistor $V_x$ in a series combination of resistors with total resistance $R_T$ and total voltage $V_T$ is given by $V_x=\frac{R_x}{R_T}V_T$. The percentage of the total voltage across a resistor $R_x$ is $P = \frac{R_x}{R_T}\times100\%$.

Step2: Calculate percentage for problem 37

Given $R_T = 560\Omega$ and $R_x=27\Omega$. Then $P=\frac{27}{560}\times 100\%$.
$P=\frac{2700}{560}\% \approx 4.82\%$

Step3: Recall voltage - divider formula for problem 38(a)

The voltage between points $A$ and $B$ in a voltage - divider circuit is given by $V_{AB}=\frac{R_2}{R_1 + R_2}V_T$. Here, $R_1 = 100\Omega$, $R_2 = 56\Omega$ and $V_T=12V$.
$V_{AB}=\frac{56}{100 + 56}\times12V=\frac{56\times12}{156}V=\frac{672}{156}V\approx 4.31V$

Step4: Recall voltage - divider formula for problem 38(b)

First, find the equivalent resistance $R_{eq}=R_1+R_2+R_3=1.0k\Omega+2.2k\Omega + 3.3k\Omega=6.5k\Omega$.
The voltage between points $A$ and $B$ is $V_{AB}=\frac{R_2+R_3}{R_{eq}}V_T$. Here, $V_T = 8V$, $R_2 = 2.2k\Omega$, $R_3 = 3.3k\Omega$ and $R_{eq}=6.5k\Omega$.
$V_{AB}=\frac{2.2 + 3.3}{6.5}\times8V=\frac{5.5}{6.5}\times8V=\frac{44}{6.5}V\approx 6.77V$

Answer:

1. Approximately $4.82\%$

38(a). Approximately $4.31V$
38(b). Approximately $6.77V$