Question

A battery of internal resistance 3 ohm is connected to 20 ohm resistor and potential difference across the resistor is 10v if another resistor of 30 is connected in series with first resistor and battery is again connected to the combination

Answer 1

Please complete the question properly .

The question should have been like this :

"A battery of internal resistance 3 ohm is connected to 20 ohm resistor and potential difference across the resistor is 10v if another resistor of 30 is connected in series with first resistor and battery is again connected to the combination  . Calculate the EMF and potential difference across combination "

Let the current flowing in the circuit be i and E be the emf of battery

By applying Kirchhoff's Voltage law :

E-20i-3i=0

E=23i

But according to the question :

Voltage across 20 ohms is 10V

by ohms' law

V=IR

10 = i x 20

i= 10/20 = 1/2 A= 0.5 A

So emf of battery is E= 23 x i=23x 0.5= 11.5V

now if 30 ohms resistance is connected then :

Total resistance = 20+30+3=53 OHMS

CURRENT :

I=E/R=11.5/53=0.216A

potential drop across combination of resistors :

v=i(20+30)\v= 50x 0.216=10.48V

Answer 2

A battery of internal resistance 3 ohm is connected to 20 ohm resistor and potential difference across the resistor is 10v if another resistor of 30 is connected in series with first resistor and battery is again connected to the combination

Let Say Battery Potential = E Volt

Then Current flow would be = E /  (3 + 20) = E/ 23  Amp

Voltage across 20 Ohm Resistance = (E/23) * 20

(E/23) * 20 = 10

=> E = 11.5 Volt

Resistor of 30 Ohm is connected so Now resistance

= 3 + 20 + 30 = 53 Ohm

Current = 11.5/53  Amp

now Potential Difference of Resistor combination =  (11.5/53) * (20 + 30)

= (11.5/53) * 50

= 10.85 V