Question

The equivalent resistance between points A and

• A. $\frac{65}{2}\Omega$
• B. $\frac{45}{2}\Omega$
• C. $\frac{5}{2}\Omega$
• D. $\frac{91}{2}\Omega$

Answer 1

None of the options matched the calculated value.

Answer 2

Let's interpret the circuit diagram based on standard conventions. The circuit consists of a square-shaped arrangement of resistors with two diagonal resistors. Let the four corner nodes be: - Top-left: A - Top-right: Q - Bottom-right: R - Bottom-left: B The resistors on the perimeter of the square are: - $R_{AQ} = 2\Omega$ (between A and Q) - $R_{QR} = 15\Omega$ (between Q and R) - $R_{RB} = 40\Omega$ (between R and B) - The left side of the square has two resistors, 8Ω and 20Ω, placed one below the other. The most common interpretation for such a drawing is that they are in series, forming the left arm of the bridge. So, $R_{BA} = 8\Omega + 20\Omega = 28\Omega$ (between B and A). The diagonal resistors are: - $R_{AR} = 30\Omega$ (between A and R) - $R_{QB} = 10\Omega$ (between Q and B) This is a complex Wheatstone bridge configuration. We can solve it using Nodal Analysis. Let the potential at point B be $0V$. Let the potential at point A be $V$. Let the potentials at nodes Q and R be $V_Q$ and $V_R$ respectively. Applying Kirchhoff's Current Law (KCL) at nodes Q and R: At Node Q: (Sum of currents leaving Q is zero) Currents leaving Q are: - To A: $(V_Q - V)/2$ - To R: $(V_Q - V_R)/15$ - To B: $(V_Q - 0)/10$ $\frac{V_Q - V}{2} + \frac{V_Q - V_R}{15} + \frac{V_Q}{10} = 0$ Multiply by LCM(2, 15, 10) = 30: $15(V_Q - V) + 2(V_Q - V_R) + 3V_Q = 0$ $15V_Q - 15V + 2V_Q - 2V_R + 3V_Q = 0$ $20V_Q - 2V_R = 15V$ --- (Equation 1) At Node R: (Sum of currents leaving R is zero) Currents leaving R are: - To Q: $(V_R - V_Q)/15$ - To B: $(V_R - 0)/40$ - To A: $(V_R - V)/30$ $\frac{V_R - V_Q}{15} + \frac{V_R}{40} + \frac{V_R - V}{30} = 0$ Multiply by LCM(15, 40, 30) = 120: $8(V_R - V_Q) + 3V_R + 4(V_R - V) = 0$ $8V_R - 8V_Q + 3V_R + 4V_R - 4V = 0$ $-8V_Q + 15V_R = 4V$ --- (Equation 2) Now we have a system of two linear equations: 1) $20V_Q - 2V_R = 15V$ 2) $-8V_Q + 15V_R = 4V$ From Equation 1, express $V_R$ in terms of $V_Q$: $2V_R = 20V_Q - 15V \implies V_R = 10V_Q - \frac{15}{2}V$ Substitute this expression for $V_R$ into Equation 2: $-8V_Q + 15(10V_Q - \frac{15}{2}V) = 4V$ $-8V_Q + 150V_Q - \frac{225}{2}V = 4V$ $142V_Q = 4V + \frac{225}{2}V$ $142V_Q = \frac{8V + 225V}{2}$ $142V_Q = \frac{233V}{2}$ $V_Q = \frac{233V}{284}$ Now, calculate $V_R$: $V_R = 10 \left(\frac{233V}{284}\right) - \frac{15}{2}V$ $V_R = \frac{2330V}{284} - \frac{15 \times 142V}{284}$ $V_R = \frac{2330V - 2130V}{284}$ $V_R = \frac{200V}{284} = \frac{50V}{71}$ The total current $I$ flowing from A to B is the sum of currents leaving node A: $I = I_{A \to Q} + I_{A \to R} + I_{A \to B}$ (where $I_{A \to B}$ is the current through the $28\Omega$ arm directly connecting A and B) $I = \frac{V - V_Q}{2} + \frac{V - V_R}{30} + \frac{V - 0}{28}$ Substitute the values of $V_Q$ and $V_R$: $I = \frac{V - \frac{233V}{284}}{2} + \frac{V - \frac{50V}{71}}{30} + \frac{V}{28}$ $I = \frac{\frac{284V - 233V}{284}}{2} + \frac{\frac{71V - 50V}{71}}{30} + \frac{V}{28}$ $I = \frac{51V}{568} + \frac{21V}{2130} + \frac{V}{28}$ $I = \frac{51V}{568} + \frac{7V}{710} + \frac{V}{28}$ To sum these fractions, find the Least Common Multiple (LCM) of the denominators (568, 710, 28). $568 = 8 \times 71$ $710 = 10 \times 71$ $28 = 4 \times 7$ LCM$(568, 710, 28) = 2^3 \times 5 \times 7 \times 71 = 8 \times 5 \times 7 \times 71 = 40 \times 497 = 19880$. $I = V \left( \frac{51 \times (19880/568)}{19880} + \frac{7 \times (19880/710)}{19880} + \frac{1 \times (19880/28)}{19880} \right)$ $I = V \left( \frac{51 \times 35}{19880} + \frac{7 \times 28}{19880} + \frac{1 \times 710}{19880} \right)$ $I = V \left( \frac{1785}{19880} + \frac{196}{19880} + \frac{710}{19880} \right)$ $I = V \left( \frac{1785 + 196 + 710}{19880} \right)$ $I = V \left( \frac{2691}{19880} \right)$ The equivalent resistance $R_{eq}$ is $\frac{V}{I}$: $R_{eq} = \frac{19880}{2691}\Omega$ Now, let's compare this calculated value with the given options: (1) $\frac{65}{2}\Omega = 32.5\Omega$ (2) $\frac{45}{2}\Omega = 22.5\Omega$ (3) $\frac{5}{2}\Omega = 2.5\Omega$ (4) $\frac{91}{2}\Omega = 45.5\Omega$ Our calculated value is $R_{eq} = \frac{19880}{2691} \approx 7.387\Omega$. This value does not match any of the provided options. This suggests a potential error in the question's diagram values or the options themselves. The final answer is $\boxed{\text{None of the options matched the calculated value.}}$