Question: In the given circuit diagram, a wire is joining points B and D. The current in this wire is:...

Question

In the given circuit diagram, a wire is joining points B and D. The current in this wire is:

Answer 1

Solution

The given circuit is a Wheatstone bridge with a 20V battery connected across points A and C, and a wire connecting points B and D. The resistances are $R_{AB} = 1\Omega$ , $R_{BC} = 2\Omega$ , $R_{AD} = 4\Omega$ , and $R_{DC} = 3\Omega$ . We need to find the current in the wire joining B and D.

Let $V_A$ , $V_B$ , $V_C$ , and $V_D$ be the potentials at points A, B, C, and D respectively. Let's set $V_C = 0V$ and $V_A = 20V$ since the battery is connected between A and C with the positive terminal at A.

Let $I_{BD}$ be the current flowing from B to D through the wire. We can use nodal analysis at points B and D.

At node B, by Kirchhoff's current law: Current entering B = Current leaving B $I_{AB} = I_{BC} + I_{BD}$ $\frac{V_A - V_B}{R_{AB}} = \frac{V_B - V_C}{R_{BC}} + I_{BD}$ $\frac{20 - V_B}{1} = \frac{V_B - 0}{2} + I_{BD}$ $2(20 - V_B) = V_B + 2 I_{BD}$ $40 - 2V_B = V_B + 2 I_{BD}$ $40 = 3V_B + 2 I_{BD}$ (Equation 1)

At node D, by Kirchhoff's current law: Current entering D = Current leaving D $I_{AD} + I_{BD} = I_{DC}$ $\frac{V_A - V_D}{R_{AD}} + I_{BD} = \frac{V_D - V_C}{R_{DC}}$ $\frac{20 - V_D}{4} + I_{BD} = \frac{V_D - 0}{3}$ $3(20 - V_D) + 12 I_{BD} = 4V_D$ $60 - 3V_D + 12 I_{BD} = 4V_D$ $60 = 7V_D - 12 I_{BD}$ (Equation 2)

The problem statement does not specify the resistance of the wire joining B and D. In such problems, it is usually assumed that the wire is ideal, meaning it has zero resistance. If the wire has zero resistance, then points B and D are at the same potential, i.e., $V_B = V_D$ . Let $V_B = V_D = V$ .

Substituting $V_B = V$ and $V_D = V$ into Equations 1 and 2: $40 = 3V + 2 I_{BD}$ (Equation 3) $60 = 7V - 12 I_{BD}$ (Equation 4)

We have a system of two linear equations with two unknowns, $V$ and $I_{BD}$ . We can solve for $I_{BD}$ . Multiply Equation 3 by 7 and Equation 4 by 3: $7 \times (40 = 3V + 2 I_{BD}) \implies 280 = 21V + 14 I_{BD}$ $3 \times (60 = 7V - 12 I_{BD}) \implies 180 = 21V - 36 I_{BD}$

Subtract the second new equation from the first new equation: $(280) - (180) = (21V + 14 I_{BD}) - (21V - 36 I_{BD})$ $100 = 14 I_{BD} + 36 I_{BD}$ $100 = 50 I_{BD}$ $I_{BD} = \frac{100}{50} = 2A$

The current in the wire joining B and D is 2A. Since we assumed the current flows from B to D and got a positive value, the direction of the current is from B to D.

We can also find the potential $V$ : From Equation 3: $40 = 3V + 2(2) \implies 40 = 3V + 4 \implies 3V = 36 \implies V = 12V$ . So, $V_B = V_D = 12V$ .