Solving Circuits

Ben Bitdiddle is trying his hand at solving circuits. For each of the circuits below, Ben has written out his reasoning, broken down into a number of steps. For each circuit:

• if Ben made a mistake, write the number of the step where Ben made his first mistake in that question, write a brief explanation of why it was a mistake, and complete the solution correctly.
• if Ben solved the circuit correctly, say so.

Answer the questions below, and upload your answers as a single PDF file. Please do not include any identifying information in your submission so that we can grade the submissions anonymously.

Throughout this problem, note that x~||~y represents the parallel combination of resistors x and y.

1) Circuit 1

In the first circuit, Ben is asked to solve for the voltage v_1 in the circuit below:

Ben takes the following steps when solving this circuit:

1. Because the 20V source is connected to the 4\Omega resistor, Ben knows that the current flowing through that resistor must be 20{\rm V} / 4\Omega = 5{\rm A}.
2. By KCL, that same current must be flowing through the 1\Omega resistor.
3. 5{\rm A} through the 1\Omega resistor means that the voltage drop across the resistor must be 5{\rm A}\cdot 1\Omega = 5{\rm V}.
4. The voltage drop across the 1\Omega resistor is exactly the v_1 Ben wanted to compute, so v_1 = 5{\rm V}.

2) Circuit 2

Ben is asked to solve for the voltage v_2 in the circuit below:

Ben takes the following steps when solving this circuit:

1. Ben notes that the two resistors are connected in parallel and decides to replace them with a single equivalent resistor whose resistance is 4\Omega || 12\Omega = 3\Omega.
2. The current through the combination is 9{\rm V} / 3\Omega = 3{\rm A}.
3. Re-expanding that combination, the current flowing through each resistor must be 3{\rm A}.
4. v_2 is the drop across the 12\Omega resistor. Because 3{\rm A} is flowing through it, v_2 = 12\Omega\cdot 3{\rm A} = 36{\rm V}.

3) Circuit 3

Ben is asked to solve for the voltage v_3 in the circuit below:

Ben takes the following steps when solving this circuit:

1. Ben notes that the two resistors are connected in series and decides to replace them with a single equivalent resistor whose resistance is 5\Omega + 1\Omega = 6\Omega.
2. The current through the combination is 30{\rm V} / 6\Omega = 5{\rm A}.
3. Re-expanding that combination, the current flowing through each resistor must be 5{\rm A}.
4. v_3 is the drop across the 1\Omega resistor. Because 5{\rm A} is flowing through it, v_3 = 1\Omega\cdot 5{\rm A} = 5{\rm V}.

4) Circuit 4

Ben is asked to solve for the voltage v_4 in the circuit below:

Ben takes the following steps when solving this circuit:

1. Ben notes that there is a 2{\rm V} drop across the total resistance on the top, which consists of the 2\Omega resistor in series with the parallel combination of 6\Omega and 10\Omega.
2. The equivalent resistance of the parallel combination of 6\Omega and 10\Omega is 6\Omega||10\Omega = 8\Omega.
3. The total resistance is, therefore, 2\Omega + 8\Omega = 10\Omega.
4. By Ohm's Law, the total current is 2{\rm V} / 10\Omega = 0.2{\rm A}, flowing left-to-right.
5. The voltage drop across the parallel combination, then, is 0.2{\rm A}\cdot 8\Omega = 1.6{\rm V}.
6. The total voltage drop v_4 is 8{\rm V} + 1.6{\rm V} = 9.6{\rm V}.

5) Circuit 5

Ben is asked to solve for v_5 in the circuit below:

Ben takes the following steps when solving this circuit:

1. Ben chooses the bottom node to be his 0{\rm V} reference node.
2. Given the values of the voltage sources, he labels the top-left node as 9{\rm V} relative to the reference potential and the middle top node as 12{\rm V} relative to the reference potential.
3. Thus, the v_5 we are interested in is 9{\rm V} - 12{\rm V} = -3{\rm V}.

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