### Related Formula For parallel branches, the potential difference V$V$ across each branch is identical: V = I_i R_i$V = I_i R_i$ According to Kirchhoff's Current Law, the total current I_texttotal$I_{\text{total}}$ is the sum of currents in all parallel branches: I_texttotal = i_1 + i_2 + i_3$I_{\text{total}} = i_1 + i_2 + i_3$ ### Core Logic Analyzing the circuit diagram: * Branch 1: Current i_1$i_1$ through 20\ Omega$20\ \Omega$ is 0.3text A$0.3\text{ A}$. * Branch 2: Contains 15\ Omega$15\ \Omega$ resistor with current i_2$i_2$. * Branch 3: Contains resistor R_1$R_1$ with current i_3$i_3$. Since the branches are in parallel, they have the same potential difference V_AB$V_{AB}$: V_AB = i_1 times 20\ Omega = 0.3text A times 20\ Omega = 6text V$V_{AB} = i_1 \times 20\ \Omega = 0.3\text{ A} \times 20\ \Omega = 6\text{ V}$ ### Step 1: Calculate Currents Current through the second branch (15\ Omega$15\ \Omega$ resistor) is: i_2 = fracV_AB15\ Omega = frac6text V15\ Omega = 0.4text A$i_2 = \frac{V_{AB}}{15\ \Omega} = \frac{6\text{ V}}{15\ \Omega} = 0.4\text{ A}$ Total current read by the ammeter is 0.9text A$0.9\text{ A}$. Thus: i_1 + i_2 + i_3 = 0.9text A$i_1 + i_2 + i_3 = 0.9\text{ A}$ 0.3text A + 0.4text A + i_3 = 0.9text A$0.3\text{ A} + 0.4\text{ A} + i_3 = 0.9\text{ A}$ 0.7text A + i_3 = 0.9text A implies i_3 = 0.2text A$0.7\text{ A} + i_3 = 0.9\text{ A} \implies i_3 = 0.2\text{ A}$ ### Step 2: Calculate R1 Now use the voltage relation for the branch containing R_1$R_1$: i_3 times R_1 = V_AB$i_3 \times R_1 = V_{AB}$ (0.2text A) times R_1 = 6text V$(0.2\text{ A}) \times R_1 = 6\text{ V}$ R_1 = frac60.2 = 30\ Omega$R_1 = \frac{6}{0.2} = 30\ \Omega$

### Pattern Recognition In parallel networks, finding the branch voltage is always the primary step. Once V = 6text V$V = 6\text{ V}$ is established, the remaining branch currents are easily found using Ohm's Law and current conservation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity