Question

Find the current (in mA) through 12 $\Omega$ resistor. The four terminals are maintained at the potentials shown.

Answer 1

1072.22 mA

Answer 2

To find the current through the 12 $\Omega$ resistor, we will use Kirchhoff's Current Law (KCL) at the central node where all four resistors meet.

1. Define the node potential: Let the potential at the central node be $V_0$.

2. Apply Kirchhoff's Current Law (KCL): According to KCL, the algebraic sum of currents leaving a node is zero. We assume current flows out of the central node $V_0$ through each resistor.

• Current through 12 $\Omega$ resistor ($I_1$): $I_1 = \frac{V_0 - 15 \, V}{12 \, \Omega}$

• Current through 6 $\Omega$ resistor ($I_2$): $I_2 = \frac{V_0 - 2 \, V}{6 \, \Omega}$

• Current through 4 $\Omega$ resistor ($I_3$): $I_3 = \frac{V_0 - (-3 \, V)}{4 \, \Omega} = \frac{V_0 + 3 \, V}{4 \, \Omega}$

• Current through 8 $\Omega$ resistor ($I_4$): $I_4 = \frac{V_0 - 4 \, V}{8 \, \Omega}$

Applying KCL at the central node: $I_1 + I_2 + I_3 + I_4 = 0$ $\frac{V_0 - 15}{12} + \frac{V_0 - 2}{6} + \frac{V_0 + 3}{4} + \frac{V_0 - 4}{8} = 0$

3. Solve for $V_0$: To eliminate the denominators, multiply the entire equation by the least common multiple (LCM) of 12, 6, 4, and 8, which is 24.

$24 \left( \frac{V_0 - 15}{12} \right) + 24 \left( \frac{V_0 - 2}{6} \right) + 24 \left( \frac{V_0 + 3}{4} \right) + 24 \left( \frac{V_0 - 4}{8} \right) = 0$

$2(V_0 - 15) + 4(V_0 - 2) + 6(V_0 + 3) + 3(V_0 - 4) = 0$

Expand the terms: $2V_0 - 30 + 4V_0 - 8 + 6V_0 + 18 + 3V_0 - 12 = 0$

Combine terms with $V_0$: $(2 + 4 + 6 + 3)V_0 = 15V_0$

Combine constant terms: $-30 - 8 + 18 - 12 = -38 + 18 - 12 = -20 - 12 = -32$

So the equation becomes: $15V_0 - 32 = 0$ $15V_0 = 32$ $V_0 = \frac{32}{15} \, V$

4. Calculate the current through the 12 $\Omega$ resistor: The current through the 12 $\Omega$ resistor is given by $I_{12\Omega} = \frac{V_{\text{high}} - V_{\text{low}}}{R}$. Since $15V > V_0 = 32/15 V \approx 2.13 V$, the current flows from the 15V terminal towards the central node.

$I_{12\Omega} = \frac{15 \, V - V_0}{12 \, \Omega}$ $I_{12\Omega} = \frac{15 - \frac{32}{15}}{12}$ $I_{12\Omega} = \frac{\frac{15 \times 15 - 32}{15}}{12}$ $I_{12\Omega} = \frac{\frac{225 - 32}{15}}{12}$ $I_{12\Omega} = \frac{\frac{193}{15}}{12}$ $I_{12\Omega} = \frac{193}{15 \times 12}$ $I_{12\Omega} = \frac{193}{180} \, A$

5. Convert the current to milliampere (mA): To convert Amperes (A) to milliamperes (mA), multiply by 1000. $I_{12\Omega} = \frac{193}{180} \times 1000 \, mA$ $I_{12\Omega} = \frac{193 \times 100}{18} \, mA$ $I_{12\Omega} = \frac{193 \times 50}{9} \, mA$ $I_{12\Omega} = \frac{9650}{9} \, mA$

$I_{12\Omega} \approx 1072.22 \, mA$

The current through the 12 $\Omega$ resistor is approximately 1072.22 mA.

Explanation of the solution:

1. Identify the common node where all resistors connect. Assign an unknown potential ($V_0$) to this node.
2. Apply Kirchhoff's Current Law (KCL) at this node: the sum of currents leaving the node is zero.
3. Express each current using Ohm's Law ($I = \frac{\Delta V}{R}$), where $\Delta V$ is the potential difference across the resistor.
4. Formulate a linear equation in terms of $V_0$.
5. Solve the equation for $V_0$.
6. Substitute the calculated $V_0$ back into the current expression for the 12 $\Omega$ resistor.
7. Convert the current from Amperes to milliamperes.

Answer:

The current through the 12 $\Omega$ resistor is approximately $1072.22 \, mA$.