CURRENT BYU 7

Physics Solution – Question 7

Solution to Question 7

1. Circuit Analysis & Topology

We analyze the circuit based on the provided polarity annotations.

Polarity Check (Figure 2):

  • Battery A: Positive (+) on Left.
  • Battery B: Positive (+) on Left. (Same orientation as A).
  • Battery C: Positive (+) on Right. (Opposite orientation to A and B).

Implication: At the Right Junction (let’s call it node $J_R$), the Positive terminals of Battery A and Battery C are connected. The Negative terminal of Battery B is connected to this same junction.

+ – – + Figure 1: Series Connection A B C $I = 3.0$ A Figure 2: Mixed Connection A B C $I_A$ $I_B$ $I_C$ Junction R

Figure 2 Diagram matches annotation: A and B are Left(+). C is Right(+).
Result: A and C push current towards the Right Junction. B draws current from it.

2. Solving Figure 1 (Series)

Three identical batteries in series. $$ \frac{3\mathcal{E}}{3r} = \frac{\mathcal{E}}{r} = 3.0 \text{ A} \quad \text{— (1)} $$

3. Solving Figure 2 (Mixed)

We use Nodal Analysis. Let the potential at the Left Junction be $0\text{ V}$ and the potential at the Right Junction be $V$.

  • Branch A: Connected via the outer loop. Battery A is oriented Left(+), Right(-). The Left side (+) connects to the Right Junction (via the long wire). The Right side (-) connects to the Left Junction.
    Analysis: Battery A acts to raise the potential of the Right Junction. It pushes current into the node. $$ I_{A,in} = \frac{\mathcal{E} – V}{r} $$
  • Branch C: Connected between the junctions. Battery C is oriented Right(+), Left(-). The Positive connects to the Right Junction.
    Analysis: Battery C also acts to raise the potential of the Right Junction. It pushes current into the node. $$ I_{C,in} = \frac{\mathcal{E} – V}{r} $$
  • Branch B: Connected between the junctions. Battery B is oriented Left(+), Right(-). The Negative connects to the Right Junction.
    Analysis: Battery B opposes the potential at the Right Junction. Current flows from the node into B (Reverse direction through the branch). $$ I_{B,out} = \frac{V – (-\mathcal{E})}{r} = \frac{V + \mathcal{E}}{r} $$

Apply Kirchhoff’s Current Law (KCL) at Right Junction:

Current In = Current Out

$$ I_A + I_C = I_B $$ $$ \frac{\mathcal{E} – V}{r} + \frac{\mathcal{E} – V}{r} = \frac{V + \mathcal{E}}{r} $$ $$ 2\mathcal{E} – 2V = V + \mathcal{E} $$ $$ \mathcal{E} = 3V \implies V = \frac{\mathcal{E}}{3} $$

Calculate Currents:

Substitute $V = \mathcal{E}/3$ and use the known ratio $\mathcal{E}/r = 3.0\text{ A}$:

  • $$ I_A = \frac{\mathcal{E} – \mathcal{E}/3}{r} = \frac{2}{3} \left(\frac{\mathcal{E}}{r}\right) = \frac{2}{3} (3.0) = 2.0 \text{ A} $$
  • $$ I_C = \frac{\mathcal{E} – \mathcal{E}/3}{r} = \frac{2}{3} \left(\frac{\mathcal{E}}{r}\right) = \frac{2}{3} (3.0) = 2.0 \text{ A} $$
  • $$ I_B = \frac{\mathcal{E}/3 + \mathcal{E}}{r} = \frac{4}{3} \left(\frac{\mathcal{E}}{r}\right) = \frac{4}{3} (3.0) = 4.0 \text{ A} $$

Verification: $2.0 + 2.0 = 4.0$. The physics holds.

Final Answer

The currents through the batteries are:

Battery A: 2.0 A
Battery B: 4.0 A
Battery C: 2.0 A

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