Questions: Consider the circuit shown, where V1=1.8 V, V2=2.40 V, R1=1.7 k Ohm, R2=1.6 k Ohm, and R3=1.5 k Ohm. - Part (a) What is the current through resistor R1 in milliamperes? I1=0.8750 mA Part (b) What is the current through resistor R2 in milliamperes? I2= square mA

Consider the circuit shown, where V1=1.8 V, V2=2.40 V, R1=1.7 k Ohm, R2=1.6 k Ohm, and R3=1.5 k Ohm.
- Part (a) What is the current through resistor R1 in milliamperes?
I1=0.8750 mA

Part (b)
What is the current through resistor R2 in milliamperes?
I2= square mA

Transcript text: Consider the circuit shown, where $V_{1}=1.8 \mathrm{~V}, V_{2}=2.40 \mathrm{~V}, R_{1}=1.7 \mathrm{k} \Omega, R_{2}=1.6 \mathrm{k} \Omega$, and $R_{3}=1.5$ $\mathrm{k} \Omega$. - Part (a) What is the current through resistor $R_{1}$ in milliamperes? \[ I_{1}=0.8750 \mathrm{~mA} \] Part (b) What is the current through resistor $R_{2}$ in milliamperes? \[ I_{2}=\square \mathrm{mA} \]

Solution

Solution Steps

Step 1: Identify the given values

We are given the following values:

$ V_1 = 1.8 \, \text{V} $
$ V_2 = 2.4 \, \text{V} $
$ R_1 = 1.7 \, \text{k} \Omega $
$ R_2 = 1.6 \, \text{k} \Omega $
$ R_3 = 1.5 \, \text{k} \Omega $
$ I_1 = 0.8750 \, \text{mA} $

Step 2: Calculate the voltage drop across $ R_1 $

Using Ohm's Law, the voltage drop across $ R_1 $ is: \[ V_{R1} = I_1 \times R_1 \] Substituting the given values: \[ V_{R1} = 0.8750 \, \text{mA} \times 1.7 \, \text{k} \Omega = 0.8750 \times 1.7 = 1.4875 \, \text{V} \]

Step 3: Determine the voltage at the node between $ R_1 $ and $ R_2 $

The voltage at the node between $ R_1 $ and $ R_2 $ is: \[ V_{\text{node}} = V_1 - V_{R1} \] Substituting the values: \[ V_{\text{node}} = 1.8 \, \text{V} - 1.4875 \, \text{V} = 0.3125 \, \text{V} \]

Step 4: Calculate the current through $ R_2 $

Using Ohm's Law, the current through $ R_2 $ is: \[ I_2 = \frac{V_{\text{node}}}{R_2} \] Substituting the values: \[ I_2 = \frac{0.3125 \, \text{V}}{1.6 \, \text{k} \Omega} = \frac{0.3125}{1600} = 0.0001953 \, \text{A} = 0.1953 \, \text{mA} \]