Question

Find the current through the 10 $\Omega$ resistor shown in figure

• A. zero
• B. 1A
• C. 2A
• D. 5A

Answer 1

Answer: (A)

zero

Answer 2

To find the current through the 10 Ω resistor, we will use Nodal Analysis.

1. Define Nodes and Potentials: Let's label the nodes in the circuit.

• Let the bottom wire be the reference node (ground), so its potential $V_A = 0V$.
• The 4.5V battery is connected between node A and node B, with the positive terminal at B. Therefore, $V_B = V_A + 4.5V = 0V + 4.5V = 4.5V$.
• The 3V battery is connected between node A and node E, with the positive terminal at E. Therefore, $V_E = V_A + 3V = 0V + 3V = 3V$.
• Let $V_C$ be the potential at node C (between 3Ω, 6Ω, and 10Ω resistors).
• Let $V_D$ be the potential at node D (between 10Ω and 1Ω resistors).

2. Apply Kirchhoff's Current Law (KCL) at Nodes C and D: KCL states that the sum of currents leaving a node is zero.

At Node C: Currents leaving node C flow through the 3Ω, 6Ω, and 10Ω resistors.

• Current through 3Ω resistor (to B): $I_{CB} = \frac{V_C - V_B}{3} = \frac{V_C - 4.5}{3}$
• Current through 6Ω resistor (to A): $I_{CA} = \frac{V_C - V_A}{6} = \frac{V_C - 0}{6} = \frac{V_C}{6}$
• Current through 10Ω resistor (to D): $I_{CD} = \frac{V_C - V_D}{10}$

Sum of currents leaving node C = 0: $\frac{V_C - 4.5}{3} + \frac{V_C}{6} + \frac{V_C - V_D}{10} = 0$

To eliminate denominators, multiply the entire equation by the least common multiple of 3, 6, and 10, which is 30: $10(V_C - 4.5) + 5(V_C) + 3(V_C - V_D) = 0$ $10V_C - 45 + 5V_C + 3V_C - 3V_D = 0$ $18V_C - 3V_D = 45 \quad \text{(Equation 1)}$

At Node D: Currents leaving node D flow through the 10Ω and 1Ω resistors.

• Current through 10Ω resistor (to C): $I_{DC} = \frac{V_D - V_C}{10}$
• Current through 1Ω resistor (to E): $I_{DE} = \frac{V_D - V_E}{1} = \frac{V_D - 3}{1}$

Sum of currents leaving node D = 0: $\frac{V_D - V_C}{10} + \frac{V_D - 3}{1} = 0$

To eliminate denominators, multiply the entire equation by 10: $(V_D - V_C) + 10(V_D - 3) = 0$ $V_D - V_C + 10V_D - 30 = 0$ $-V_C + 11V_D = 30 \quad \text{(Equation 2)}$

3. Solve the System of Equations: We have two linear equations with two unknowns ($V_C$ and $V_D$):

1. $18V_C - 3V_D = 45$
2. $-V_C + 11V_D = 30$

From Equation 2, express $V_C$ in terms of $V_D$: $V_C = 11V_D - 30$

Substitute this expression for $V_C$ into Equation 1: $18(11V_D - 30) - 3V_D = 45$ $198V_D - 540 - 3V_D = 45$ $195V_D = 45 + 540$ $195V_D = 585$ $V_D = \frac{585}{195}$ $V_D = 3V$

Now, substitute the value of $V_D$ back into the expression for $V_C$: $V_C = 11(3) - 30$ $V_C = 33 - 30$ $V_C = 3V$

4. Calculate the Current through the 10 Ω Resistor: The current through the 10 Ω resistor is given by Ohm's Law: $I_{10\Omega} = \frac{V_C - V_D}{10}$. $I_{10\Omega} = \frac{3V - 3V}{10}$ $I_{10\Omega} = \frac{0}{10}$ $I_{10\Omega} = 0A$

The current through the 10 Ω resistor is zero.