Question

section 5 - 6 kirchhoffs voltage law

1. the following voltage drops are measured across three resistors in series: 5.5 v, 8.2 v, and 12.3 v. what is the value of the source voltage to which these resistors are connected?
2. five resistors are in series with a 20 v source. the voltage drops across four of the resistors are 1.5 v, 5.5 v, 3 v, and 6 v. how much voltage is dropped across the fifth resistor?
3. determine the unspecified voltage drop(s) in each circuit of figure 5 - 76. show how to connect a voltmeter to measure each unknown voltage drop.
4. in the circuit of figure 5 - 77, determine the resistance of (r_{4}).
5. find (r_{1}), (r_{2}), and (r_{3}) in figure 5 - 78.

Explanation:

Step1: Recall Kirchhoff's Voltage Law

In a closed - loop circuit, the sum of the voltage sources equals the sum of the voltage drops.

Step2: Solve problem 30

The voltage drops across the three resistors are $V_1 = 5.5V$, $V_2=8.2V$, and $V_3 = 12.3V$. The voltage of the source $V_s$ to which these resistors are connected is the sum of the voltage drops. So, $V_s=V_1 + V_2+V_3=5.5 + 8.2+12.3=26V$.

Step3: Solve problem 31

The voltage of the source $V_s = 20V$. The voltage drops across four resistors are $V_1 = 1.5V$, $V_2 = 5.5V$, $V_3=3V$, and $V_4 = 6V$. Let the voltage drop across the fifth resistor be $V_5$. By Kirchhoff's Voltage Law, $V_s=V_1 + V_2+V_3+V_4+V_5$. Then $V_5=V_s-(V_1 + V_2+V_3+V_4)=20-(1.5 + 5.5+3+6)=4V$.

Step4: Solve problem 32 (circuit a)

The voltage of the source $V = 15V$. The known voltage drops are $V_1 = 2V$, $V_2=0.5V$, $V_3 = 1.5V$, and $V_4 = 3.2V$. Let the unknown voltage drop be $V_x$. By Kirchhoff's Voltage Law, $V=V_1 + V_2+V_3+V_4+V_x$. So, $V_x=V-(V_1 + V_2+V_3+V_4)=15-(2 + 0.5+1.5+3.2)=7.8V$. To measure the unknown voltage drop, connect the voltmeter in parallel across the component with the unknown voltage drop.

Step5: Solve problem 32 (circuit b)

Let the voltage of the source be $V_s$. The resistors are in series, and the ratio of the resistances is $R:4R:3R:2R$. The voltage drops are proportional to the resistances. Let the voltage drop across the first resistor be $V_1 = 8V$. The total resistance $R_{total}=R + 4R+3R+2R = 10R$. The voltage drop across the second resistor $V_2$ (unknown), using the ratio of resistances, $\frac{V_2}{V_1}=\frac{4R}{R}$, so $V_2 = 32V$. To measure the unknown voltage drop, connect the voltmeter in parallel across the corresponding resistor.

Answer:

1. $26V$
2. $4V$
3. (a) $7.8V$, connect voltmeter in parallel across the component with unknown voltage drop; (b) $32V$, connect voltmeter in parallel across the corresponding resistor.