Question

Unbalanced wheatstone bridge.

① find $R_{eq}$ of circuit ② Current passes through battery ③ $V_c - V_D$ ④ current through 3$\Omega$

Answer 1

1. $R_{eq} = \frac{13}{9} \Omega$
2. Current passes through battery = $\frac{90}{13} \text{ A}$
3. $V_C - V_D = -\frac{30}{13} \text{ V}$
4. Current through 3Ω = $\frac{10}{13} \text{ A}$ (from D to C)

Answer 2

The circuit given is an unbalanced Wheatstone bridge. We will use Kirchhoff's laws to find the potentials at nodes C and D, and then calculate the required quantities.

Let the potential at the positive terminal of the battery be $V_P = 10V$ and at the negative terminal be $V_Q = 0V$. The resistors are connected as follows:

• $R_{PC} = 2\Omega$
• $R_{CQ} = 1\Omega$
• $R_{PD} = 1\Omega$
• $R_{DQ} = 2\Omega$
• $R_{CD} = 3\Omega$

Let $V_C$ and $V_D$ be the potentials at points C and D, respectively.

1. Apply Kirchhoff's Current Law (KCL) at nodes C and D:

• At node C: Current entering C from P equals current leaving C towards D and Q. $\frac{V_P - V_C}{R_{PC}} = \frac{V_C - V_D}{R_{CD}} + \frac{V_C - V_Q}{R_{CQ}}$ $\frac{10 - V_C}{2} = \frac{V_C - V_D}{3} + \frac{V_C - 0}{1}$ Multiply by 6 to clear denominators: $3(10 - V_C) = 2(V_C - V_D) + 6V_C$ $30 - 3V_C = 2V_C - 2V_D + 6V_C$ $30 = 11V_C - 2V_D$ (Equation 1)

• At node D: Current entering D from P and C equals current leaving D towards Q. $\frac{V_P - V_D}{R_{PD}} + \frac{V_C - V_D}{R_{CD}} = \frac{V_D - V_Q}{R_{DQ}}$ $\frac{10 - V_D}{1} + \frac{V_C - V_D}{3} = \frac{V_D - 0}{2}$ Multiply by 6 to clear denominators: $6(10 - V_D) + 2(V_C - V_D) = 3V_D$ $60 - 6V_D + 2V_C - 2V_D = 3V_D$ $60 + 2V_C - 8V_D = 3V_D$ $2V_C - 11V_D = -60$ (Equation 2)

2. Solve the system of linear equations for $V_C$ and $V_D$: Equation 1: $11V_C - 2V_D = 30$ Equation 2: $2V_C - 11V_D = -60$

Multiply Equation 1 by 11 and Equation 2 by 2: $121V_C - 22V_D = 330$ $4V_C - 22V_D = -120$

Subtract the second modified equation from the first: $(121V_C - 4V_C) - (22V_D - 22V_D) = 330 - (-120)$ $117V_C = 450$ $V_C = \frac{450}{117} = \frac{150}{39} = \frac{50}{13} \text{ V}$

Substitute $V_C$ back into Equation 1: $11\left(\frac{50}{13}\right) - 2V_D = 30$ $\frac{550}{13} - 2V_D = 30$ $2V_D = \frac{550}{13} - 30 = \frac{550 - 390}{13} = \frac{160}{13}$ $V_D = \frac{80}{13} \text{ V}$

Now, let's answer the specific questions:

③ Find $V_C - V_D$ $V_C - V_D = \frac{50}{13} - \frac{80}{13} = -\frac{30}{13} \text{ V}$

④ Current through 3Ω resistor Current $I_{CD} = \frac{V_C - V_D}{R_{CD}} = \frac{-30/13}{3} = -\frac{10}{13} \text{ A}$ The negative sign indicates that the current flows from D to C. The magnitude of the current through the 3Ω resistor is $\frac{10}{13} \text{ A}$.

② Current passes through battery The total current from the battery is the sum of currents entering nodes C and D from P. Current $I_{PC} = \frac{V_P - V_C}{R_{PC}} = \frac{10 - 50/13}{2} = \frac{(130 - 50)/13}{2} = \frac{80/13}{2} = \frac{40}{13} \text{ A}$ Current $I_{PD} = \frac{V_P - V_D}{R_{PD}} = \frac{10 - 80/13}{1} = \frac{(130 - 80)/13}{1} = \frac{50}{13} \text{ A}$ Total current from battery $I_{battery} = I_{PC} + I_{PD} = \frac{40}{13} + \frac{50}{13} = \frac{90}{13} \text{ A}$

① Find $R_{eq}$ of circuit The equivalent resistance is the total voltage divided by the total current. $R_{eq} = \frac{V_{battery}}{I_{battery}} = \frac{10}{90/13} = 10 \times \frac{13}{90} = \frac{130}{90} = \frac{13}{9} \Omega$