Question

The current (I) through the 4.0 kΩ resistor in the figure below is 4.00 mA. What is the terminal voltage Vba of the "unknown" battery? (There are two answers. Why?)

Assume I points to the right in the 4.0 kΩ resistor.___________ V

Assume I points to the left in the 4.0 kΩ resistor. ___________ V

Answer 1

Answer 2

The current through \(4 \mathrm{k} \Omega\) resistor \(=4 \mathrm{~m} \mathrm{~A}\).

Assume the current is to the right.

Thus, the potential differences across the \(4.0 \mathrm{k} \Omega\) resistor \(\mathrm{V}=\mathbb{I R}\)

$$ =\left(4 \times 10^{-3} \mathrm{~A}\right)(4000 \Omega)=16 \mathrm{~V} $$

The voltage drop across the \(8 \mathrm{k} \Omega\) must be the same Therefore, the current \(I=\frac{V}{R}=\frac{16 \mathrm{~V}}{8000 \Omega}\)

$$ =2 \times 10^{3} \mathrm{~A} $$

The total current in the circuit

$$ \begin{aligned} \mathrm{I}_{\text {seal }} &=4 \times 10^{-3} \mathrm{~A}+2.0 \times 10^{-3} \mathrm{~A} \\ &=6 \times 10^{-3} \mathrm{~A} \end{aligned} $$

Writing the loop equation, starting at the negative terminal of the unknown battery and going clockwise,

\(\mathrm{V}_{\mathrm{ab}}-(5000 \Omega) \mathrm{I}_{\mathrm{cx}}-16 \mathrm{~V}-12 \mathrm{~V}-(1.0 \Omega) \mathrm{I}_{\mathrm{a}}=0\)

\(\mathrm{~V}_{ab}=28.0 \mathrm{~V}+(5001 \Omega)\left(6 \times 10^{-9} \mathrm{~A}\right)\)

\(=58.006 \mathrm{~V}\)

If the current is to the left, the voltage drop across the parallel combination of resistors is still \(16 \mathrm{~V},\) but with opposite orientation. Writing the loop equation,

\(\begin{aligned} \mathrm{V}_{ab } &=+(5000 \Omega) \mathrm{I}_{\mathrm{sa}}+16 \mathrm{~V}-12 \mathrm{~V}+(1.0 \Omega) \mathrm{I}_{\mathrm{ax}}=0 \\ \mathrm{~V}_{\mathrm{ab}} &=-4.0 \mathrm{~V}-(5001 \Omega)\left(6 \times 10^{-3} \mathrm{~A}\right) \\ &=-34.006 \mathrm{~V} \end{aligned}\)