A dc source of emf ,E_1 = 100,V and internal resistance r = 0.5,Omega , a storage battery of emf E

Q: A $dc$ source of $emf \,E_1 = 100\,V$ and internal resistance $r = 0.5\,\Omega ,$ a storage battery of emf $E_2 = 90\,V$ and an external resistance $R$ are connected as shown in figure. For what value of $R$ no current will pass through the battery ? ................ $\Omega$

c $\frac{100}{R+r}=\frac{90}{R}$ $\Rightarrow \frac{R+r}{R}=\frac{10}{9}$ $\Rightarrow \quad 1+\frac{0.5}{\mathrm{R}}=\frac{10}{9}$ $\Rightarrow \quad \frac{0.5}{\mathrm{R}}=\frac{1}{9}$ $\therefore R=4.5 \,\Omega$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

A $dc$ source of $emf \,E_1 = 100\,V$ and internal resistance $r = 0.5\,\Omega ,$ a storage battery of emf $E_2 = 90\,V$ and an external resistance $R$ are connected as shown in figure. For what value of $R$ no current will pass through the battery ? ................ $\Omega$

A$5.5$
B$3.5$
C$4.5$
D$2.5$

JEE MAIN 2013, Medium

STD 11 Science
NEET
STD 12 - 3. current electricity
Physics

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