Question

In the circuit shown in figure, the currents $I_1, I_2$ and $I_3$ are shown in diagram below. Then,

• A. $I_1 = \frac{98}{77}A$
• B. $I_2 = \frac{212}{77}A$
• C. $I_3 = \frac{60}{77}A$
• D. $I_1 + I_2 = \frac{128}{77}A$

Answer 1

None of the options are correct. The calculated values are: $I_1 = \frac{37}{73} A$, $I_2 = \frac{109}{73} A$, and $I_3 = \frac{95}{73} A$.

Answer 2

1. Define Node Potentials:

Let the potential of the bottom wire (common ground) be $V_0 = 0V$. Based on the battery connections:

• Potential at node A (positive terminal of 8V battery) is $V_A = 8V$.

• Potential at node C (positive terminal of 6V battery) is $V_C = 6V$.

Let the potential at node B be $V_B$. Let the potential at node D (junction of the two 1Ω resistors in the middle branch, and where $I_3$ originates) be $V_D$.

2. Express Currents in terms of Node Potentials:

• Current $I_2$ (from A to D through 1Ω): $I_2 = \frac{V_A - V_D}{1} = \frac{8 - V_D}{1}$

• Current $I_1$ (from D to C through 1Ω): $I_1 = \frac{V_D - V_C}{1} = \frac{V_D - 6}{1}$

• Current $I_3$ (from D to 0V through 5Ω): $I_3 = \frac{V_D - 0}{5} = \frac{V_D}{5}$

• Current from A to B through 2Ω: $I_{AB} = \frac{V_A - V_B}{2} = \frac{8 - V_B}{2}$

• Current from B to C through 3Ω: $I_{BC} = \frac{V_B - V_C}{3} = \frac{V_B - 6}{3}$

• Current from B to D through 1Ω: $I_{BD} = \frac{V_B - V_D}{1} = V_B - V_D$

3. Apply Kirchhoff's Current Law (KCL):

At Node B:

Sum of currents entering B = Sum of currents leaving B

$I_{AB} = I_{BC} + I_{BD}$

$\frac{8 - V_B}{2} = \frac{V_B - 6}{3} + (V_B - V_D)$

Multiply by 6 to eliminate denominators:

$3(8 - V_B) = 2(V_B - 6) + 6(V_B - V_D)$

$24 - 3V_B = 2V_B - 12 + 6V_B - 6V_D$

$24 - 3V_B = 8V_B - 12 - 6V_D$

$36 = 11V_B - 6V_D$ (Equation 1)

At Node D:

Sum of currents entering D = Sum of currents leaving D

$I_2 + I_{BD} = I_1 + I_3$

$\frac{8 - V_D}{1} + (V_B - V_D) = \frac{V_D - 6}{1} + \frac{V_D}{5}$

Multiply by 5 to eliminate denominators:

$5(8 - V_D) + 5(V_B - V_D) = 5(V_D - 6) + V_D$

$40 - 5V_D + 5V_B - 5V_D = 5V_D - 30 + V_D$

$40 + 5V_B - 10V_D = 6V_D - 30$

$70 = 16V_D - 5V_B$ (Equation 2)

4. Solve the System of Equations:

We have the system:

1. $11V_B - 6V_D = 36$
2. $-5V_B + 16V_D = 70$

Multiply Equation 1 by 5: $55V_B - 30V_D = 180$

Multiply Equation 2 by 11: $-55V_B + 176V_D = 770$

Add the two modified equations:

$(55V_B - 55V_B) + (-30V_D + 176V_D) = 180 + 770$

$146V_D = 950$

$V_D = \frac{950}{146} = \frac{475}{73} V$

Substitute $V_D$ back into Equation 1:

$11V_B - 6\left(\frac{475}{73}\right) = 36$

$11V_B = 36 + \frac{2850}{73}$

$11V_B = \frac{36 \times 73 + 2850}{73} = \frac{2628 + 2850}{73} = \frac{5478}{73}$

$V_B = \frac{5478}{11 \times 73} = \frac{498}{73} V$

5. Calculate the Currents $I_1, I_2, I_3$:

$I_1 = V_D - 6 = \frac{475}{73} - 6 = \frac{475 - 6 \times 73}{73} = \frac{475 - 438}{73} = \frac{37}{73} A$

$I_2 = 8 - V_D = 8 - \frac{475}{73} = \frac{8 \times 73 - 475}{73} = \frac{584 - 475}{73} = \frac{109}{73} A$

$I_3 = \frac{V_D}{5} = \frac{1}{5} \times \frac{475}{73} = \frac{95}{73} A$

6. Compare with Options:

(A) $I_1 = \frac{98}{77}A$. Our $I_1 = \frac{37}{73}A$. (Incorrect)

(B) $I_2 = \frac{212}{77}A$. Our $I_2 = \frac{109}{73}A$. (Incorrect)

(C) $I_3 = \frac{60}{77}A$. Our $I_3 = \frac{95}{73}A$. (Incorrect)

(D) $I_1 + I_2 = \frac{128}{77}A$. Our $I_1 + I_2 = \frac{37}{73} + \frac{109}{73} = \frac{146}{73} = 2A$. (Incorrect)

Based on the calculations, none of the provided options are correct.