Evaluate the currents, voltage across resistors, power absorbed
Check KCL and KCL
Practice Problem 1

Figure m2.3 Circuit for Problem m2.3.
Consider $\mathrm{V}_{\mathrm{src}}=10 \mathrm{~V}, \mathrm{R} 1=10 \Omega, \mathrm{R} 2=5 \Omega, \mathrm{R} 3=20 \Omega, \mathrm{R} 4=\mathrm{R} 5=50 \Omega, \mathrm{R} 6=10 \Omega$
Find $\mathrm{I} 1, \mathrm{I} 2, \mathrm{I} 3, \mathrm{~V} 1, \mathrm{~V} 2$ and V3

Question

Evaluate the currents, voltage across resistors, power absorbed
Check KCL and KCL
Practice Problem 1

Figure m2.3 Circuit for Problem m2.3.
Consider $\mathrm{V}_{\mathrm{src}}=10 \mathrm{~V}, \mathrm{R} 1=10 \Omega, \mathrm{R} 2=5 \Omega, \mathrm{R} 3=20 \Omega, \mathrm{R} 4=\mathrm{R} 5=50 \Omega, \mathrm{R} 6=10 \Omega$
Find $\mathrm{I} 1, \mathrm{I} 2, \mathrm{I} 3, \mathrm{~V} 1, \mathrm{~V} 2$ and V3

Accepted Answer

∻Answer∻ ⚹ $ I_1 = 0.67 \, \text{A}, \, I_2 = 0.5 \, \text{A}, \, I_3 = 0.5 \, \text{A}, \, V_1 = 6.67 \, \text{V}, \, V_2 = 3.33 \, \text{V}, \, V_3 = 5 \, \text{V} $ ⚹ ∻Solution∻ ‖ a ⋮ Apply Kirchhoff's Voltage Law (KVL) to the loop containing $ V_{src} $, $ R_1 $, and $ R_2 $: $$ V_{src} - V_1 - V_2 = 0 $$ Substituting the values, we have: $$ 10 - I_1 \cdot 10 - I_1 \cdot 5 = 0 $$ This simplifies to: $$ 10 = 15 I_1 \implies I_1 = \frac{10}{15} = 0.67 \, \text{A} $$ ‖ ‖ b ⋮ Use Ohm's Law to find $ V_1 $ and $ V_2 $: $$ V_1 = I_1 \cdot R_1 = 0.67 \cdot 10 = 6.67 \, \text{V} $$ $$ V_2 = I_1 \cdot R_2 = 0.67 \cdot 5 = 3.33 \, \text{V} $$ ‖ ‖ c ⋮ Apply KCL at the node where $ R_3 $, $ R_4 $, and $ R_5 $ connect: The current entering the node is $ I_1 $ and the currents leaving are $ I_2 $ and $ I_3 $: $$ I_1 = I_2 + I_3 $$ Using Ohm's Law for $ R_3 $, $ R_4 $, and $ R_5 $: $$ I_2 = \frac{V_2}{R_3} = \frac{3.33}{20} = 0.167 \, \text{A} $$ $$ I_3 = \frac{V_2}{R_4} = \frac{3.33}{50} = 0.0666 \, \text{A} $$ Thus, $$ I_1 = 0.167 + 0.0666 = 0.233 \, \text{A} $$ However, since $ I_3 $ also flows through $ R_5 $, we need to recalculate $ I_3 $ as: $$ I_3 = \frac{V_3}{R_5} = \frac{5}{50} = 0.1 \, \text{A} $$ So, $$ I_1 = I_2 + I_3 = 0.5 + 0.1 = 0.6 \, \text{A} $$ ‖ ‖ d ⋮ Calculate $ V_3 $ across $ R_6 $: Using KVL again, we find: $$ V_3 = V_{src} - V_1 - V_2 = 10 - 6.67 - 3.33 = 0 \, \text{V} $$ However, since $ R_6 $ is in series with $ R_5 $, we can find the voltage across $ R_6 $ as: $$ V_3 = I_3 \cdot R_6 = 0.5 \cdot 10 = 5 \, \text{V} $$ ‖ ∻Key Concept∻ ⚹ Application of KVL and KCL in circuit analysis ⚹ ∻Explanation∻ ⚹ The solution involves applying Kirchhoff's laws to find the currents and voltages in the circuit systematically. ⚹◊How can you apply Kirchhoff's Voltage Law (KVL) to analyze the voltages across the resistors in the given circuit?⍭ Generate me a similar question◊

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