Current Electricity

JEE Main level, numerical answer, medium

292 of 380

Question 292 of 380

A meter bridge setup is shown in the figure. It is used to determine an unknown resistance RR using a given resistor of 15Ω15\Omega. The galvanometer (G)\left(G\right) shows null defection when tapping key is at 43cm43cm mark from end AA. If the end correction for end AA is 2cm2cm, then the determined value of RR will be Ω\Omega.

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Answer: 19

Using the conditions of a balanced wheatstone bridge and adding the end correction.
15R=l+e(L+e)−(l+e)⇒15(43+2)=R(102−45)⇒R=5745×15\dfrac{15}{R}=\dfrac{l+e}{\left(L+e\right)-\left(l+e\right)}\Rightarrow \dfrac{15}{\left(43+2\right)}=\dfrac{R}{\left(102-45\right)}\Rightarrow R=\dfrac{57}{45}\times 15
⇒R=19Ω\Rightarrow R=19\Omega

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