# NEET Physics Current Electricity Questions Solved

In the given figure, battery E is balanced on 55 cm length of potentiometer wire but when a resistance of 10 Ω is connected in parallel with the battery then it balances on 50 cm length of the potentiometer wire then internal resistance r of the battery is

(1) 1 Ω

(2) 3 Ω

(3) 10 Ω

(4) 5 Ω

(1) $r=\left(\frac{{l}_{1}-{l}_{2}}{{l}_{2}}\right)×R\text{'}⇒r=\left(\frac{55-50}{50}\right)×10=1\Omega$

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