4_example_KVL_01

4_example_KVL_01 - KIRCHOFF’S VOLTAGE LAW: EXAMPLE 1...

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Unformatted text preview: KIRCHOFF’S VOLTAGE LAW: EXAMPLE 1 GIVEN: Consider the circuit shown, where R1=20Ω R2=40Ω V1=20V REQUIRED: (a) The current through R1. (b) The current through R2. (c) The current leaving the voltage source. SOLUTION: (a) First, we identify the loops in the circuit. As shown below, we can choose any two of the three loops. Kirchoff’s Voltage Law: The sum of the voltage drops around any closed loop is zero. Applying KVL to Loop 1: Applying Ohm’s Law to R1 yields . . . Energy is always dissipated by resistors. Voltage decreases in the direction of positive current flow. When applying KVL, assume voltage decreases across a resistor in the assumed direction of current flow (and increases opposite the direction of current flow). (b) To find IR2, we can use either Loop 2 or Loop 3. Let’s apply KVL to Loop 2. Applying Ohm’s Law: (c) First determine the equivalent resistance of the circuit: Applying Ohm’s Law: DISCUSSION: • • • Kirchhoff’s Voltage Law is a useful tool for circuit analysis. The voltage drops across both resistors were equal even though the currents were different. The voltage drop is ALWAYS the same across two resistors in parallel. Notice that IR1 + IR2 = I. This means that current is conserved. We will learn later that this is an application of Kirchhoff’s Current Law (KCL). ...
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This note was uploaded on 02/08/2011 for the course ENGR 120 taught by Professor Hall during the Winter '08 term at LA Tech.

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