Marianopolis 203-SN2-RE tutoring

Use one voltage reading to find two branch currents

This paper-and-pencil circuit model connects a measurement with Ohm’s law and charge conservation. Assume steady direct current, ideal wires and an ideal 6.0 V source. The ideal voltmeter draws no current; real meters have a finite input resistance.

Connect a 3.0 kΩ resistor and a 6.0 kΩ resistor in parallel between nodes A and B. The source’s positive terminal is at A. A voltmeter with its positive lead at A and negative lead at B reads +6.0 V. Find the conventional current in each resistor and the current supplied by the source.

1. 1. Read the same two nodes

   The meter reads $V_A-V_B$, so A is 6.0 V above B. Both resistors connect directly between these same nodes. Each has a 6.0 V drop from A to B; the voltage is not divided between parallel branches. Choose $I_1$ and $I_2$ from A to B.

2. 2. Calculate each branch separately

   Use $I=V_{AB}/R$, where $V_{AB}$ means $V_A-V_B$. Convert kiloohms to ohms before calculating: 1 kΩ = 1000 Ω. The positive results follow our arrows, from A to B.

   $$\begin{aligned}I_1&=\frac{6.0\,\mathrm V}{3.0\times10^3\,\Omega}\\&=2.0\times10^{-3}\,\mathrm A\\&=2.0\,\mathrm{mA},\end{aligned}$$$$\begin{aligned}I_2&=\frac{6.0\,\mathrm V}{6.0\times10^3\,\Omega}\\&=1.0\,\mathrm{mA}.\end{aligned}$$

3. 3. Balance the current and check the voltage

   At A, the source current splits between the two resistors. The ideal voltmeter adds no current, so $I_{\text{source}}=I_1+I_2$ = 3.0 mA. The smaller resistance carries the larger current. Check each branch: (2.0 mA)(3.0 kΩ) = 6.0 V and (1.0 mA)(6.0 kΩ) = 6.0 V. Each resistor therefore returns the same voltage reading. Units agree: V/Ω = A and mA × kΩ = V.

Try two checks