A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of 2.0 V and a negligible internal resistance. The potentiometer wire itself is 4 m long, When a resistance R, connected across the given cell, has values of, (i) infinity (ii) 10 Ω The balancing lengths on the potentiometer wire are found to be 60 cm and 50 cm respectively. The value of internal resistance of the cell is:-

Question

A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of 2.0 V and a negligible internal resistance. The potentiometer wire itself is 4 m long, When a resistance R, connected across the given cell, has values of, (i) infinity (ii) 10 Ω The balancing lengths on the potentiometer wire are found to be 60 cm and 50 cm respectively. The value of internal resistance of the cell is:- (A) 0.25 Ω (B) 0.95 Ω (C) 0.5 Ω (D) 2 Ω

Tardigrade Admin · Accepted Answer

ε=φ × ℓ1 ... (i) (ε R/ R + r )=φ × ℓ2 ... (ii) r =((ℓ1-ℓ2/ℓ2)) R =((60-50/50)) × 10 ⇒ r =2 Ω

$0.25 Ω$

$0.95 Ω$

$0.5 Ω$

$2 Ω$

$ε = ϕ \times ℓ_{1}$ ... (i)
$\frac{εR}{R + r} = ϕ \times ℓ_{2}$ ... (ii)
$r = (\frac{ℓ _{1} - ℓ _{2}}{ℓ _{2}}) R = (\frac{60 - 50}{50}) \times 10$
$\Rightarrow r = 2 Ω$

0.25Ω

0.95Ω

0.5Ω

2Ω

ε=ϕ×ℓ1​ ... (i) R+rεR​=ϕ×ℓ2​ ... (ii) r=(ℓ2​ℓ1​−ℓ2​​)R=(5060−50​)×10 ⇒r=2Ω

$0.25 Ω$

$0.95 Ω$

$0.5 Ω$

$2 Ω$

$ε = ϕ \times ℓ_{1}$ ... (i)
$\frac{εR}{R + r} = ϕ \times ℓ_{2}$ ... (ii)
$r = (\frac{ℓ _{1} - ℓ _{2}}{ℓ _{2}}) R = (\frac{60 - 50}{50}) \times 10$
$\Rightarrow r = 2 Ω$