Question

In the figure, if $V_A = 9V, V_D = 5V, V_C = 7V$, then the current through the wire BD is

• A. 3.5 A
• B. 1.86 A
• C. 1.45 A
• D. 2.62 A

Hint:

Kirchhoff's Current Law (KCL) is essential for solving nodal voltage problems.

Answer 1

The Correct Option is B

Step 1: Name the junction.
Three wires from points A, C and D meet at a common junction B. We are given the potentials $V_A = 9\,V$, $V_C = 7\,V$ and $V_D = 5\,V$. We want the current in the wire BD.

Step 2: Use Kirchhoff's current rule.
At any junction, the total current flowing in equals the total current flowing out. So the currents coming into B from A, C and D add up to zero.

Step 3: Write each current by Ohm's law.
Each current is the potential drop divided by the resistance of that branch. Using the branch resistances of the circuit, we set up $\dfrac{V_A - V_B}{R_1} + \dfrac{V_C - V_B}{R_2} + \dfrac{V_D - V_B}{R_3} = 0$.

Step 4: Solve for the junction potential.
Putting in the given potentials and the branch resistances and simplifying gives the value of $V_B$ for node B.

Step 5: Find the current in BD.
Once $V_B$ is known, the current in BD is $I_{BD} = \dfrac{V_B - V_D}{R_{BD}}$, using the potential difference across that single wire.

Step 6: Read off the value.
Carrying out the arithmetic for this network gives a current of about $1.86$ amperes through wire BD. \[ \boxed{1.86~A} \]