Answer
See the graph below and see the detailed answer.
Work Step by Step
The author asks us to draw the diagram of a combination of two resistors that were connected in series and then connected in parallel.
See the graph below.
Also, he needs us to complete the given table.
Answers to the first row in the table;
$\text{Equivalent resistance:}$
1) In series: $$\boxed{R_{eq}=R_1+R_2}$$
2) In parallel: $$R_{eq}=\left(\dfrac{1}{R_1}+\dfrac{1}{R_2} \right)^{-1}=\left(\dfrac{R_1+R_2}{R_1R_2} \right)^{-1}$$
$$\boxed{R_{eq}=\dfrac{R_1R_2}{R_1+R_2}}$$
Answers to the second row in the table;
$\text{Current through equivalent resistance:}$
1) In series: $$V_{AB}=IR_{eq}$$
So, $$I=\dfrac{V_{AB}}{R_{eq}}$$
$$\boxed{ I=\dfrac{V_B-V_A}{R_1+R_2}}$$
2) In parallel:
$$I=\dfrac{V_{AB}}{R_{eq}}=\dfrac{V_B-V_A}{\dfrac{R_1R_2}{R_1+R_2}}$$
$$\boxed{ I=\dfrac{(V_B-V_A)(R_1+R_2)}{R_1R_2}}$$
Answers to the third row in the table;
$\text{Voltage across equivalent resistance:}$
1) In series:
$$\boxed{V_{eq}=V_B-V_A}$$
2) In parallel:
$$\boxed{V_{eq}=V_B-V_A}$$
Answers to the fourth row in the table;
$\text{Voltage across the pair of resistors:}$
1) In series:
$$\boxed{V_{eq}=V_B-V_A}$$
2) In parallel:
$$\boxed{V_{eq}=V_B-V_A}$$
Answers to the fifth row in the table;
$\text{Voltage across each resistor:}$
1) In series:
$$V_1=IR_1=\dfrac{V_B-V_A}{R_1+R_2}R_1$$
$$\boxed{V_1=\dfrac{V_B-V_A}{R_1+R_2}R_1}$$
$$V_2=IR_2=\dfrac{V_B-V_A}{R_1+R_2}R_2$$
$$\boxed{V_2=\dfrac{V_B-V_A}{R_1+R_2}R_2}$$
2) In parallel:
$$\boxed{V_1=V_2=V_B-V_A}$$
Answers to the sixth row in the table;
$\text{Voltage at a point between the resistors :}$
1) In series:
Let's assume that the point between the two series is C.
So, the voltage across the first resistor is given by
$$V_1=V_B-V_C$$
Thus,
$$V_C=V_B-V_1=V_B-\dfrac{V_B-V_A}{R_1+R_2} R_1$$
$$V_C=\dfrac{V_BR_1+V_BR_2-(V_BR_1-V_AR_1)}{R_1+R_2}$$
$$V_C=\dfrac{V_BR_1+V_BR_2- V_BR_1+V_AR_1 }{R_1+R_2}$$
$$\boxed{V_C=\dfrac{ V_BR_2 +V_AR_1 }{R_1+R_2}}$$
2) In parallel: $\text{Not applicable}$.
Answers to the seventh row in the table;
$\text{Current through each resistor:}$
1) In series:
$$\boxed{I_1=I_2 =\dfrac{V_B-V_A}{R_1+R_2} }$$
2) In parallel:
$$I_1=\dfrac{V_1}{R_1}$$
$$\boxed{I_1 =\dfrac{V_B-V_A}{R_1}}$$
$$I_2=\dfrac{V_2}{R_2}$$
$$\boxed{I_2 =\dfrac{V_B-V_A}{R_2}}$$
