From the combination of resistors with resistance values R_1=R_2=R_3=5\ Omega$R_{1}=R_{2}=R_{3}=5\ \Omega$ and R_4=10\ Omega$R_{4}=10\ \Omega$, which of the following combination is the best circuit to get an equivalent resistance of 6\ Omega$6\ \Omega$?

### Related Formula P = fracV^2R_eq$$P = \frac{V^2}{R_{eq}}$$ ### Core Logic To find total power, collapse the entire external circuit and battery internal resistance into a single equivalent resistance R_eq$R_{eq}$ across the battery's ideal terminals. ### Step 1: Equivalent Resistance Calculation Analyzing the diagram: The circuit simplifies to an equivalent resistance R_eq$R_{eq}$ combining the parallel/series elements along with the internal resistance r = 2/3 \, Omega$r = 2/3 \, \Omega$. The final simplified equivalent resistance of the entire system calculates to: R_eq = frac43 \, Omega$$R_{eq} = \frac{4}{3} \, \Omega$$ ### Step 2: Calculate Power P = fracV^2R_eq$$P = \frac{V^2}{R_{eq}}$$ P = frac2^24/3$$P = \frac{2^2}{4/3}$$ P = frac44/3 = 3 text W$$P = \frac{4}{4/3} = 3 \text{ W}$$ ### Pattern Recognition Whenever "entire circuit" power is asked, include the battery's internal resistance inside R_eq$R_{eq}$ so you can use P = E^2 / R_total$P = E^2 / R_{total}$ directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity

### Related Formula R_T = R_0(1 + alpha Delta T)$$R_T = R_0(1 + \alpha \Delta T)$$ ### Step 1: Series Combination Let base resistance at 0^circ textC$0^\circ \text{C}$ be R$R$. For series: R_texteq = R_1 + R_2$$R_{\text{eq}} = R_1 + R_2$$ (2R)[1 + alpha_texteq,s Delta T] = R(1 + alpha_1 Delta T) + R(1 + alpha_2 Delta T)$$(2R)[1 + \alpha_{\text{eq,s}} \Delta T] = R(1 + \alpha_1 \Delta T) + R(1 + \alpha_2 \Delta T)$$ 2 + 2alpha_texteq,s Delta T = 2 + (alpha_1 + alpha_2)Delta T$$2 + 2\alpha_{\text{eq,s}} \Delta T = 2 + (\alpha_1 + \alpha_2)\Delta T$$ alpha_texteq,s = fracalpha_1 + alpha_22$$\alpha_{\text{eq,s}} = \frac{\alpha_1 + \alpha_2}{2}$$ ### Step 2: Parallel Combination For parallel at 0^circ textC$0^\circ \text{C}$, R_texteq,0 = R/2$R_{\text{eq,0}} = R/2$. R_texteq,p = fracR_1 R_2R_1 + R_2$$R_{\text{eq,p}} = \frac{R_1 R_2}{R_1 + R_2}$$ fracR2 [1 + alpha_texteq,p Delta T] = fracR^2 (1 + alpha_1 Delta T)(1 + alpha_2 Delta T)R(2 + (alpha_1 + alpha_2)Delta T)$$\frac{R}{2} [1 + \alpha_{\text{eq,p}} \Delta T] = \frac{R^2 (1 + \alpha_1 \Delta T)(1 + \alpha_2 \Delta T)}{R(2 + (\alpha_1 + \alpha_2)\Delta T)}$$ frac12 (1 + alpha_texteq,p Delta T) = frac1 + (alpha_1 + alpha_2)Delta T2left(1 + fracalpha_1 + alpha_22Delta Tright)$$\frac{1}{2} (1 + \alpha_{\text{eq,p}} \Delta T) = \frac{1 + (\alpha_1 + \alpha_2)\Delta T}{2\left(1 + \frac{\alpha_1 + \alpha_2}{2}\Delta T\right)}$$ Using binomial expansion for small Delta T$\Delta T$: 1 + alpha_texteq,p Delta T approx [1 + (alpha_1 + alpha_2)Delta T] left[ 1 - fracalpha_1 + alpha_22Delta T right]$$1 + \alpha_{\text{eq,p}} \Delta T \approx [1 + (\alpha_1 + \alpha_2)\Delta T] \left[ 1 - \frac{\alpha_1 + \alpha_2}{2}\Delta T \right]$$ 1 + alpha_texteq,p Delta T approx 1 + (alpha_1 + alpha_2)Delta T - fracalpha_1 + alpha_22Delta T$$1 + \alpha_{\text{eq,p}} \Delta T \approx 1 + (\alpha_1 + \alpha_2)\Delta T - \frac{\alpha_1 + \alpha_2}{2}\Delta T$$ alpha_texteq,p Delta T = fracalpha_1 + alpha_22Delta T$$\alpha_{\text{eq,p}} \Delta T = \frac{\alpha_1 + \alpha_2}{2}\Delta T$$ alpha_texteq,p = fracalpha_1 + alpha_22$$\alpha_{\text{eq,p}} = \frac{\alpha_1 + \alpha_2}{2}$$ ### Pattern Recognition For two identical base resistances, the effective temperature coefficient is simply the arithmetic mean of their individual coefficients, regardless of whether they are in series or parallel. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity

### Related Formula R_textparallel = frac1sum frac1R_i$$R_{\text{parallel}} = \frac{1}{\sum \frac{1}{R_i}}$$ ### Core Logic

By carefully identifying the nodes, we can see that a 6\,Omega$6\,\Omega$ resistor in the middle branch is short-circuited by a direct zero-resistance wire path across it.

Removing the short-circuited 6\,Omega$6\,\Omega$ resistor simplifies the circuit into three identical branches connected directly between the terminals A$A$ and B$B$.

### Step 2: Equivalent Calculation The simplified circuit consists of three identical 3\,Omega$3\,\Omega$ resistors in parallel. R_texteq = 3 times frac13 = 1\,Omega$$R_{\text{eq}} = 3 \times \frac{1}{3} = 1\,\Omega$$ ### Pattern Recognition Always trace nodes directly connected by straight wires (zero resistance). Any resistor with both ends connecting to the exact same electrical node is shorted out and can be erased. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity