Chapter 28, Problem 28.11P

A battery with ε = 6.00 V and no internal resistance supplies current to the circuit shown in Figure P27.9. When the double-throw switch S is open as shown in the figure, the current in the battery is 1.00 mA. When the switch is closed in position a, the current in the battery is 1.20 mA. When the switch is closed in position b, the current in the battery is 2.00 mA. Find the resistances (a) R₁, (b) R₂, and (c) R₃.

Figure P27.9 Problems 9 and 10.

To determine

The value of the resistance $R_1$.

Answer to Problem 28.11P

The value of the resistance $R_1$ is $1.00\text{k}\Omega$.

Explanation of Solution

Given information: Emf across the battery is $6.00\text{V}$, current in the battery when the switch S is open is $1.00\text{mA}$, current in the battery when the switch is closed in position $a$ is $1.20\text{mA}$, current in the battery when the switch is closed in position $b$ is $2.00\text{mA}$.

Explanation:

When the switch S is open, then the three resistors $R_1,R_2\text{and}{R}_3$ are in series.

Formula to calculate the equivalent resistance across the circuit, when the switch S is open.

$\begin{array}{c}V=I_1{R}_{\text{eq1}}\\ R_{\text{eq1}}=\frac{V}{I_1}\end{array}$ (1)

Here,

$R_{\text{eq1}}$ is the equivalent resistance across the circuit, when the switch S is open.

$V$ is the voltage across the battery.

$I_1$ is the current in the battery when the switch S is open.

As the total emf across the battery is equal to the voltage across the battery.

$\epsilon =V$

Here,

$\epsilon$ is the total emf across the battery.

Substitute $\epsilon$ for $V$ in equation (1) to find $R_{\text{eq1}}$,

$R_{\text{eq1}}=\frac{\epsilon }{I_1}$ (2)

Formula to calculate the equivalent resistance across the circuit, when the switch S is open.

$R_{\text{eq1}}=R_1+R_2+R_3$ (3)

Here,

$R_1$ is the value of resistance 1.

$R_2$ is the value of resistance 2.

$R_3$ is the value of resistance 3.

Substitute $\frac{\epsilon }{I_1}$ for $R_{\text{eq1}}$ in equation (3),

$R_1+R_2+R_3=\frac{\epsilon }{I_1}$ (4)

Substitute $6.00\text{V}$ for $\epsilon$, $1.00\text{mA}$ for $I_1$ in equation (4) to find $R_1+R_2+R_3$,

$\begin{array}{c}{R}_1+R_2+R_3=\frac{6.00\text{V}}{1.00\text{mA}\times \frac{1\text{A}}{1000\text{mA}}}\\ =6000\Omega \end{array}$ (5)

When the switch is closed in position $a$, then the three resistors $R_1,R_P\text{and}{R}_3$ are in series.

From equation (2), formula to calculate the equivalent resistance across the circuit, when the switch is closed in position $a$,

$R_{\text{eq2}}=\frac{\epsilon }{I_2}$ (6)

Here,

$R_{\text{eq2}}$ is the equivalent resistance across the circuit, when the switch is closed in position $a$.

$I_2$ is the current in the battery when the switch is closed in position $a$.

Formula to calculate the resistance when the resistors are connected in parallel.

$R_P=\frac{R_2{R}_2}{R_2+R_2}$

From equation (3), formula to calculate the equivalent resistance across the circuit, when the switch is closed in position $a$.

$R_{\text{eq2}}=R_1+R_P+R_3$ (7)

Substitute $\frac{R_2{R}_2}{R_2+R_2}$ for $R_P$ in equation (7),

$R_{\text{eq2}}=R_1+\frac{R_2{R}_2}{R_2+R_2}+R_3$ (8)

Substitute $\frac{\epsilon }{I_2}$ for $R_{\text{eq2}}$ in equation (8),

$R_1+\frac{R_2{R}_2}{R_2+R_2}+R_3=\frac{\epsilon }{I_2}$ (9)

Substitute $6.00\text{V}$ for $\epsilon$, $1.20\text{mA}$ for $I_2$ in equation (9) to find $R_1+\frac{R_2{R}_2}{R_2+R_2}+R_3$,

$\begin{array}{c}{R}_1+\frac{R_2{R}_2}{R_2+R_2}+R_3=\frac{6.00\text{V}}{1.20\text{mA}\times \frac{1\text{A}}{1000\text{mA}}}\\ R_1+\frac{R_2}{2}+R_3=5000\Omega \end{array}$ (10)

When the switch is closed in position $b$, then the $R_3$ resistor is short circuited and the two resistors $R_1\text{and}{R}_2$ are in series.

