Chapter 4, Problem 72E

Consider the LED circuit containing a red, green, and blue LED as shown in Fig. 4.89. The LEDs behave much like a voltage source resulting in the circuit in Fig. 4.89, where the light output from each LED will be proportional to the current flowing through the LED. (a) Calculate the current flowing through each LED (I_Red, I_Green, and I_Blue) if R₁ = R₂ = R₃ = 100 Ω. (b) Determine the resistor values R₁, R₂, and R₃ needed to ensure that the LEDs each have a current of 4 mA flowing through them.

FIGURE 4.89

To determine

Find the current flowing through the each LED in the circuit of Figure 4.89.

Answer to Problem 72E

The current flowing through the each LED in the circuit are $I_{Red}=16.96\text{mA}$, $I_{Green}=11.64\text{mA}$, and $I_{Blue}=1.493\text{mA}$.

Explanation of Solution

Given data:

Refer to Figure 4.89 in the textbook.

Calculation:

The given circuit is redrawn as shown in Figure 1.

Apply Kirchhoff’s voltage law for loop current $i_1$ in Figure 1.

$\begin{array}{l}50i_1+100\left(i_1-i_2\right)=5-1.8\\ 50i_1+100i_1-100i_2=3.2\end{array}$

$150i_1-100i_2=3.2$        (1)

Apply Kirchhoff’s voltage law for loop current $i_2$ in Figure 1.

$\begin{array}{l}10i_2+100\left(i_2-i_1\right)+100\left(i_2-i_3\right)=-2.2+1.8\\ 10i_2+100i_2-100i_1+100i_2-100i_3=-0.4\end{array}$

$-100i_1+210i_2-100i_3=-0.4$        (2)

Apply Kirchhoff’s voltage law for loop current $i_3$ in Figure 1.

$\begin{array}{l}10i_3+100i_3+100\left(i_3-i_2\right)=-3.2+2.2\\ 10i_3+100i_3+100i_3-100i_2=-1\end{array}$

$-100i_2+210i_3=-1$        (3)

Rearrange the equation (3) as follows,

$\begin{array}{l}100i_2=210i_3+1\\ i_2=\frac{210}{100}{i}_3+\frac{1}{100}\end{array}$

$i_2=2.1i_3+0.01$        (4)

Substitute equation (4) in equation (1).

$\begin{array}{l}150i_1-100\left(2.1i_3+0.01\right)=3.2\\ 150i_1-210i_3-1=3.2\\ 150i_1=3.2+210i_3+1\\ 150i_1=4.2+210i_3\end{array}$

Reduce the equation as follows,

$i_1=\frac{4.2}{150}+\frac{210}{150}{i}_3$

$i_1=0.028+1.4i_3$        (5)

Substitute equation (4), (5) in equation (2) to find the current $i_3$ in amperes.

$\begin{array}{l}-100\left(0.028+1.4i_3\right)+210\left(2.1i_3+0.01\right)-100i_3=-0.4\\ -2.8-140i_3+441i_3+2.1-100i_3=-0.4\\ -140i_3+441i_3-100i_3=-0.4-2.1+2.8\\ 201i_3=0.3\end{array}$

Reduce the equation as follows,

$\begin{array}{c}{i}_3=\frac{0.3}{201}\\ =1.493\times {10}^{-3}\text{A}\\ =1.493\text{mA}\left\{\because 1\text{mA}={10}^{-3}\right\}\end{array}$

Substitute $1.493\times {10}^{-3}$ for $i_3$ in equation (5) to find the current $i_1$ in amperes.

$\begin{array}{c}{i}_1=0.028+1.4\left(1.493\times {10}^{-3}\right)\\ =0.03009\times {10}^3\times {10}^{-3}\text{A}\\ =30.09\text{mA}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Substitute $1.493\times {10}^{-3}$ for $i_3$ in equation (4) to find the current $i_2$ in amperes.

$\begin{array}{c}{i}_2=2.1\left(1.493\times {10}^{-3}\right)+0.01\\ =0.01313\times {10}^3\times {10}^{-3}\text{A}\\ =13.13\text{mA}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

In Figure 1, the current $I_{Red}$ flowing through LED is calculated as follows.

$I_{Red}=i_1-i_2$        (6)

Substitute $30.09\text{mA}$ for $i_1$ and $13.13\text{mA}$ for $i_2$ to find the current $I_{Red}$ in amperes.

$\begin{array}{c}{I}_{Red}=30.09\text{mA}-13.13\text{mA}\\ =16.96\text{mA}\end{array}$

In Figure 1, the current $I_{Green}$ flowing through LED is calculated as follows.

$I_{Green}=i_2-i_3$        (7)

Substitute $1.493\text{mA}$ for $i_3$ and $13.13\text{mA}$ for $i_2$ to find the current $I_{Green}$ in amperes.

$\begin{array}{c}{I}_{Green}=13.13\text{mA}-1.493\text{mA}\\ =11.64\text{mA}\end{array}$

In Figure 1, the current $I_{Blue}$ flowing through LED is calculated as follows.

$I_{Blue}=i_3$        (8)

Substitute $1.493\text{mA}$ for $i_3$ to find the current $I_{Blue}$ in amperes.

