Verified Solution • 2025–26 Session moderate QJEEM24Ph7e19: JEE Main 2024 Physics - Class 12 JEEM24Ph7e19 Solution: In the given circuit, the current flowing through the… In the given circuit, the current flowing through the resistance 20Ω20 \Omega20Ω is 0.3 A0.3 \mathrm{~A}0.3 A , while the ammeter reads 0.9 A0.9 \mathrm{~A}0.9 A . The value of R1\mathrm{R}_1R1 is _________ Ω\OmegaΩ . Step-by-Step Verified Solution Given, i1=0.3 A,i1+i2+i3=0.9 A\mathrm{i}_1=0.3 \mathrm{~A}, \mathrm{i}_1+\mathrm{i}_2+\mathrm{i}_3=0.9 \mathrm{~A}i1=0.3 A,i1+i2+i3=0.9 A So, VAB=i1×20Ω=20×0.3 V=6 V\mathrm{V}_{\mathrm{AB}}=\mathrm{i}_1 \times 20 \Omega=20 \times 0.3 \mathrm{~V}=6 \mathrm{~V}VAB=i1×20Ω=20×0.3 V=6 V amp;i2=6 V15Ω=25 Aamp;i1+i2+i3=910 Aamp;310+25+i3=910amp;710+i3=910amp;i3=0.2 Aamp;So,i3×R1=6 Vamp;(0.2)R1=6amp;R1=60.2=30Ω\begin{aligned}& \mathrm{i}_2=\frac{6 \mathrm{~V}}{15 \Omega}=\frac{2}{5} \mathrm{~A} \\& \mathrm{i}_1+\mathrm{i}_2+\mathrm{i}_3=\frac{9}{10} \mathrm{~A} \\& \frac{3}{10}+\frac{2}{5}+\mathrm{i}_3=\frac{9}{10} \\& \frac{7}{10}+\mathrm{i}_3=\frac{9}{10} \\& \mathrm{i}_3=0.2 \mathrm{~A} \\& \mathrm{So}, \mathrm{i}_3 \times \mathrm{R}_1=6 \mathrm{~V} \\& (0.2) \mathrm{R}_1=6 \\& \mathrm{R}_1=\frac{6}{0.2}=30 \Omega\end{aligned}amp;i2=15Ω6 V=52 Aamp;i1+i2+i3=109 Aamp;103+52+i3=109amp;107+i3=109amp;i3=0.2 Aamp;So,i3×R1=6 Vamp;(0.2)R1=6amp;R1=0.26=30Ω Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring
