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QJEEM19Phb91e: JEE Main 2019 Physics - Class 12 JEEM19Phb91e Solution: The galvanometer deflection, when key K 1 is closed but K 2 is…

The galvanometer deflection, when key K1 is closed but K2 is open, equals θ\theta 0 (see figure). On closing K2 also and adjusting R2 to 5 Ω\Omega , the deflection in galvanometer becomes θ05.{{{\theta _0}} \over 5}. . The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery] :

JEE Main 2019 (Online) 12th January Morning Slot Physics - Current Electricity Question 255 English

Multiple Choice Options

  • A 5 Ω\Omega
  • B 25 Ω\Omega
  • C 12 Ω\Omega
  • D 22 Ω\Omega

Step-by-Step Verified Solution

case I :

ig = E220+Rg{E \over {220 + {R_g}}} = C θ\theta 0     . . .(i)

Case II :

ig = (E220+5Rg5+Rg)\left( {{E \over {220 + {{5{R_g}} \over {5 + {R_g}}}}}} \right) ×\times 5(Rg+5){5 \over {\left( {{R_g} + 5} \right)}} = Cθ05{{C{\theta _0}} \over 5}       . . .(ii)

⇒\Rightarrow    5E225Rg+1100{{5E} \over {225{R_g} + 1100}} = Cθ05{{C{\theta _0}} \over 5}       . . .(ii)

E220+Rg=Cθ{E \over {220 + {R_g}}} = C\theta       . . .(i)

⇒\Rightarrow    225Rg+11001100+5Rg=5{{225{R_g} + 1100} \over {1100 + 5{R_g}}} = 5

⇒\Rightarrow   5500 + 25Rg = 225Rg + 1100

200Rg = 4400

Rg = 22 Ω\Omega

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