$I_{Blue}=1.493\text{mA}$

Conclusion:

Thus, the current flowing through the each LED in the circuit are $I_{Red}=16.96\text{mA}$, $I_{Green}=11.64\text{mA}$, and $I_{Blue}=1.493\text{mA}$.

To determine

Find the resistance values of $R_1$, $R_2$, and $R_3$ in the circuit of Figure 4.89.

Answer to Problem 72E

The resistance values of $R_1$, $R_2$, and $R_3$ in the circuit are $650\Omega$, $530\Omega$, and $270\Omega$ respectively.

Explanation of Solution

Given data:

The current,

$I_{Red}=I_{Green}=I_{Blue}=4\text{mA}$

Calculation:

The given circuit is redrawn as shown in Figure 2.

Refer to Part (a),

Substitute $4\text{mA}$ for $I_{Blue}$ in equation (8) to find the current $i_3$ in amperes.

$i_3=4\text{mA}$

Substitute $4\text{mA}$ for $I_{Green}$ and $4\text{mA}$ for $i_3$ in equation (7) to find the current $i_2$ in amperes.

$\begin{array}{l}4\text{mA}=i_2-4\text{mA}\\ i_2=4\text{mA}+4\text{mA}\\ i_2=8\text{mA}\end{array}$

Substitute $4\text{mA}$ for $I_{Red}$ and $8\text{mA}$ for $i_2$ in equation (6) to find the current $i_1$ in amperes.

$\begin{array}{l}4\text{mA}=i_1-8\text{mA}\\ i_1=4\text{mA}+8\text{mA}\\ i_1=12\text{mA}\end{array}$

Apply Kirchhoff’s voltage law for loop current $i_1$ in Figure 2.

$50i_1+R_1{I}_{Red}=5-1.8$

Substitute $12\text{m}$ for $i_1$ and $4\text{m}$ for $I_{Red}$ to find the resistance $R_1$.

$\begin{array}{l}50\left(12\text{m}\right)+R_1\left(4\text{m}\right)=5-1.8\\ 50\left(12\times {10}^{-3}\right)+R_1\left(4\times {10}^{-3}\right)=3.2\left\{\because 1\text{m}={10}^{-3}\right\}\\ R_1\left(4\times {10}^{-3}\right)=3.2-50\left(12\times {10}^{-3}\right)\\ R_1=\frac{3.2-50\left(12\times {10}^{-3}\right)}{4\times {10}^{-3}}\end{array}$

Reduce the equation as follows,

$R_1=650\Omega$

Apply Kirchhoff’s voltage law for loop current $i_2$ in Figure 2.

$\begin{array}{l}10i_2+R_1\left(i_2-i_1\right)+R_2{I}_{Green}=-2.2+1.8\\ \end{array}$

Substitute $12\text{m}$ for $i_1$, $8\text{m}$ for $i_2$, $650$ for $R_1$, and $4\text{m}$ for $I_{Green}$ to find the resistance $R_2$.

$\begin{array}{l}10\left(8\text{m}\right)+650\left(8\text{m}-12\text{m}\right)+R_2\left(4\text{m}\right)=-2.2+1.8\\ 10\left(8\times {10}^{-3}\right)+650\left(8\times {10}^{-3}-12\times {10}^{-3}\right)+R_2\left(4\times {10}^{-3}\right)=-0.4\left\{\because 1\text{m}={10}^{-3}\right\}\\ 10\left(8\times {10}^{-3}\right)-650\left(4\times {10}^{-3}\right)+R_2\left(4\times {10}^{-3}\right)=-0.4\\ R_2\left(4\times {10}^{-3}\right)=-0.4-10\left(8\times {10}^{-3}\right)+650\left(4\times {10}^{-3}\right)\end{array}$

Reduce the equation as follows,

$\begin{array}{c}{R}_2=\frac{-0.4-10\left(8\times {10}^{-3}\right)+650\left(4\times {10}^{-3}\right)}{4\times {10}^{-3}}\\ =530\Omega \end{array}$

Apply Kirchhoff’s voltage law for loop current $i_3$ in Figure 2.

$10I_{Blue}+R_3{I}_{Blue}+R_2\left(i_3-i_2\right)=-3.2+2.2$

Substitute $4\text{m}$ for $i_3$, $8\text{m}$ for $i_2$, $530$ for $R_2$, and $4\text{m}$ for $I_{Blue}$ to find the resistance $R_3$.

$\begin{array}{l}10\left(4\text{m}\right)+R_3\left(4\text{m}\right)+530\left(4\text{m}-8\text{m}\right)=-3.2+2.2\\ 10\left(4\times {10}^{-3}\right)+R_3\left(4\times {10}^{-3}\right)+530\left(-4\times {10}^{-3}\right)=-1\left\{\because 1\text{m}={10}^{-3}\right\}\\ R_3\left(4\times {10}^{-3}\right)=-1-530\left(-4\times {10}^{-3}\right)-10\left(4\times {10}^{-3}\right)\\ R_3=\frac{-1-530\left(-4\times {10}^{-3}\right)-10\left(4\times {10}^{-3}\right)}{4\times {10}^{-3}}\end{array}$

Reduce the equation as follows,

$R_3=270\Omega$

Conclusion:

Thus, the resistance values of $R_1$, $R_2$, and $R_3$ in the circuit are $650\Omega$, $530\Omega$, and $270\Omega$ respectively.