When a resistor of 11 Omega is connected in s | vedclass

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

Question

(C) The current $i$ drawn from a cell is given by $i = \frac{E}{R+r}$,where $E$ is the electromotive force,$R$ is the external resistance,and $r$ is t

Vedclass · Accepted Answer

$2.5$

Vedclass · Answer

(C) The current $i$ drawn from a cell is given by $i = \frac{E}{R+r}$,where $E$ is the electromotive force,$R$ is the external resistance,and $r$ is the internal resistance.Case $I$: When $R_1 = 11 \ \Omega$,$i_1 = 0.5 \ A$.$0.5 = \frac{E}{11+r} \implies E = 0.5(11+r) \quad (i)$Case $II$: When $R_2 = 5 \ \Omega$,the current increases by $0.4 \ A$,so $i_2 = 0.5 + 0.4 = 0.9 \ A$.$0.9 = \frac{E}{5+r} \implies E = 0.9(5+r) \quad (ii)$Equating $(i)$ and $(ii)$:$0.5(11+r) = 0.9(5+r)$$5.5 + 0.5r = 4.5 + 0.9r$$5.5 - 4.5 = 0.9r - 0.5r$$1.0 = 0.4r$$r = \frac{1.0}{0.4} = 2.5 \ \Omega$.

When a resistor of $11 \Omega$ is connected in series with an electric cell,the current flowing in it is $0.5 \ A$. Instead,when a resistor of $5 \Omega$ is connected to the same electric cell in series,the current increases by $0.4 \ A$. The internal resistance of the cell is (in $\Omega$)

Similar Questions

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+4x+1=0$,then $(\alpha+\beta)^{-1}+(\beta+\gamma)^{-1}+(\gamma+\alpha)^{-1}$ is equal to

What is the decreasing order of reactivity amongst the following compounds towards aromatic electrophilic substitution?
$(i)$ Chlorobenzene
$(ii)$ Benzene
$(iii)$ Anilinium chloride
$(iv)$ Toluene

$A$ vessel contains $1$ mole of $O_2$ gas (molar mass $32$) at a temperature $T$. The pressure of the gas is $P$. An identical vessel containing $1$ mole of $He$ gas (molar mass $4$) at a temperature $2T$ has a pressure of:

$A$ pair of perpendicular lines passes through the origin and also through the points of intersection of the curve $x^2+y^2=4$ with $x+y=a$,where $a>0$. Then $a$ is equal to

If $EAN$ of central metal cation $M^{2+}$ in a non-chelating complex is $36$ and atomic number of metal $M$ is $26$,then the number of monodentate ligands in this complex is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module