### Related Formula For parallel resistors: R_textparallel = fracR_a R_bR_a + R_b$R_{\text{parallel}} = \frac{R_a R_b}{R_a + R_b}$ Total equivalent resistance in series: R_texteq = R_1 + R_textparallel + R_4$R_{\text{eq}} = R_1 + R_{\text{parallel}} + R_4$ Total current from source: I = fracVR_texteq$I = \frac{V}{R_{\text{eq}}}$ ### Core Logic Analyzing the circuit network: * R_1 = 2\ Omega$R_1 = 2\ \Omega$ * R_2 = 4\ Omega$R_2 = 4\ \Omega$ and R_3 = 4\ Omega$R_3 = 4\ \Omega$ are in parallel. * R_4 = 1\ Omega$R_4 = 1\ \Omega$ Calculate the equivalent resistance of the parallel combination: R_textparallel = fracR_2 times R_3R_2 + R_3 = frac4 times 44 + 4 = 2\ Omega$R_{\text{parallel}} = \frac{R_2 \times R_3}{R_2 + R_3} = \frac{4 \times 4}{4 + 4} = 2\ \Omega$ ### Step 1: Calculate Total Equivalent Resistance and Current Total equivalent resistance is: R_texteq = R_1 + R_textparallel + R_4 = 2 + 2 + 1 = 5\ Omega$R_{\text{eq}} = R_1 + R_{\text{parallel}} + R_4 = 2 + 2 + 1 = 5\ \Omega$ Total circuit current is: I = fracVR_texteq = frac10text V5\ Omega = 2text A$I = \frac{V}{R_{\text{eq}}} = \frac{10\text{ V}}{5\ \Omega} = 2\text{ A}$ ### Step 2: Determine Current in R3 The total current of 2text A$2\text{ A}$ enters the parallel branch of R_2$R_2$ and R_3$R_3$. Since R_2 = R_3 = 4\ Omega$R_2 = R_3 = 4\ \Omega$, the current divides equally between them: I_R_3 = I times fracR_2R_2 + R_3 = 2 times frac48 = 1text A$I_{R_3} = I \times \frac{R_2}{R_2 + R_3} = 2 \times \frac{4}{8} = 1\text{ A}$

### Pattern Recognition Equal parallel resistors split current exactly down the middle. Once total current is found to be 2text A$2\text{ A}$, the parallel branches share it as 1text A$1\text{ A}$ each without further math. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity