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QJEEM16Phbd2e: JEE Main 2016 Physics - Class 12 JEEM16Phbd2e Solution: A 50 resistance is connected to a battery of 5 V. A…

A 50 Ω\Omega resistance is connected to a battery of 5 V. A galvanometer of resistance 100 Ω\Omega is to be used as an ammeter to measure current through the resistance, for this a resistance rs is connected to the galvanometer. Which of the following connections should be employed if the measured current is within 1% of thecurrent without the ammeter in the circuit ?

Multiple Choice Options

  • A rs = 0.5 Ω\Omega in parallel with the galvanometer
  • B rs = 0.5 Ω\Omega in series with the galvanometer
  • C rs = 1 Ω\Omega in series with galvanometer
  • D rs =1 Ω\Omega in parallel with galvanometer

Step-by-Step Verified Solution

JEE Main 2016 (Online) 9th April Morning Slot Physics - Current Electricity Question 277 English Explanation

Current in the circuit without ammeter

I == 550=0.1{5 \over {50}} = 0.1 A

∴\therefore    With ammeter current == 0.1 ×\times 99100{{99} \over {100}} = 0.099 A

With ammeter equivalent resistance,

Req = 50 + 100 rs100+rs{{100\,{r_s}} \over {100 + {r_s}}}

∴\therefore    0.099 = 550+100 rs100+rs{5 \over {50 + {{100\,{r_s}} \over {100 + {r_s}}}}}

⇒\Rightarrow    50 + 100 rs100+rs{{100\,{r_s}} \over {100 + {r_s}}} = 50.099{5 \over {0.099}}

⇒\Rightarrow     100 rs100+rs=0.5{{100\,{r_s}} \over {100 + {r_s}}} = 0.5

⇒\Rightarrow    100 rs = 50 + 0.5rs

⇒\Rightarrow    99.5 rs = 50

⇒\Rightarrow    rs == 5099.5=0.5 Ω{{50} \over {99.5}} = 0.5\,\Omega

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