Ten identical batteries each of emf 2 V are connected in series to a 8 Ω resistor. If the current in the circuit is 2 A, then the internal resistance of each battery is

Question

Ten identical batteries each of emf 2 V are connected in series to a 8 Ω resistor. If the current in the circuit is 2 A, then the internal resistance of each battery is (A) 0.2 Ω (B) 0.3 Ω (C) 0.4 Ω (D) 0.5 Ω

Tardigrade Admin · Accepted Answer

For this circuit, I=(E text eff /R text elr ) So, 2 =(20/8+10 r) 16+20 r =20 20 r =20-16 r =(4/20) r =(1/5)=0.2 Ω

Q. Ten identical batteries each of emf $2 V$ are connected in series to a $8 Ω$ resistor. If the current in the circuit is $2 A$ , then the internal resistance of each battery is

$0.2 Ω$

$0.3 Ω$

$0.4 Ω$

$0.5 Ω$

$1 Ω$

For this circuit,
$I = \frac{E _{eff}}{R _{elr}}$
So, $2 = \frac{20}{8 + 10 r}$
$16 + 20 r = 20$
$20 r = 20 - 16$
$r = \frac{4}{20}$
$r = \frac{1}{5} = 0.2 Ω$

Q. Ten identical batteries each of emf 2V are connected in series to a 8Ω resistor. If the current in the circuit is 2A, then the internal resistance of each battery is

0.2Ω

0.3Ω

0.4Ω

0.5Ω

1Ω

For this circuit, I=Relr ​Eeff ​​ So, 2=8+10r20​ 16+20r=20 20r=20−16 r=204​ r=51​=0.2Ω