#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# SEE THE ATTACHMENTAccording to the formula V=E-Ir (E is the emf of cell with internal resistance r), we say that whenever we increase the current there is decrease in potential difference and vice versa.BUT THERE IS A TANGLED DOUBT IN MY MINDV=E-IrV+Ir= EIR+Ir=ELet we increase the current I somehow (by increasing the frequency of electrons) to I’ such that I’>IBut since current remain same in series. The result will be like…I’R + I’r= EV+I’r=EV=E-I’rBut V=I’R which increased rather than decreasingWhich is not possibleMY VIEWSI think V=E-IrI1R=E-I2rI1R=E+(-I2r)I1 is the current flowing through the wire while I2 is the current flowing inside the cell which is moving in the opposite direction (i.e. negative to positive) We actually alter the amount of current inside the cell by some special method which results to the change in overall voltage of the cell.AM I RIGHT?.   IF WRONG PLEASE PUT UP YOUR VIEWS.

Neeti
571 Points
6 years ago
the current is same in both, the external resistor and battery. the formula basically is E= IR + Ir means TOTAL voltage supplied by the cell or battery is equal to voltage drop across the external resistance plus voltage drop across the cell due to the internal resistance of the battery. IR being the voltage drop across the external resistance and Ir being the voltage drop across the cell due to internal resistance of the cell.
feel free to ask follow-up questions :)