From equation (2), formula to calculate the equivalent resistance across the circuit, when the switch is closed in position $b$,

$R_{\text{eq3}}=\frac{\epsilon }{I_3}$ (11)

Here,

$R_{\text{eq3}}$ is the equivalent resistance across the circuit, when the switch is closed in position $b$.

$I_3$ is the current in the battery when the switch is closed in position $a$.

From equation (3), formula to calculate the equivalent resistance across the circuit, when the switch is closed in position $b$.

$R_{\text{eq3}}=R_1+R_2$ (12)

Substitute $\frac{\epsilon }{I_3}$ for $R_{\text{eq3}}$ in equation (12),

$R_1+R_2=\frac{\epsilon }{I_3}$ (13)

Substitute $6.00\text{V}$ for $\epsilon$, $2.00\text{mA}$ for $I_3$ in equation (3) to find $R_1+R_2$,

$\begin{array}{c}{R}_1+R_2=\frac{6.00\text{V}}{2.00\text{mA}\times \frac{1\text{A}}{1000\text{mA}}}\\ =3000\Omega \end{array}$ (14)

Subtract equation (14) from (5) to find $R_3$,

$\begin{array}{c}{R}_1+R_2+R_3-\left(R_1+R_2\right)=6000\Omega -3000\Omega \\ R_3=3000\Omega \\ =3.00\text{k}\Omega \end{array}$ (15)

Thus, the value of the resistance $R_3$ is $3.00\text{k}\Omega$.

Subtract equation (14) from (10) to find $R_2$,

$\begin{array}{c}{R}_1+\frac{R_2}{2}+R_3-\left(R_1+R_2\right)=5000\Omega -3000\Omega \\ R_3-\frac{R_2}{2}=2000\Omega \\ =2.00\text{k}\Omega \end{array}$ (16)

Substitute $3.00\text{k}\Omega$ for $R_{\text{3}}$ in equation (16) to find $R_2$,

$\begin{array}{c}3.00\text{k}\Omega -\frac{R_2}{2}=2.00\text{k}\Omega \\ \frac{R_2}{2}=1.00\text{k}\Omega \\ R_2=2.00\text{k}\Omega \end{array}$ (17)

Thus, the value of the resistance $R_2$ is $2.00\text{k}\Omega$.

Substitute $3.00\text{k}\Omega$ for $R_{\text{3}}$, $2.00\text{k}\Omega$ for $R_2$ in equation (5) to find $R_1$,

$\begin{array}{c}{R}_1+2.00\text{k}\Omega +3.00\text{k}\Omega =6000\Omega \\ R_1=6000\Omega \times \frac{1\text{k}\Omega }{1000\Omega }-2.00\text{k}\Omega -3.00\text{k}\Omega \\ R_1=1.00\text{k}\Omega \end{array}$

Thus, the value of the resistance $R_1$ is $1.00\text{k}\Omega$.

Conclusion:

Therefore, the value of the resistance $R_1$ is $1.00\text{k}\Omega$.

To determine

The value of the resistance $R_2$.

Answer to Problem 28.11P

The value of the resistance $R_2$ is $2.00\text{k}\Omega$.

Explanation of Solution

Given information: Emf across the battery is $6.00\text{V}$, current in the battery when the switch S is open is $1.00\text{mA}$, current in the battery when the switch is closed in position $a$ is $1.20\text{mA}$, current in the battery when the switch is closed in position $b$ is $2.00\text{mA}$.

Explanation:

From part (a) equation (17), the value of resistance $R_2$ is $2.00\text{k}\Omega$.

Thus, the value of the resistance $R_2$ is $2.00\text{k}\Omega$.

Conclusion:

Therefore, the value of the resistance $R_2$ is $2.00\text{k}\Omega$.

To determine

The value of the resistance $R_3$.

Answer to Problem 28.11P

The value of the resistance $R_3$ is $3.00\text{k}\Omega$.

Explanation of Solution

Given information: Emf across the battery is $6.00\text{V}$, current in the battery when the switch S is open is $1.00\text{mA}$, current in the battery when the switch is closed in position $a$ is $1.20\text{mA}$, current in the battery when the switch is closed in position $b$ is $2.00\text{mA}$.

Explanation:

From part (a) equation (15), the value of resistance $R_3$ is $3.00\text{k}\Omega$.

Thus, the value of the resistance $R_3$ is $3.00\text{k}\Omega$.

Conclusion:

Therefore, the value of the resistance $R_3$ is $3.00\text{k}\Omega$.