Questions: Given the following circuit where V5=5, IT=20 mA, and P2=42 mW, find P1, the power dissipated in resistor R1.

Transcript text: Given the following circuit where $V_{5}=5, I_{T}=20 \mathrm{~mA}$, and $P_{2}=42 \mathrm{~mW}$, find $P_{1}$, the power dissipated in resistor $R_{1}$.

Solution

Solution Steps

Step 1: Identify Given Values

$ V_s = 5 \, V $
$ I_T = 20 \, mA $
$ P_2 = 42 \, mW $

Step 2: Calculate Resistance $ R_2 $

Using the power formula $ P = \frac{V^2}{R} $, we can rearrange to find $ R $: \[ R_2 = \frac{V_s^2}{P_2} = \frac{5^2}{42 \times 10^{-3}} = \frac{25}{0.042} \approx 595.24 \, \Omega \]

Step 3: Calculate Current Through $ R_2 $

Using Ohm's Law $ I = \frac{V}{R} $: \[ I_2 = \frac{V_s}{R_2} = \frac{5}{595.24} \approx 8.4 \, mA \]

Step 4: Calculate Current Through $ R_1 $

Since the total current $ I_T $ is the sum of the currents through $ R_1 $ and $ R_2 $: \[ I_1 = I_T - I_2 = 20 \, mA - 8.4 \, mA = 11.6 \, mA \]

Step 5: Calculate Power Dissipated in $ R_1 $

Using the power formula $ P = I^2 R $ and Ohm's Law $ V = IR $: \[ P_1 = V_s \times I_1 = 5 \, V \times 11.6 \, mA = 5 \times 0.0116 = 0.058 \, W = 58 \, mW \]

Final Answer

\[ P_1 = 58 \, mW \]

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