What will be the ratio of the equivalent elec | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the two cells have electromotive forces $E_1$ and $E_2$ and internal resistances $r_1$ and $r_2$.For series connection,the equivalent emf is $

Vedclass · Accepted Answer

$2$:$1$

Vedclass · Answer

(B) Let the two cells have electromotive forces $E_1$ and $E_2$ and internal resistances $r_1$ and $r_2$.For series connection,the equivalent emf is $E_s = E_1 + E_2$.For parallel connection,the equivalent emf is $E_p = \frac{E_1r_2 + E_2r_1}{r_1 + r_2}$.Assuming the two cells are identical with $E_1 = E_2 = E$ and $r_1 = r_2 = r$:In series,$E_s = E + E = 2E$.In parallel,$E_p = \frac{Er + Er}{r + r} = \frac{2Er}{2r} = E$.The ratio of equivalent emf in series to parallel is $\frac{E_s}{E_p} = \frac{2E}{E} = 2:1$.

What will be the ratio of the equivalent electromotive force (emf) of two cells connected in series to the equivalent emf of two cells connected in parallel?

Similar Questions

Two batteries,one of e.m.f. $12 \text{ V}$ and internal resistance $2 \Omega$ and other of e.m.f. $6 \text{ V}$ and internal resistance $1 \Omega$,are connected as shown in the figure. What will be the reading of the voltmeter '$V$' (in $\text{ V}$)?

Explain cell,emf,and internal resistance. Derive the relation between potential difference,emf,and internal resistance.

In the given diagram,an ideal voltmeter $V$ reads $1.45\,V$. Then the relation between $r_1$ and $r_2$ is:

$EMF$ is most closely related to

Vedclass Test Series

Exam Paper Generator

Online Exam Module