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QJEEM20Ph4447: JEE Main 2020 Physics - Class 12 JEEM20Ph4447 Solution: The series combination of two batteries, both of the same emf…

The series combination of two batteries, both of the same emf 10 V, but different internal resistance of 20 Ω\Omega and 5 Ω\Omega , is connected to the parallel combination of two resistors 30 Ω\Omega and R Ω\Omega . The voltage difference across the battery of internal resistance 20 Ω\Omega is zero, the value of R (in Ω\Omega ) is : _______

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JEE Main 2020 (Online) 8th January Evening Slot Physics - Current Electricity Question 223 English Explanation

10 – I × 20 = 0

⇒\Rightarrow I = 0.5A

Also,

I = 10+1020+5+30R30+R{{10 + 10} \over {20 + 5 + {{30R} \over {30 + R}}}}

⇒\Rightarrow 12{1 \over 2} = 2025+30R30+R{{20} \over {25 + {{30R} \over {30 + R}}}}

⇒\Rightarrow 40 = 25 + 30R30+R{{{30R} \over {30 + R}}}

⇒\Rightarrow 15 = 30R30+R{{{30R} \over {30 + R}}}

⇒\Rightarrow 30 + R = 2R

⇒\Rightarrow R = 30 Ω\Omega

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